General Topology

Frรฉdรฉric Latrรฉmoliรจre
December 21, 2014

1 Topological Spaces and continuous functions

1.1 The category of Topological spaces

Definition 1.1.

Let E be a set. A topology ๐’ฏ on E is a subset of ๐’ซโข(E) such that:

  1. (Top1)

    Eโˆˆ๐’ฏ and โˆ…โˆˆ๐’ฏ.

  2. (Top2)

    For all U,Vโˆˆ๐’ฏ one has UโˆฉVโˆˆ๐’ฏ.

  3. (Top3)

    For all ๐’ฐโŠ†๐’ฏ one has โ‹ƒ๐’ฐโŠ†๐’ฏ.

A pair (E,๐’ฏ) is a topological space when E is a set and ๐’ฏ is a topology on E. Moreover, a subset U of E is called open, if Uโˆˆ๐’ฏE, and closed, if โˆEโขUโˆˆ๐’ฏE.

The section on fundamental examples collects many examples of topologies. Note that {โˆ…,E} and ๐’ซโข(E) both are obvious topologies on any set E, but of course the examples in the next section are more substantial. For now, we will work out a few basic properties of topologies.

Proposition 1.2.

Let E be a (nonempty) set, (F,๐’ฏF) be a topological space, and f:Eโ†’F a function. Define:

๐’ฏโข(f)={f-1โข(V)โˆฃVโˆˆ๐’ฏF}โข.

Then (E,๐’ฏโข(f)) is a topological space. One calls ๐’ฏโข(f) the initial topology for f or the topology induced by f.

Proof.

By construction, f-1โข(F)=E and f-1โข(โˆ…)=โˆ… so โˆ…,Eโˆˆ๐’ฏโข(f). Let U1,U2โˆˆ๐’ฏโข(f). Then by definition, there exists V1,V2โˆˆ๐’ฏF such that Ui=f-1โข(Vi) for i=1,2. Thus:

U1โˆฉU2=f-1โข(V1)โˆฉf-1โข(V2)=f-1โข(V1โˆฉV2)โข.

Hence by definition, U1โˆฉU2โˆˆ๐’ฏF. Last, let ๐’ฐโŠ‚๐’ฏโข(f). By definition, each element of ๐’ฐ is the preimage by f of some open set VU in (F,๐’ฏF). We set:

๐’ฑ={Vโˆˆ๐’ฏF:โˆƒUโˆˆ๐’ฐโขU=f-1โข(V)}โข.

Now xโˆˆโ‹ƒ๐’ฐ if and only if there exists Uโˆˆ๐’ฐ such that xโˆˆU. Now, Uโˆˆ๐’ฐ if and only if there exists Vโˆˆ๐’ฑ such that U=f-1โข(V). Hence xโˆˆโ‹ƒ๐’ฐ if and only if there exists Vโˆˆ๐’ฑ such that xโˆˆf-1โข(V), which in turn is equivalent to xโˆˆf-1โข(โ‹ƒ๐’ฑ). Hence:

โ‹ƒ๐’ฐ=f-1โข(โ‹ƒ๐’ฑ)

and therefore, we have shown that ๐’ฏโข(f) is a topology on E. โˆŽ

Definition 1.3.

Let (E,๐’ฏE) and (F,๐’ฏF) be two topological spaces. Let f:Eโ†’F be a function. We say that f is continuous on E when:

โˆ€Vโˆˆ๐’ฏFโขf-1โข(V)โˆˆ๐’ฏEโข.
Remark 1.4.

It is immediate that f:Eโ†’F is continuous if and only if for all closed subset CโŠ†F, the set f-1โข(C) is closed as well.

Remark 1.5.

It is immediate by definition that any constant function is continuous. It is also obvious that the identity function on any topological space is continuous.

Proposition 1.6.

Let (E,๐’ฏE), (F,๐’ฏF) and (G,๐’ฏG) be three topological spaces. Let f:Eโ†’F and g:Fโ†’G. If f and g are continuous, so is gโˆ˜f.

Proof.

Let Vโˆˆ๐’ฏG. Then g-1โข(V)โˆˆ๐’ฏF by continuity of g. Hence f-1โข(g-1โข(V))โˆˆ๐’ฏE by continuity of f. โˆŽ

Definition 1.7.

The category whose objects are topological spaces and morphisms are continuous functions is the category of topological spaces.

Definition 1.8.

An isomorphism in the category of topological spaces is called an homeomorphism.

Remark 1.9.

An homeomorphism is by definition a continuous map which is also a bijection and whose inverse function is also continuous. There are continuous bijections which are not homeomorphisms (see the chapter Fundamental Examples).

1.2 Order on Topologies on a given set

The initial topology ๐’ฏโข(f) induced by a function f:Eโ†’F is a subset of the topology on E if and only if f is continuous. This motivates the following definition.

Definition 1.10.

Let E be a set. Let ๐’ฏ1 and ๐’ฏ2 be two topologies on E. We say that ๐’ฏ1 is finer than ๐’ฏ2 and ๐’ฏ2 is coarser than ๐’ฏ1 when ๐’ฏ2โŠ†๐’ฏ1.

Of course, inclusion induces an order relation on topologies on a given set. A remarkable property is that any nonempty subset of the ordered set of topologies on a given set always admits a greatest lower bound.

Theorem 1.11.

Let E be a set. Let ๐’ฏ be a nonempty set of topologies on E. Then the set:

๐’ฏ=โ‹€๐’ฏ={UโŠ†E:โˆ€ฯƒโˆˆ๐’ฏโขUโˆˆฯƒ}

is a topology on E and it is the greatest lower bound of ๐’ฏ (where the order between topology is given by inclusion).

Proof.

We first show that ๐’ฏ is a topology. By definition, for all ฯƒโˆˆ๐’ฏ, since ฯƒ is a topology on E, we have E,โˆ…โˆˆฯƒ. Hence โˆ…,Eโˆˆ๐’ฏ. Now, let U,Vโˆˆ๐’ฏ. Let ฯƒโˆˆ๐’ฏ be arbitrary. By definition of ๐’ฏ, we have Uโˆˆฯƒ and Vโˆˆฯƒ. Therefore, UโˆฉVโˆˆฯƒ since ฯƒ is a topology. Since ฯƒ was arbitrary in ๐’ฏ, we conclude that UโˆฉVโˆˆ๐’ฏ by definition. The same proof can be made for unions. Let ๐’ฐโŠ†๐’ฏ. Let ฯƒโˆˆ๐’ฏ be arbitrary. By definition of ๐’ฏ, we have ๐’ฐโŠ†ฯƒ. Since ฯƒ is a topology, โ‹ƒ๐’ฐโˆˆฯƒ. Hence, as ฯƒ was arbitrary, โ‹ƒ๐’ฐโˆˆ๐’ฏ. So ๐’ฏ is a topology on E. By construction, ๐’ฏโŠ†ฯƒ for all ฯƒโˆˆ๐’ฏ, so ๐’ฏ is a lower bound for ๐’ฏ. Assume given a new topology ฯ on E such that ฯโŠ†ฯƒ for all ฯƒ in ๐’ฏ. Let Uโˆˆฯ. Then for all ฯƒโˆˆ๐’ฏ we have Uโˆˆฯƒ. Hence by definition Uโˆˆ๐’ฏ. So ฯโŠ†๐’ฏ and thus ๐’ฏ is the greatest lower bound of ๐’ฏ. โˆŽ

Corollary 1.12.

Let E be a set and (F,๐’ฏF) be a topological space. The smallest topology which makes a function f:Eโ†’F continuous is ๐’ฏโข(f).

Proof.

Let ๐’ฏ be the set of all topologies on E such that f is continuous. By definition, ๐’ฏโข(f) is a lower bound of ๐’ฏ. Moreover, ๐’ฏโข(f)โˆˆ๐’ฏ. Hence it is the greatest lower bound. โˆŽ

We can use Theorem (1.11) to define other interesting topologies. Note that trivially ๐’ซโข(E) is a topology, so given any AโŠ†๐’ซโข(E) for some set E, there is at least one topology contaning A. From this:

Definition 1.13.

Let E be a set. Let A be a subset of ๐’ซโข(E). The greatest lower bound of the set:

๐’ฏA={ฯƒโขย is a topology onย โขE:AโŠ†ฯƒ}

is the smallest topology on E containing A. We call it the topology induced by A on E and we denote it ๐’ฏโข(A).

We note that since ๐’ฏโข(A) belongs to ๐’ฏA by construction, it is indeed the smallest topology containing A.

In general, we will want a better description of the topology induced by a particular set than the general intersection above. This is not always possible, but the concept of basis allows one to obtain useful descriptions of topologies.

1.3 Basis

When inducing a topology from a family โ„ฌ of subsets of some set E, the fact that โ„ฌ enjoys the following property greatly simplifies our understanding of ๐’ฏโข(โ„ฌ).

Definition 1.14.

Let E be a set. A topological basis on E is a subset โ„ฌ of ๐ŸE such that:

  1. 1.

    E=โ‹ƒโ„ฌ,

  2. 2.

    For all B,Cโˆˆโ„ฌ such that BโˆฉCโ‰ โˆ…, we have:

    โˆ€xโˆˆBโˆฉCโขโˆƒDโˆˆโ„ฌโขxโˆˆDโˆงDโŠ†BโˆฉCโข.

The main purpose for this definition stems from the following theorem:

Theorem 1.15.

Let E be some set. Let โ„ฌ be a topological basis on E. Then the topology induced by โ„ฌ is:

๐’ฏโข(โ„ฌ)={โ‹ƒ๐’ฐ:๐’ฐโŠ†โ„ฌ}โข.
Proof.

Denote, for this proof, the set {โ‹ƒ๐’ฐ:๐’ฐโŠ†โ„ฌ}, by ฯƒ, and let us abbreviate ๐’ฏโข(โ„ฌ) by ๐’ฏ. We wish to prove that ๐’ฏ=ฯƒ. First, note that โ„ฌโŠ†ฯƒ by construction. By definition, โ„ฌโŠ†๐’ฏ and since ๐’ฏ is a topology, it is closed under arbitrary unions. Hence ฯƒโŠ†๐’ฏ. To prove the converse, it is sufficient to show that ฯƒ is a topology. As it contains โ„ฌ and ๐’ฏ is the smallest such topology, this will provide us with the inverse inclusion. By definition, โ‹ƒโˆ…=โˆ… and thus โˆ…โˆˆฯƒ. By assumption, since โ„ฌ is a basis, E=โ‹ƒโ„ฌ so Eโˆˆฯƒ. As the union of unions of elements in โ„ฌ is a union of elements in โ„ฌ, ฯƒ is closed under abritrary unions. Now, let U,V be elements of โ„ฌ. If UโˆฉV=โˆ… then UโˆฉVโˆˆฯƒ. Assume that U and V are not disjoints. Then by definition of a basis, for all xโˆˆUโˆฉV there exists Wxโˆˆโ„ฌ such that xโˆˆWx and WxโŠ†UโˆฉV. So:

UโˆฉV=โ‹ƒxโˆˆUโˆฉVWx

and therefore, by definition, UโˆฉVโˆˆฯƒ. We conclude that the intersection of two arbitary elements in ฯƒ is again in ฯƒ by using the distributivity of the union with respect to the intersection. โˆŽ

The typical usage of this theorem is the following corollary. We shall say that a basis โ„ฌ on a set E is a basis for a topology ๐’ฏ on E when the smallest topology containing โ„ฌ is ๐’ฏ.

Corollary 1.16.

Let โ„ฌ be a topological basis for a topology ๐’ฏ on E. A subset U of E is in ๐’ฏ if and only if for any xโˆˆU there exists Bโˆˆโ„ฌ such that xโˆˆB and BโŠ†U.

Proof.

We showed that any open set for the topology ๐’ฏ is a union of elements in โ„ฌ: hence if xโˆˆU for Uโˆˆ๐’ฏ then there exists Bโˆˆโ„ฌ such that xโˆˆB and BโŠ†U. Conversely, if U is some subset of E such that for all xโˆˆU there exists Bxโˆˆโ„ฌ such that xโˆˆBx and BxโŠ†U then U=โ‹ƒxโˆˆUBx and thus Uโˆˆ๐’ฏ. โˆŽ

As a basic application, we show that:

Corollary 1.17.

Let (E,๐’ฏE) and (F,๐’ฏF) be two topological spaces. Let โ„ฌ be a basis for the topology ๐’ฏE. Let f:Eโ†’F. Then f is continuous on E if and only if:

โˆ€Vโˆˆ๐’ฏFโขโˆ€xโˆˆf-1โข(V)โขโˆƒBโˆˆโ„ฌโขxโˆˆBโˆงBโŠ‚f-1โข(V)โข.
Corollary 1.18.

Let (E,๐’ฏE) and (F,๐’ฏF) be two topological spaces. Let โ„ฌ be a basis for the topology ๐’ฏF. Let f:Eโ†’F. Then f is continuous on E if and only if:

โˆ€Vโˆˆโ„ฌโขf-1โข(V)โˆˆ๐’ฏEโข.
Proof.

By definition, continuity of f implies 1.18. Conversely, assume 1.18 holds. Let Vโˆˆ๐’ฏF. Then there exists ๐’ฐโŠ†โ„ฌ such that V=โ‹ƒ๐’ฐ. Now by assumption, f-1โข(B)โˆˆ๐’ฏE for all Bโˆˆ๐’ฐ and thus f-1โข(V)=โ‹ƒBโˆˆ๐’ฐf-1โข(B)โˆˆ๐’ฏE since ๐’ฏE is a topology. โˆŽ

We leave to the reader to write the statement when both E and F have a basis.

1.4 Interior and Closure

Given a topological space (E,๐’ฏ), an arbitrary subset A of E may be neither open nor closed. It is useful to find closest โ€œapproximationsโ€ for A which are either open or closed. The proper notions are as follows.

Proposition 1.19.

Let (E,๐’ฏ) be a topological space. Let AโŠ†E. There exists a (necessarily unique) largest open set in E contained in A and a (necessarily unique) smallest closed set in E containing A (where the order is given by inclusion).

Proof.

Let ๐’ฐ={Uโˆˆ๐’ฏ:UโŠ†A}. Then since ๐’ฏ is a topology, โ‹ƒ๐’ฐโˆˆ๐’ฏ. Moreover, by definition, if xโˆˆโ‹ƒ๐’ฐ then there exists Uโˆˆ๐’ฐ such that xโˆˆU and since UโŠ†A we conclude xโˆˆA. Therefore โ‹ƒ๐’ฐ is by construction the largest open subset of E contained in A.

The reasonning is similar for the smallest closed set. Namely, let โ„ฑ={FโŠ‚E:โˆEโขFโˆˆ๐’ฏโˆงAโŠ‚F}. Since closed sets are the complements of open sets, it follows that arbitrary intersections of closed sets is closed. Thus โ‹‚โ„ฑ is a closed set, and it is the smallest containing A by construction. โˆŽ

Definition 1.20.

Let (E,๐’ฏ) be a topological space. Let AโŠ†E. The interior of A is the largest open subset of E contained in A denoted by AฬŠ. The closure of A is the smallest closed subset of E containing A and is denoted by Aยฏ.

Thus we always have AฬŠโŠ†AโŠ†Aยฏ. Moreover:

Proposition 1.21.

Let (E,๐’ฏ) be a topological space. Let A,BโŠ†E. Then:

  1. 1.

    If AโŠ†B then AฬŠโŠ†BฬŠ and AยฏโŠ†Bยฏ ,

  2. 2.

    AโˆฉBโžฬŠ=AฬŠโˆฉBฬŠ ,

  3. 3.

    AโˆชBยฏ=AยฏโˆชBยฏ .

Proof.

If AโŠ†B then AฬŠโŠ†AโŠ†B. Since AฬŠ is open and a subset of B, it is contained in BฬŠ by definition. Similarly, if AโŠ†B then AโŠ‚Bยฏ. Now, Bยฏ is closed and contains A so it must contained the smallest closed subset containing A, namely Aยฏ.

Note that AโˆฉBโŠ‚A so AโˆฉBโžฬŠโŠ‚AฬŠ. Similarly with B, so AโˆฉBโžฬŠโŠ†AฬŠโˆฉBฬŠ. On the other hand, AฬŠโˆฉBฬŠ is an open set (as the intersection of two open sets), and since AฬŠโŠ‚A and BฬŠโŠ‚B, it is included in AโˆฉB. Hence it is included in the largest open subset contained in AโˆฉB, which completes this proof.

The same reasonning can be applied to the last assertion of this proposition. โˆŽ

The main theorem about closures is the following.

Theorem 1.22.

Let (E,๐’ฏ) be a topological space. Let AโŠ†E. Then xโˆˆAยฏ if and only if for all Vโˆˆ๐’ฏ such that xโˆˆV we have VโˆฉAโ‰ โˆ….

Proof.

Assume first that xโˆ‰Aยฏ. Then xโˆˆโˆEโขAยฏ and โˆEโขAยฏ is open by definition; moreover by construction โˆEโขAยฏโˆฉA=โˆ… since AโŠ†Aยฏ so โˆEโขAยฏโŠ†โˆEโขA. Conversely, let xโˆˆE and assume that there exists Vโˆˆ๐’ฏ such that xโˆˆV and yet VโˆฉA=โˆ…. Then AโŠ†โˆEโขV and โˆEโขV is closed, so by definition AยฏโŠ†โˆEโขV, hence xโˆ‰Aยฏ. โˆŽ

In general, one can thus write:

Aยฏ={xโˆˆE:โˆ€Vโˆˆ๐’ฏโขxโˆˆVโŸนVโˆฉAโ‰ โˆ…}โข.

The following is immediate, so we omit the proof.

Corollary 1.23.

Let (E,๐’ฏ) be a topological space and โ„ฌ a basis for ๐’ฏ. Let AโŠ†E. Then xโˆˆAยฏ if and only if for all Vโˆˆโ„ฌ such that xโˆˆV we have VโˆฉAโ‰ โˆ….

We can use the main theorem about closure to show that:

Proposition 1.24.

Let (E,๐’ฏ) be a topological space. Let AโŠ†E. Then โˆEโขAยฏ=โˆEโขAฬŠ and โˆEโขAโžฬŠ=โˆEโขAยฏ.

Proof.

We shall only prove that โˆEโขAยฏ=โˆEโขAฬŠ as both assertions are proved similarly. Note first that AฬŠโŠ†A so โˆEโขAโŠ†โˆEโขAฬŠ and โˆEโขAฬŠ is closed. Hence โˆEโขAยฏโŠ†โˆEโขAฬŠ. Conversely, let xโˆˆโˆEโขAฬŠ. Let Vโˆˆ๐’ฏ such that xโˆˆV. Assume VโˆฉโˆEโขA=โˆ…. Then VโŠ†A so, as V open, VโŠ†AฬŠ, which contradicts xโˆˆโˆEโขAฬŠ. So VโˆฉโˆEโขAโ‰ โˆ…, and thus xโˆˆโˆEโขAยฏ as desired. โˆŽ

1.5 Characterizations of Continuity

Theorem 1.25.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Let f:Eโ†’F be given. Then f is continuous if and only if for all AโŠ†E, fโข(Aยฏ)โŠ†fโข(A)ยฏ.

Proof.

Assume first that f is continuous. Let AโŠ†E. Since f is continuous, f-1โข(fโข(A)ยฏ) is closed, and by definition of images and preimages,

AโŠ†f-1โข(fโข(A)ยฏ)

so AยฏโŠ†f-1โข(fโข(A)ยฏ) as desired.


Let us now assume that for any AโŠ†E, we have fโข(Aยฏ)โŠ†fโข(A)ยฏ. Then let CโŠ†F be a closed set. Let D=f-1โข(C). By assumption, DยฏโŠ†f-1โข(fโข(Dยฏ))โŠ†f-1โข(fโข(D)ยฏ)โŠ†f-1โข(Cยฏ)=f-1โข(C)=D. So DยฏโŠ†D, hence D=Dยฏ so D is closed. Hence f is continous. โˆŽ

Remark 1.26.

In the next chapter on fundamental examples, one will encounter many topological spaces where singletons are not open. Note that in those cases, if f:eโ†’F is constant, then fโข(E)ฬŠ would be empty, and thus we can not expect fโข(EฬŠ)โŠ†fโข(E)โžฬŠ in general for continuous functions.

Theorem 1.27.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Let f:Eโ†’F be given. Then f is continuous if and only if for all AโŠ†F, we have f-1โข(AฬŠ)โŠ†f-1โข(A)โžฬŠ.

Proof.

Assume first that f is continuous. Let AโŠ†F. Then f-1โข(AฬŠ)โˆˆ๐’ฏE since f is continuous. Since f-1โข(AฬŠ)โŠ†f-1โข(A), we conclude that f-1โข(AฬŠ)โŠ†f-1โข(A)โžฬŠ (by maximality of f-1โข(A)โžฬŠ among all open sets in E contained in A).

Conversely, assume that for all AโŠ†F, f-1โข(AฬŠ)โŠ†f-1โข(A)โžฬŠ. Let Vโˆˆ๐’ฏF. Then by assumption:

f-1โข(V)=f-1โข(VฬŠ)โŠ†f-1โข(V)โžฬŠโŠ†f-1โข(V)

so f-1โข(V)=f-1โข(V)โžฬŠ i.e. f-1โข(V)โˆˆ๐’ฏE. Since Vโˆˆ๐’ฏF was arbitrary, f is continuous. โˆŽ

Theorem 1.28.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Let f:Eโ†’F be given. Then f is continuous if and only if for all AโŠ†F we have f-1โข(A)ยฏโŠ†f-1โข(Aยฏ).

Proof.

Assume first that f is continuous. Let AโŠ†F. Then AโŠ†Aยฏ so f-1โข(A)โŠ†f-1โข(Aยฏ). Since f is continuous, then f-1โข(Aยฏ) is closed, and thus by minimality of f-1โข(A)ยฏ, we have f-1โข(A)ยฏโŠ†f-1โข(Aยฏ).

Conversely, assume that for all AโŠ†F we have f-1โข(A)ยฏโŠ†f-1โข(Aยฏ). Let CโŠ†F be closed. Then:

f-1โข(C)ยฏโŠ†f-1โข(Cยฏ)=f-1โข(C)

so f-1โข(C)=f-1โข(C)ยฏ i.e. f-1โข(C) is closed. Since C was an abritrary closed subset of F, the function f is continuous. โˆŽ

1.6 Topologies defined by functions

Continuity was phrased by stating the topology induced by the function is coarser than the topology on the domain. We can extend this idea to define topologies. Note that if E is endowed with the indiscrete topology ๐ŸE then any function, to any topological space, is continuous. Hence the following definition can be made:

Definition 1.29.

Let E be a set. Let (F,๐’ฏF) be a topological space. Let โ„ฑ be a nonempty set of functions from E to F. The smallest topology on E such that all the functions in โ„ฑ are continuous is the initial topology induced by โ„ฑ on E. We denote it by ๐’ฏโข(โ„ฑ).

Proposition 1.30.

Let E be a set. Let (F,๐’ฏ) be a topological space. Let โ„ฑ be a nonempty set of functions from E to F. Let:

โ„ฌ={B(f1,U1),โ€ฆ,(fn,Un):f1,โ€ฆ,fnโˆˆโ„ฑ,U1,โ€ฆ,Unโˆˆ๐’ฏF}

where

B(f1,U1),โ€ฆ,(fn,Un)=f1-1โข(U1)โˆฉโ€ฆโˆฉfn-1โข(Un)

for all n-tuple (f1,โ€ฆ,fn) of functions in โ„ฑ and U1,โ€ฆ,Unโˆˆ๐’ฏF. Then โ„ฌ is a basis for the initial topology induced by โ„ฑ.

Proof.

Note that for all fโˆˆโ„ฑ, we must have ๐’ฏโข(f)โŠ†๐’ฏโข(โ„ฑ) by defintion of continuity. Hence, if ๐’ฏ is any topology on E such that all the functions in โ„ฑ are continuous, then โ„ฌโŠ†๐’ฏ. In particular, ๐’ฏโข(โ„ฌ)โŠ†๐’ฏโข(โ„ฑ). On the other hand, by construction, every function in โ„ฑ is continuous for ๐’ฏโข(โ„ฌ), so ๐’ฏโข(โ„ฑ)โŠ†๐’ฏโข(โ„ฌ) by definition of the initial topology. Hence ๐’ฏโข(โ„ฑ)=๐’ฏโข(โ„ฌ).

It remains to show that โ„ฌ is a basis on E. By definition, note that Eโˆˆโ„ฌ since E=f-1โข(F) for any fโˆˆโ„ฑ. Now, note that by definition, the intersection of two elements in โ„ฌ is still in โ„ฌ, so โ„ฌ is a trivially a basis. โˆŽ

The main two theorems regarding initial topologies describe its universal property:

Theorem 1.31.

Let E be a set. Let (F,๐’ฏF) and (D,๐’ฏD) be topological spaces. Let โ„ฑ be a nonempty set of functions from E to F. A function f:Dโ†’E is continuous when E is endowed with the initial topology ๐’ฏโข(โ„ฑ) if and only if gโˆ˜f is continuous for all gโˆˆโ„ฑ.

Proof.

If f is continuous, then for all gโˆˆโ„ฑ the function gโˆ˜f is continuous, as the composition of two continuous functions.

Conversely, assume that gโˆ˜f is continuous for all gโˆˆโ„ฑ. Let g1,โ€ฆ,gnโˆˆโ„ฑ and V1,โ€ฆ,Vnโˆˆ๐’ฏF and set:

W=g1-1โข(V1)โˆฉโ€ฆโˆฉgn-1โข(Vn)โข.

Now, since g1โˆ˜f,โ€ฆ,gnโˆ˜f are all continuous by assumption, f-1โข(g1-1โข(V1)),โ€ฆ,f-1โข(gnโข(Vn)) are all in ๐’ฏD. Hence f-1โข(W)โˆˆ๐’ฏD. Since W was an arbitrary set in a basis for the final topology on E, f is continuous. โˆŽ

Theorem 1.32.

Let E be a set. Let (F,๐’ฏF) and (D,๐’ฏD) be topological spaces. Let โ„ฑ be a nonempty set of functions from E to F. The initial topology for โ„ฑ is the unique topology such that, given any topological space (D,๐’ฏD) and given any function f:Dโ†’E, then f is continuous if and only if gโˆ˜f is continuous for all gโˆˆโ„ฑ.

Proof.

We showed that the initial topology for โ„ฑ has the desired property. Conversely, assume that the given property holds for some topology ๐’ฏ on E. In particular, let f:Eโ†’E be the identity, seen as a function from the topological space (E,๐’ฏโข(โ„ฑ)) into the space (E,๐’ฏ). Using the specified universal property, we see that f is continuous, so ๐’ฏโข(โ„ฑ) is finer than ๐’ฏ. Conversely, note that f-1 is also continuous for the same reason, so ๐’ฏ is finer than ๐’ฏโข(โ„ฑ), so these two topologies agree. This completes the proof of this universal property. โˆŽ

Last, we can obtain a basis for initial topologies in a natural way:

Theorem 1.33.

Let E be a set, (F,๐’ฏF) be a topological space, โ„ฑ a nonempty set of functions from E to F. Let โ„ฌ be a basis for ๐’ฏF. Then the set:

๐’ž={f1-1โข(B1)โˆฉโ€ฆโˆฉfn-1โข(Bn):f1,โ€ฆ,fnโˆˆโ„ฑ,B1,โ€ฆ,Bnโˆˆโ„ฌ}

is a basis for the initial topology for โ„ฑ.

Proof.

This is a quick computation. โˆŽ

The dual notion of initial topology is final topology:

Definition 1.34.

Let F be a set. Let โ„ฑ be a set of triplets (E,๐’ฏ,f) where (E,๐’ฏ) is a topological space, and f:Eโ†’F is a function. The final topology ๐’ฏโข(โ„ฑ) on F is the smallest topology such that all fuctions f such that (E,๐’ฏ,f)โˆˆโ„ฑ are continuous.

In the following chapters, we will see that the product topology, the trace topology and the norm topology are examples of initial topologies. The quotient topology is an example of a final topology.

2 Fundamental Examples

This chapter provides various examples of topological spaces which will be used all along these notes and are often at the core of the subject.

2.1 Trivial Topologies

Definition 2.1.

Let E be a set. The topology {โˆ…,E} is the indiscrete topology on E. The topology ๐ŸE is the discrete topology on E.

Proposition 2.2.

Let (F,๐’ฏ) be a topological space. All functions from (E,๐ŸE) into (F,๐’ฏ) are continuous. The only continuous functions from (E,{โˆ…,E}) to F are constant if (F,๐’ฏ) is T1.

Proof.

When E is given the discrete topology, then for all open subsets V of F one has f-1โข(V) open in E. On the other hand, if E is given the indiscrete topology and (F,๐’ฏ) is T1 then assume f takes two values, l and k. Then โˆFโข{l} is open, so f-1โข(โˆFโข{l}) must be open, and as it is nonempty (it contains f-1โข({k})), it is all of E, which is absurd. โˆŽ

2.2 Order topologies

Definition 2.3.

Let (E,โ‰ค) be a linearly ordered set. Let โˆž,-โˆž be two symbols not in E. Define Eยฏ=Eโˆช{-โˆž,โˆž} and extend โ‰ค to Eยฏ by setting for all x,yโˆˆEยฏ:

xโ‰คyโ‡”(xโˆˆEโˆงyโˆˆEโˆงxโ‰คy)โˆจ(x=-โˆž)โˆจ(y=โˆž).

Let < be the relation โ‰คโฃโˆฉโฃโ‰  (i.e. x<yโ‡”(xโ‰คyโˆงxโ‰ y)) on Eยฏ. Let a,bโˆˆEยฏ. We define the set (a,b)={xโˆˆE:a<xโˆงx<b}.

Definition 2.4.

Let (E,โ‰ค) be a linearly ordered set. Then the topology induced by the set:

โ„E={(a,b):a,bโˆˆEยฏ}

is the order topology on E.

Proposition 2.5.

The set โ„E is a basis for the order topology on E.

Proof.

Since E is linearly ordered, so is Eยฏ. It is immediate that (a,b)โˆฉ(c,d)=(aโ€ฒ,dโ€ฒ) if aโ€ฒ is the largest of a and c and dโ€ฒ is the smallest of b and d. โˆŽ

Remark 2.6.

The default topology on โ„ is the order topology.

Definition 2.7.

Let (E,โ‰ค) be an ordered set, and a,bโˆˆE. We define [a,b]={xโˆˆE:aโ‰คxโˆงxโ‰คb}.

2.3 Trace Topologies

Proposition 2.8.

Let (E,๐’ฏE) be a topological space. Let AโŠ†E. Then the trace topology on A induced by ๐’ฏE is the topology:

๐’ฏAE={UโˆฉA:Uโˆˆ๐’ฏE}โข.
Proof.

It is a trivial exercise to show that the above definition indeed gives a topology on A. โˆŽ

Just as easy is the following observation:

Proposition 2.9.

Let (E,๐’ฏ) be a topological space. Let โ„ฌ be a basis for ๐’ฏ. Let AโŠ†E. Then the set {BโˆฉA:Bโˆˆโ„ฌ} is a basis for the trace topology on A induced by ๐’ฏ.

Proof.

Trivial exercise. โˆŽ

Remark 2.10.

The default topology on โ„•, โ„ค and โ„š is the trace topology induced by the order topology on โ„. Since {n}=(n-12,n+12)โˆฉโ„ค for all nโˆˆโ„ค, we see that the natural topologies form โ„• and โ„ค are in fact the discrete topology.

Remark 2.11.

Let AโŠ†E, where (E,๐’ฏ) is a topological space. Let i:Aโ†’E be the inclusion map. Then the trace topology is the initial topology for {i}, i.e. ๐’ฏโข(i).

2.4 Product Topologies

Definition 2.12.

Let I be some nonempty set. Let us assume given a family (Ei,๐’ฏi)iโˆˆI of topological spaces. A basic open set of the cartesian product โˆiโˆˆIEi is a set of the form โˆiโˆˆIUi where {iโˆˆI:Uiโ‰ Ei} is finite and for all iโˆˆI, we have Uiโˆˆ๐’ฏi.

Definition 2.13.

Let I be some nonempty set. Let us assume given a family (Ei,๐’ฏi)iโˆˆI of topological spaces. The product topology on โˆiโˆˆIEi is the smallest topology containing all the basic open sets.

Proposition 2.14.

Let I be some nonempty set. Let us assume given a family (Ei,๐’ฏi)iโˆˆI of topological spaces. The collection of all basic open sets is a basis on the set โˆiโˆˆIEi.

Proof.

Trivial exercise. โˆŽ

Remark 2.15.

The product topology is not just the basic open sets on the cartesian products: there are many more open sets!

Proposition 2.16.

Let I be some nonempty set. Let us assume given a family (Ei,๐’ฏi)iโˆˆI of topological spaces. The product topology on โˆiโˆˆIEi is the initial topology for the the set {pi:iโˆˆI} where pi:โˆjโˆˆIEjโ†’Ei is the canonical surjection for all iโˆˆI.

Proof.

Fix iโˆˆI. Let Vโˆˆ๐’ฏEi. By definition, pi-1โข(V)=โˆjโˆˆIUj where Uj=Ej for jโˆˆIโˆ–{i}, and Ui=V. Hence pi-1โข(V) is open in the product topology. As V was an arbitrary open subset of Ei, the map pi is continuous by definition. Hence, as i was arbitrary in I, the initial topology for {pi:iโˆˆI} is coarser than the product topology.

Conversely, note that the product topology is generated by {pi-1โข(V):iโˆˆI,Vโˆˆ๐’ฏEi}, so it is coarser than the initial topology for {pi:iโˆˆI}. This concludes this proof. โˆŽ

Corollary 2.17.

Let I be some nonempty set. Let us assume given a family (Ei,๐’ฏi)iโˆˆI of topological spaces. Let ๐’ฏ be the product topology on F=โˆiโˆˆIEi. Let (D,๐’ฏD) be a topological space. Then f:Dโ†’F is continuous if and only if piโˆ˜f is continuous from (D,๐’ฏD) to (Ei,๐’ฏEi) for all iโˆˆI, where pi is the canonical surjection on Ei for all iโˆˆI.

Proof.

We simply applied the fundamental property of initial topologies. โˆŽ

Remark 2.18.

The box topology on the cartesian product is the smallest topology containing all possible cartesian products of open sets. It is finer than the product topology in general. Since the product topology is the coarsest topology which makes the canonical projections continuous, it is the preferred one on cartesian products. Of course, both agree on finite products.

Remark 2.19.

The product topology is the default topology on a cartesian product of topological spaces.

2.5 Metric spaces

Definition 2.20.

Let E be a set. A function d:Eร—Eโ†’[0,โˆž) is a distance on E when:

  1. 1.

    For all x,yโˆˆE, we have dโข(x,y)=0 if and only if x=y,

  2. 2.

    For all x,yโˆˆE we have dโข(x,y)=dโข(y,x),

  3. 3.

    For all x,y,zโˆˆE we have dโข(x,y)โ‰คdโข(x,z)+dโข(z,y).

Definition 2.21.

A pair (E,d) is a metric space when E is a set and d a distance on E.

The following is often useful:

Proposition 2.22.

Let (E,d) be a metric space. Let x,y,zโˆˆE. Then:

|dโข(x,y)-dโข(x,z)|โ‰คdโข(y,z)โข.
Proof.

Since dโข(x,y)โ‰คdโข(x,z)+dโข(z,y) we have dโข(x,y)-dโข(x,z)โ‰คdโข(z,y)=dโข(y,z). Since dโข(x,z)โ‰คdโข(x,y)+dโข(y,z) we have dโข(x,z)-dโข(x,y)โ‰คdโข(y,z). Hence the proposition holds. โˆŽ

Definition 2.23.

Let (E,d) be a metric space. Let xโˆˆE and rโˆˆ(0,โˆž)โŠ†โ„. The open ball of center x and radius r in (E,d) is the set:

Bโข(x,r)={yโˆˆE:dโข(x,y)<r}โข.
Definition 2.24.

Let (E,d) be a metric space. The metric topology on E induced by d is the smallest topology containing all the open balls of E.

Theorem 2.25.

Let (E,d) be a metric space. The set of all open balls on E is a basis for the metric topology on E induced by d.

Proof.

It is enough to show that the set of all open balls is a basis. By definition, E=โ‹ƒxโˆˆEBโข(x,1). Now, let us be given Bโข(x,rx) and Bโข(y,ry) for some x,yโˆˆE and rx,ry>0. If the intersection of these two balls is empty, we are done; let us assume that there exists zโˆˆBโข(x,rx)โˆฉBโข(y,ry). Let ฯ be the smallest of rx-dโข(x,z) and ry-dโข(y,z). Let wโˆˆBโข(z,ฯ). Then:

dโข(x,w)โ‰คdโข(x,z)+dโข(z,w)<dโข(x,z)+rx-dโข(x,z)=rx

so wโˆˆBโข(x,rx). Similarly, wโˆˆBโข(y,ry). Hence, Bโข(z,ฯ)โŠ†Bโข(x,rx)โˆฉBโข(y,ry) as desired. โˆŽ

The following theorem shows that metric topologies are minimal in the sense of making the distance functions continuous.

Theorem 2.26.

Let (E,d) be a metric space. For all xโˆˆE, the function yโˆˆEโ†ฆdโข(x,y) is continuous on E for the metric topology. Moreover, the metric topology is the smallest topology such that all the functions in the set {yโ†ฆdโข(x,y):xโˆˆE} are continuous.

Proof.

Fix xโˆˆE. It is sufficient to show that the preimage of [0,r) and (r,โˆž) by dx:yโˆˆEโ†ฆdโข(x,y) is open in the metric topology of E, where rโ‰ฅ0 is arbitrary. Indeed, these intervals form a basis for the topology of [0,โˆž). Let rโ‰ฅ0 be given. Then dx-1โข([0,r))=Bโข(x,r) by definition, so it is open. Moreover, it shows that the minimal topology making all these maps continuous must indeed contain the metric topology. Now, let yโˆˆE such that dโข(x,y)>r. Let ฯ=dโข(x,y)-r>0. Then if dโข(w,y)<ฯ for some wโˆˆE then:

dโข(x,y)โ‰คdโข(x,w)+dโข(w,y)โขย soย โขdโข(x,y)-dโข(w,y)โ‰คdโข(x,w)

so dโข(x,w)>r. Hence

Bโข(y,ฯ)โŠ‚dx-1โข((r,โˆž))

for all yโˆˆdx-1โข((r,โˆž)). Therefore, dx-1โข((r,โˆž)) is open, as desired, and our proposition is proven. โˆŽ

Remark 2.27.

The topology on [0,โˆž) is the trace topology on [0,โˆž) induced by the usual, i.e. the order topology on โ„.

Remark 2.28.

The metric topology is the default topology on a metric space.

There are more examples of continuous functions between metric spaces. More precisely, a natural category for metric spaces consists of metric spaces and Lipschitz maps as arrows, defined as follows:

Definition 2.29.

Let (E,dE), (F,dF) be metric spaces. A function f:Eโ†’F is k-Lipschitz for kโˆˆ[0,โˆž) if:

โˆ€x,yโˆˆEโขdFโข(fโข(x),fโข(y))โ‰คkโขdEโข(x,y)โข.
Definition 2.30.

Let (E,dE), (F,dF) be metric spaces. Let f:Eโ†’F be a Lipschitz function. Then the Lipschitz constant of f is defined by:

Lipโข(f)=supโก{dFโข(fโข(x),fโข(y))dEโข(x,y):x,yโˆˆE,xโ‰ y}โข.
Remark 2.31.

Lipโข(f)=0 if and only if f is constant.

Proposition 2.32.

Let (E,dE), (F,dF) be metric spaces. If f:Eโ†’F is a Lipschitz function, then it is continuous.

Proof.

Assume f is nonconstant (otherwise the result is trivial). Let k be the Lipschitz constant for f. Let yโˆˆF and ฯต>0. Let xโˆˆf-1โข(Bโข(y,ฯต)). Let zโˆˆE such that dEโข(x,z)<ฮดx=ฯต-dโข(fโข(x),y)k (note that the upper bound is nonzero).

(2.1) dFโข(fโข(z),y) โ‰ค dFโข(fโข(z),fโข(x))+dFโข(fโข(x),y)
(2.2) โ‰ค kโขdEโข(x,z)+dFโข(fโข(x),y)
(2.3) < ฯต-dFโข(fโข(x),y)+dFโข(fโข(x),y)=ฯต.

Hence f-1โข(Bโข(y,ฯต))=โ‹ƒxโˆˆf-1โข(Bโข(y,ฯต))Bโข(x,ฮดx). So f is continuous. โˆŽ

Remark 2.33.

The proof of continuity for Lipshitz maps can be simplified: it is a consequence of the squeeze theorem. We refer to the chapter on metric spaces for this.

Remark 2.34.

Using Lipshitz maps as morphisms for a category of metric spaces is natural. Another, more general type of morphisms, would be uniform continuous maps, which are discussed in the compact space chapter.

2.6 Co-Finite Topologies

A potential source for counter-examples, the family of cofinite topologies is easily defined:

Proposition 2.35.

Let E be a set. Let:

๐’ฏcofโข(E)={โˆ…}โˆช{UโŠ‚E:โˆEโขUโขย is finiteย }โข.

Then ๐’ฏcofโข(E) is a topology on E.

Proof.

By definition, โˆ…โˆˆ๐’ฏcofโข(E). Moreover, โˆEโขE=โˆ… which is finite, so Eโˆˆ๐’ฏcofโข(E). Let U,Vโˆˆ๐’ฏcofโข(E). If U or V is empty then UโˆฉV=โˆ… so UโˆฉVโˆˆ๐’ฏcofโข(E). Otherwise, โˆEโข(UโˆฉV)=โˆEโขUโˆชโˆโขV which is finite, since by definition โˆEโขU and โˆEโขV are finite. Hence UโˆฉVโˆˆ๐’ฏcofโข(E). Last, let ๐’ฐโŠ†๐’ฏcofโข(E). Again, if ๐’ฐ={โˆ…} then โ‹ƒ๐’ฐ=โˆ…โˆˆ๐’ฏcofโข(E). Let us now assume that ๐’ฐ contains at least one nonempty set V. Then:

โˆEโขโ‹ƒ๐’ฐ=โ‹‚{โˆEโขU:Uโˆˆ๐’ฐ}โŠ†โˆEโขVโข.

Since โˆEโขV is finite by definition, so is โ‹ƒ๐’ฐ, which is therefore in ๐’ฏcofโข(E). This completes our proof. โˆŽ

2.7 The one-point compactification of โ„•

Limits of sequences is a central tool in topology and this section introduces the natural topology for this concept. The general notion of limit is the subject of the next chapter.

Definition 2.36.

Let โˆž be some symbol not found in โ„•. We define โ„•ยฏ to be โ„•โˆช{โˆž}.

Proposition 2.37.

The set:

๐’ฏโ„•ยฏ={UโŠ†โ„•ยฏ:(UโŠ†โ„•)โˆจ(โˆžโˆˆUโˆงโˆโ„•Uย is finite)}

is a topology on โ„•ยฏ.

Proof.

By definition, โˆ…โŠ†โ„• so โˆ…โˆˆ๐’ฏโ„•ยฏ. Moreover โˆโ„•ยฏโขโ„•ยฏ=โˆ… which has cardinal 0 so โ„•ยฏโˆˆ๐’ฏโ„•ยฏ. Let U,Vโˆˆ๐’ฏโ„•ยฏ. If either U or V is a subset of โ„• then UโˆฉV is a subset of โ„• so UโˆฉVโˆˆ๐’ฏโ„•ยฏ. Othwiwse, โˆžโˆˆUโˆฉV. Yet โˆโ„•ยฏโข(UโˆฉV)=โˆโ„•ยฏโขUโˆชโˆโ„•ยฏโขV which is finite as a finite union of finite sets. Hence UโˆฉVโˆˆ๐’ฏโ„•ยฏ again.

Last, assume that ๐’ฐโŠ†๐’ฏโ„•ยฏ. Of course, โˆžโˆˆโ‹ƒ๐’ฐ if and only if โˆžโˆˆU for some Uโˆˆ๐’ฐ. So, if โˆžโˆ‰โ‹ƒ๐’ฐ then โ‹ƒ๐’ฐโˆˆ๐’ฏโ„•ยฏ by definition. If, on the other hand, โˆžโˆˆโ‹ƒ๐’ฐ, then there exists Uโˆˆ๐’ฐ with โˆโ„•ยฏโขU finite. Now, โˆโ„•ยฏโขโ‹ƒ๐’ฐ=โ‹‚{โˆโ„•ยฏโขV:Vโˆˆ๐’ฐ}โŠ†โˆโ„•ยฏโขU so it is finite, and thus again โ‹ƒ๐’ฐโˆˆ๐’ฏโ„•ยฏ. โˆŽ

3 Limits

3.1 Topological Separation and Hausdorff spaces

The general definition of topology allows for examples where elements of a topological space, seen as a set, can not be distinguished from each others (for instance if the topology is indiscrete). When points can be topologically differentiated, a topology is in some sense separated. There are many axioms, or definitions, of separability, and we will use the most common and intuitive: namely, the notion of Hausdorff spaces. We do however present a few basic notions in this section which are weaker than Hausdorff separation, as such spaces are certainly common in mathematics.

Definition 3.1.

Let (E,๐’ฏ) be a topological space. We say that ๐’ฏ is T0 when given any two points x,yโˆˆE with xโ‰ y, there exists an open set Uโˆˆ๐’ฏ such that either xโˆˆU,yโˆ‰U or yโˆˆU,xโˆ‰U.

Thus a space is T0 when there are enough open sets to separate the points, i.e. when the set of all open sets containing one point is not the same as the set of all open sets containing a different point. However, this notion is not symmetric. The following definition add that the separation property should be symmetric:

Definition 3.2.

Let (E,๐’ฏ) be a topological space. We say that ๐’ฏ is T1 when given any two points x,yโˆˆE with xโ‰ y, there exists two open sets U,Vโˆˆ๐’ฏ such that xโˆˆU,yโˆ‰U and yโˆˆV,xโˆ‰V.

A key advantage of T1 separation is:

Proposition 3.3.

Let (E,๐’ฏ) be a topological space. Then ๐’ฏ is T1 if and only if for all xโˆˆE the set {x} is closed.

Proof.

Assume ๐’ฏ is T1. Fix xโˆˆE. Let yโˆˆEโˆ–{x}. Then there exists Uyโˆˆ๐’ฏ such that xโˆ‰Uy and yโˆˆUy. We can thus write:

โˆEโข{x}=โ‹ƒyโˆˆEโˆ–{x}Uy

which shows that โˆEโข{x} is open as desired.

Assume now that all singletons are closed. Let x,yโˆˆE with xโ‰ y. Then xโˆˆโˆEโข{y} and yโˆˆโˆEโข{x}, i.e. ๐’ฏ is T1. โˆŽ

Note that T0 is not enough for the above result.

Corollary 3.4.

Let (E,๐’ฏ) be a topological space. (E,๐’ฏ) is T1 if and only if for all xโˆˆE we have {x}=โ‹‚{Uโˆˆ๐’ฏ:xโˆˆU}.

Proof.

Assume (E,๐’ฏ) is T1. Let S=โ‹‚{Uโˆˆ๐’ฏ:xโˆˆU}. Let yโˆˆS. If yโ‰ x then there exists Uโˆˆ๐’ฏ such that xโˆˆU and yโˆ‰U. This is a contradiction. So S={x}.

Assume now that for all xโˆˆE we have {x}=โ‹‚{Uโˆˆ๐’ฏ:xโˆˆU}. Let x,yโˆˆE such that xโ‰ y. Then:

โ‹‚{Uโˆˆ๐’ฏ:xโˆˆU}โŠˆโ‹‚{Uโˆˆ๐’ฏ:yโˆˆU}

, so there exists Uโˆˆ๐’ฏ such that xโˆˆU and yโˆ‰U. Similarly:

โ‹‚{Uโˆˆ๐’ฏ:yโˆˆU}โŠˆโ‹‚{Uโˆˆ๐’ฏ:xโˆˆU}

so there exists Uโˆˆ๐’ฏ such that yโˆˆU and xโˆ‰U. Hence (E,๐’ฏ) is T1. โˆŽ

Example 3.5.

The indiscrete topology is not T0.

Example 3.6.

Let E be an infinite set (for instance E=โ„), endowed with the cofinite topology ๐’ฏ=๐’ฏcofโข(E). By defintion, {x} is closed for all xโˆˆE so (E,๐’ฏ) is T1. We make a useful observation. Let f:Eโ†’โ„ be a continuous map, where the codomain is endowed with the usual order topology. Assume f is not constant: then there exists x,yโˆˆE such that fโข(x)โ‰ fโข(y). Without loss of generality, we assume fโข(x)<fโข(y). Let r=12โข(fโข(y)-fโข(x)). Let U=(fโข(x)-r,fโข(x)+r) and V=(fโข(y)-r,fโข(y)+r). Then UโˆฉV=โˆ… and U,V are open sets in โ„ such that xโˆˆU and yโˆˆV. Since f is continuous, f-1โข(U) and f-1โข(V) are open in E, i.e. are cofinite. Since U and V are disjoint, we conclude that f-1โข(U) is in the complement of f-1โข(V), which is finite. Since the complement of f-1โข(U) is finite as well, we conclude that E is finite, which is a contradiction. So f is a constant.

The cofinite topology on infinite set example shows that T1 still allows for counter intuitive situations. We also saw that we could find two disjoint open sets in โ„ containing given distinct points: this stronger property is the separation axiom we will focus our attention to.

Definition 3.7.

Let (E,๐’ฏ) be a topological space. We say that (E,๐’ฏ) is a Hausdorff space (or T2) when for any x,yโˆˆE such that xโ‰ y there exists U,Vโˆˆ๐’ฏ such that UโˆฉV=โˆ…, xโˆˆU and yโˆˆV.

Proposition 3.8.

If (E,๐’ฏ) is Hausdorff, then it is T1.

Proof.

This result holds by definition. โˆŽ

Example 3.9.

Let (E,โ‰ค) be a linearly ordered set and let ๐’ฏ be the associated order topology on E. Then (E,๐’ฏ) is Hausdorff. Indeed, let x,yโˆˆE with xโ‰ y; without loss of generality we assume that x<y. Then if there exists zโˆˆE such that x<z<y then xโˆˆ(-โˆž,z) and yโˆˆ(z,โˆž), where both intervals are disjoint and open. Otherwise, (-โˆž,y)โˆฉ(x,โˆž)=โˆ… and xโˆˆ(-โˆž,y),yโˆˆ(x,โˆž).

Example 3.10.

In particular, โ„ is Hausdorff.

Example 3.11.

Let (E,๐’ฏ) be a Hausdorff space. Then by definition, it is immediate that the trace topology induced on any subset of E is also Hausdorff.

Example 3.12.

In particular, the topology on โ„š is Hausdorff.

Example 3.13.

The discrete topology on any set is always Hausdorff.

Example 3.14.

In particular, the topology on โ„ค and โ„• is Hausdorff.

Example 3.15.

Let (E,d) be a metric space. Then (E,๐’ฏd) is Hausdorff. Indeed, let x,yโˆˆE such that xโ‰ y. Let r=12โขdโข(x,y) and note that r>0 by definition of a distance. Let zโˆˆBโข(x,r). Then:

dโข(x,y)โ‰คdโข(x,z)+dโข(z,y)โขsoโขr<dโข(z,y)

and by symmetry, if dโข(w,y)<r then dโข(w,x)>r. Hence Bโข(x,r)โˆฉBโข(y,r)=โˆ….

Example 3.16.

Let (Ei,๐’ฏi)iโˆˆI be some family of Hausdorff topological spaces. Then the cartesian product with the product topology is Hausdorff. Indeed, let ๐ฑ=(xi)iโˆˆI and ๐ฒ=(yi)iโˆˆI such that ๐ฑโ‰ ๐ฒ. By definition, there exists i0โˆˆI such that xi0โ‰ yi0. Since Ei0 is Hausdorff, there exists Ui0,Vi0 such that xi0โˆˆUi0, yi0โˆˆVi0 and Ui0โˆฉVi0=โˆ…. Define the two families:

๐ฎ:iโˆˆIโ†ฆ{Eiย ifย iโ‰ i0,Ui0otherwise,

and

๐ฏ:iโˆˆIโ†ฆ{Eiย ifย iโ‰ i0,Vi0otherwise,

and set U=โˆiโˆˆI๐ฎi and V=โˆiโˆˆI๐ฏi. Then by construction, ๐ฑโˆˆU, ๐ฒโˆˆV and UโˆฉV=โˆ….

The following result is a characterization of Hausdorff separation.

Theorem 3.17.

Let (E,๐’ฏ) be a topological space. Then (E,๐’ฏ) is Hausdorff if and only if ฮ”={(x,x):xโˆˆE} is closed in E2 for the product topology.

Proof.

Assume E is Hausdorff. Let (x,y)โˆˆโˆE2โขฮ” (so xโ‰ y!). Then there exists two disjoint open sets U and V in E such that xโˆˆU and yโˆˆV. If (z,z)โˆˆUร—V then zโˆˆU and zโˆˆV which is impossible since U and V are disjoint. So Uร—V, which is open in the product topology, contains (x,y) by definition, and is a subset of โˆE2โขฮ”. Hence โˆE2โขฮ” is open.

Conversely, assume โˆE2โขฮ” is open. Let (x,y)โˆˆโˆE2โขฮ”. Since basic open sets form a basis for the product topology, there exists U,V open in E such that (x,y)โˆˆUร—V and Uร—VโŠ†โˆE2โขฮ”. Now, by definition, (Uร—V)โˆฉฮ”=โˆ… so as above, UโˆฉV=โˆ…, as desired. So (E,๐’ฏ) is Hausdorff. โˆŽ

Hausdorff spaces have a nice relation with continuous maps as well.

Proposition 3.18.

Let (E,๐’ฏE) be a topological space. Let (F,๐’ฏF) be a Hausdorff topological space. Let f:Eโ†’F and g:Eโ†’F be given.

  1. 1.

    If f is continuous, then the kernel kerโข(f)={(x,y)โˆˆE2:fโข(x)=fโข(y)} of f is closed.

  2. 2.

    If f is continuous, then the graph graph(f)={(x,f(x))โˆˆEร—F} of f is closed.

  3. 3.

    If f and g are continuous, then the equalizer eqโข(f,g)={xโˆˆE:fโข(x)=gโข(x)} of f,g is closed.

Proof.

Since (F,๐’ฏF) is Hausdorff, the set ฮ”={(x,x):xโˆˆF} is closed.

We first prove that the kernel of f is closed when f is continuous. Let:

ฮบ:Eร—E โŸถ Fร—F
(x,y) โŸผ (fโข(x),fโข(y))โข.

The function (x,y)โˆˆEร—Eโ†ฆx is continuous since the product topology is the initial topology for the canonical surjections, and f is continuous by assumption, so (x,y)โˆˆEร—Eโ†ฆfโข(x) is continuous as the composition of two continuous functions. Similarly, (x,y)โˆˆEร—Eโ†ฆfโข(y) is continuous. Hence, by the universal property of the initial topology, ฮบ is continuous.

Now, by definition, kernelโข(f)=ฮบ-1โข(ฮ”) and thus it is closed.

We now prove that the graph of f is closed when f is continuous. Let:

ฮด:Eร—F โŸถ Fร—F
(x,y) โŸผ (fโข(x),y)โข.

By a similar argument as for ฮบ, the map ฮด is continuous since (x,y)โˆˆEร—Fโ†ฆy and (x,y)โˆˆEร—Fโ†ฆx are continuous for the product topology, and f is continuous, so by composition of continuous functions, (x,y)โˆˆEร—Fโ†ฆfโข(x) is continuous. Hence by the universal property of the product topology (seen again as the initial topology for the canonical surjections), ฮด is continuous.

Now graphโข(f)=ฮด-1โข(ฮ”) so the graph of f is closed.

Last, we show the equalizer of f and g is closed when both f and g are continuous. Let:

ฮท:E โŸถ Fร—F
x โŸผ (fโข(x),gโข(x))โข.

By assumption, f and g are continuous, so by the universal property of the product topology, ฮท is continuous. So eqโข(f,g)=ฮท-1โข(ฮ”) is closed. โˆŽ

Proposition 3.19.

Let (E,๐’ฏE) and (F,๐’ฏF) be two topological spaces. Let f:Eโ†’F.

  1. 1.

    If f open (i.e. Uโˆˆ๐’ฏEโŸนfโข(U)โˆˆ๐’ฏF ) and a surjection, and if the kernel of f is closed then (F,๐’ฏF) is Hausdorff.

  2. 2.

    If f is a continuous, open surjection, then the kernel of f is closed if and only if (F,๐’ฏF) is Hausdorff.

Proof.

Let x,yโˆˆF with xโ‰ y. Since f is surjective, there exists w,zโˆˆE such that fโข(w)=x and fโข(z)=y. Since xโ‰ y, (w,z) is in the complement of the kernel of f, which is open by assumption. Since {Uร—V:U,Vโˆˆ๐’ฏE} is a basis for the product topology on E2, there exists Uw,Uzโˆˆ๐’ฏE such that (w,z)โˆˆUwร—Uz and Uwร—Uz is a subset of the complement of the kernel of f. Let tโˆˆfโข(Uw) and rโˆˆfโข(Uz). If fโข(r)=fโข(t) then (t,r) lies in the kernel of f and in Uwร—Uz, which is a contradiction. Hence Vx=fโข(Uw) and Vy=fโข(Uz) are disjoint. Since f is open, they are open. By construction, xโˆˆVx and yโˆˆVy. So (F,๐’ฏF) is Hausdorff.

Assume now that f is a continuous, open surjection. We just proved that since f is an open surjection, if its kernel is closed then (F,๐’ฏF) is closed. Conversely, if f is continuous and (F,๐’ฏF) is Hausdorff then its kernel is closed. Hence the equivalence stated. โˆŽ

Remark 3.20.

We offer an alternative proof of the first assertion of the previous proposition when f is assumed to be open and bijective. We keep the notations used in that proposition.

Assume that f is an open bijection and kerโข(f) is closed. Since f is bijective, it has a right inverse g:Fโ†’E. If U is open in E then g-1โข(U)=fโข(U) is open, so g is continuous. Moreover, letting ฮ”={(y,y):yโˆˆF} and since f is surjective, we have:

ฮ”=fโข(kernelโข(f))=g-1โข(kernelโข(f))

which is therefore closed by assumption, as the preimage of the closed set kernelโข(f) by the continuous function g. Hence (F,๐’ฏF) is Hausdorff.

Remark 3.21.

An open map may not be continuous. For instance, a nonconstant map from {0,1} with the indiscrete topology to {0,1} with the discrete topology is always open but never continuous.

3.2 Limits along a set

Dealing with continuity can be complicated from the original definition. It is easier to introduce limits and the notion of continuity at a point. We first introduce a piece of vocabulary:

Definition 3.22.

Let (E,๐’ฏE) be a topological space. Let aโˆˆE. The set of open neighborhoods of a in ๐’ฏE is the set ๐’ฑ๐’ฏEโข(a) of all Uโˆˆ๐’ฏE such that aโˆˆU.

Definition 3.23.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Assume (F,๐’ฏF) is a Hausdorff space. Let AโŠ†E. Let aโˆˆAยฏ and lโˆˆF. Let f:Eโ†’F. We say that f has limit l at a along A when:

โˆ€Vโˆˆ๐’ฑ๐’ฏFโข(l)โขโˆƒUโˆˆ๐’ฑ๐’ฏEโข(a)โขfโข(UโˆฉA)โŠ†Vโข.
Remark 3.24.

If aโˆ‰Aยฏ then the notion is silly, since we could find one open set U containing a and so that UโˆฉA=โˆ… in the above definition.

Proposition 3.25.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Assume (F,๐’ฏF) is a Hausdorff space. Let AโŠ†E. Let aโˆˆAยฏ and lโˆˆF. Let f:Eโ†’F and g:Eโ†’F. Assume that fโข(x)=gโข(x) for all xโˆˆA. Then f has limit l at a along A if and only if g has limit l at a along A.

Proof.

Simply observe that the definition of limit involves only f(โ‹…โˆฉA) and g(โ‹…โˆฉA). โˆŽ

We thus can define without ambiguity the limit of a partially defined function at a point in the closure of its domain:

Definition 3.26.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Assume AโŠ†E is nonempty and f:Aโ†’F. Then f has limit l at aโˆˆAยฏ along A if any extension of f to E has limit l at a along A.

Proposition 3.27.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Assume (F,๐’ฏF) is a Hausdorff space. Let BโŠ†AโŠ†E. Let aโˆˆBยฏ and lโˆˆF. Let f:Eโ†’F. Then if f has limit l at a along A then f has limit l at a along B.

Proof.

First, since BโŠ†B, we have BยฏโŠ†Aยฏ, so aโˆˆBยฏ implies that aโˆˆAยฏ, hence the notion of limits are well-defined. Let Vโˆˆ๐’ฑ๐’ฏFโข(l). Since limxโ†’a,xโˆˆAโกfโข(x)=l, there exists Uโˆˆ๐’ฑ๐’ฏEโข(a) such that fโข(UโˆฉA)โŠ†V. Since BโŠ†A we have fโข(UโˆฉB)โŠ†fโข(UโˆฉA)โŠ†V as desired. โˆŽ

Proposition 3.28.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Assume (F,๐’ฏF) is a Hausdorff space. Let AโŠ†E. Let aโˆˆAยฏ and lโˆˆF. Let f:Eโ†’F. If f has a limit at a along A then this limit is unique and denoted by limxโ†’a,xโˆˆAโกfโข(x).

Proof.

Assume f has limit l and lโ€ฒ with lโ‰ lโ€ฒ at a along A. Since F is Hausdorff, there exists V,Vโ€ฒ open in F such that lโˆˆV and lโ€ฒโˆˆVโ€ฒ. Then by definition of limits, there exists U,Uโ€ฒ open in E so that aโˆˆU, aโˆˆUโ€ฒ and fโข(UโˆฉA)โŠ†V and fโข(Uโ€ฒโˆฉA)โŠ†Vโ€ฒ. So fโข(AโˆฉUโˆฉUโ€ฒ)โŠ†VโˆฉVโ€ฒ=โˆ…. This is absurd since aโˆˆUโˆฉUโ€ฒ and aโˆˆAยฏ so AโˆฉUโˆฉUโ€ฒโ‰ โˆ…. โˆŽ

Proposition 3.29.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Assume (F,๐’ฏF) is a Hausdorff space. Let AโŠ†E. Let aโˆˆAยฏ and lโˆˆF. Let f:Eโ†’F. If f has a limit l at a along A then lโˆˆfโข(A)ยฏ.

Proof.

Let l=limxโ†’a,xโˆˆAโกfโข(x). Let Vโˆˆ๐’ฏF such that lโˆˆV. Then by definition of limit, there exists Uโˆˆ๐’ฏE such that aโˆˆU and fโข(UโˆฉA)โŠ†V. By definition, fโข(UโˆฉA)โŠ‚fโข(A). Since aโˆˆAยฏ, we have โˆ…โ‰ AโˆฉU so โˆ…โ‰ fโข(AโˆฉU)โŠ†Vโˆฉfโข(A). So lโˆˆfโข(A)ยฏ. โˆŽ

Remark 3.30.

In general, it is difficult to compute fโข(A), so the result is used as follows: if fโข(A)โŠ†B then limxโ†’a,xโˆˆAโกfโข(x)โˆˆBยฏ.

It is unpractical to check a statement for all open sets, as in general they are difficult to describe. The use of a basis makes things more amenable, and in fact it is the main role of basis.

Proposition 3.31.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces, with (F,๐’ฏF) Hausdorff. Assume that we are given a basis โ„ฌE for ๐’ฏE and a basis โ„ฌF for ๐’ฏF. Let AโŠ†E, aโˆˆAยฏ and lโˆˆF. Let f:Eโ†’F. Let us denote by โ„ฌl the subset of all Bโˆˆโ„ฌF such that lโˆˆB, and we define โ„ฌa similarly. Then f has limit l at a along A if and only if:

โˆ€Bโˆˆโ„ฌlโขโˆƒCโˆˆโ„ฌaโขfโข(CโˆฉA)โŠ‚Bโข.
Proof.

Assume 3.31. Let Vโˆˆ๐’ฏF such that lโˆˆV. By definition, there exists Bโˆˆโ„ฌl such that lโˆˆB and BโŠ†V. By 3.31 there exists CโˆˆmโขaโขtโขhโขcโขaโขlโขBa, i.e. an open set in E containing a, such that fโข(CโˆฉA)โŠ†BโŠ†V. This shows that f converges to l at a along A.

Conversely, assume that f converges to l at a along A. Let Bโˆˆโ„ฌl. By definition, B is an open set in F containing l so there exists Uโˆˆ๐’ฏE with aโˆˆU such that fโข(UโˆฉA)โŠ†B by definition of limit. There exists Cโˆˆโ„ฌE such that aโˆˆC and CโŠ†U as โ„ฌE is a basis for ๐’ฏE. Hence, fโข(CโˆฉA)โŠ‚fโข(CโˆฉU)โŠ‚C as desired. โˆŽ

Proposition 3.32.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces, with (F,๐’ฏF) T1. Let f:Eโ†’F, AโŠ†E. If aโˆˆA and if f has a limit at a along A then this limit is fโข(a).

Proof.

Assume l is the limit of f at a along A and lโ‰ fโข(a). Then since ๐’ฏF is T1, there exists Vโˆˆ๐’ฏF such that lโˆˆV,fโข(a)โˆ‰V. By definition of limit, there exists Uโˆˆ๐’ฏE with aโˆˆU and fโข(UโˆฉA)โŠ†V. Now, aโˆˆAโˆฉU so fโข(a)โˆˆfโข(AโˆฉU)โŠ†V which is absurd. โˆŽ

Proposition 3.33.

Let (E,๐’ฏE) be a topological space. Let (F,๐’ฏF) be a Hausdorff topological space. Let A,Aโ€ฒโŠ†E. Let aโˆˆAยฏโˆฉAโ€ฒยฏ. Then f has limit l at a along AโˆชAโ€ฒ if and only if f has limit l at a along A and along Aโ€ฒ.

Proof.

Note that aโˆˆAยฏโˆฉAโ€ฒยฏ implies that aโˆˆAโˆชAโ€ฒยฏ. The condition is necessary since AโŠ†AโˆชAโ€ฒ and Aโ€ฒโŠ†AโˆชAโ€ฒ. Conversely, let Vโˆˆ๐’ฑ๐’ฏFโข(l). By definition of limits along A and Aโ€ฒ there exist U,Uโ€ฒโˆˆ๐’ฑ๐’ฏEโข(a) such that fโข(AโˆฉU)โŠ†V and fโข(Aโ€ฒโˆฉUโ€ฒ)โŠ†Vโ€ฒ. Hence fโข((UโˆฉUโ€ฒ)โˆฉ(AโˆชAโ€ฒ))โŠ†V. Since UโˆฉUโ€ฒโˆˆ๐’ฑ๐’ฏEโข(a), the proof is complete. โˆŽ

Theorem 3.34.

Let (E,๐’ฏE), (F,๐’ฏF) and (G,๐’ฏG) be three topological spaces, where ๐’ฏF and ๐’ฏG are Hausdorff. Let f:Eโ†’F and g:Fโ†’G. Let AโŠ†E, aโˆˆAยฏ. Set B=fโข(A). If b is the limit of f at a along A and l is the limit of g at b along B then l is the limit of gโˆ˜f at a along A.

Proof.

Note first that bโˆˆBยฏ, as necessary to make sense of the statement of the theorem. Let Wโˆˆ๐’ฏG with lโˆˆW. There exists Vโˆˆ๐’ฏF with bโˆˆV such that gโข(VโˆฉB)โŠ†W. Now, there exists Uโˆˆ๐’ฏE such that aโˆˆU and fโข(UโˆฉA)โŠ†V. By construction, fโข(AโˆฉU)โŠ†B so fโข(AโˆฉU)โŠ†BโˆฉV. Hence, gโข(fโข(UโˆฉA))โŠ†W as desired. โˆŽ

Remark 3.35.

Beware of this theorem. Take fโข(x)=xโขsinโก(1x), gโข(x)=0 for xโ‰ 0 and gโข(0)=1, and A=B=โ„โˆ–{0} and a=b=0. Then:

0=limxโ†’0,xโ‰ 0โกfโข(x)=limyโ†’0,y-ฬธ0โกgโข(y)

yet gโˆ˜f has no limit at 0 along the set of nonzero reals.

3.3 Continuity at a point

Definition 3.36.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Let aโˆˆE. Then f is continuous at a if:

โˆ€Uโˆˆ๐’ฑ๐’ฏFโข(fโข(a))โขโˆƒUโˆˆ๐’ฑ๐’ฏEโข(a)โขfโข(U)โŠ†Vโข.

The notion of continuity at a point looks more familiar when the codomain is Hausdorff.

Proposition 3.37.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. Assume (F,๐’ฏF) is Hausdorff. Let aโˆˆE. Then f is continuous at a if:

limxโ†’a,xโˆˆEโกfโข(x)=fโข(a)

.

The main connection between continuity between spaces and continuity at a point is given in the following key theorem.

Theorem 3.38.

Let (E,๐’ฏE) and (F,๐’ฏF) be two topological spaces. The function f is continuous at every xโˆˆE if and only if f is continuous on E.

Proof.

Assume first that f is continuous at every xโˆˆE. Let Vโˆˆ๐’ฏF. Let xโˆˆf-1โข(V). Then f is continuous at x so there exists Uxโˆˆ๐’ฏE such that xโˆˆUx and fโข(Ux)โŠ†V (since fโข(x)โˆˆV). Consequently, UxโŠ†f-1โข(V). Hence:

f-1โข(V)=โ‹ƒxโˆˆf-1โข(V)Ux

and thus f-1โข(V) is open. So f is continuous on E.

Conversely, assume f is continuous on E. Let xโˆˆE. Let Vโˆˆ๐’ฏF such that fโข(x)โˆˆV. Then f-1โข(V) is open in E by continuity of f on E, and it contains x. Let U=f-1โข(V). Then fโข(U)โŠ†V. This concludes the proof. โˆŽ

3.4 Limit of Sequences

We refer to the chapter Fundamental Examples for the topology on the one point compactification โ„•ยฏ of โ„•.

Definition 3.39.

Let E be a set. A sequence in E is a function from โ„• to E.

Notation 3.40.

Sequences are denoted as families. Namely, if x is sequence in E, then we write xn for xโข(n) for nโˆˆโ„•, and we usually identify x with (xn)nโˆˆโ„•.

Remark 3.41.

By abuse of language, we will also call sequences functions from a subset {nโˆˆโ„•:nโ‰ฅN} of โ„•, for some Nโˆˆโ„•. The obvious re-indexing is left implicit.

The definition of limit for functions apply to sequences.

Definition 3.42.

Let (E,๐’ฏ) be a topological space. Let (xn)nโˆˆโ„• be a sequence in E. Then we say that (xn)nโˆˆโ„• converges to lโˆˆE when the limit of (xn)nโˆˆโ„• has limit l at โˆž along โ„• in ๐’ฏโ„•ยฏ.

Remark 3.43.

We learnt that if two functions agree on some set A inside of a topological space E, then their limits along a agree (including whether they exist!). So the previous definition is understood as follows: choose any extension of a sequence to โ„•ยฏ (i.e. pick some element xโˆž in E). Then take its limit at โˆž in โ„•ยฏ along โ„•. Then the existence and, if applicable, the value of this limit is independant of the choice the extension of our sequence. It is the limit of the sequence.

Proposition 3.44.

Let (E,๐’ฏ) be a Hausdorff topological space. A sequence (xn)nโˆˆโ„• converges to lโˆˆE if and only if:

โˆ€Vโˆˆ๐’ฑ๐’ฏEโข(l)โขโˆƒNโˆˆโ„•โขโˆ€nโ‰ฅNโขxnโˆˆVโข.
Proof.

Assume that (xn)nโˆˆโ„• has limit l. Let Vโˆˆ๐’ฑ๐’ฏEโข(l). By definition of limits and of the topology ๐’ฏโ„•ยฏ, there exists a finite set S in โ„• such that xnโˆˆV for all nโˆˆโ„•โˆ–F. Let N be the successor of the greatest element in F if F is nonempty, or 0 otherwise. Then for all nโ‰ฅN we have xnโˆˆV as desired.

The converse is obvious. โˆŽ

Definition 3.45.

A subsequence of a sequence (xn)nโˆˆโ„• in a set E is a sequence of the form (xฯ•โข(n))nโˆˆโ„• for some strictly increasing function ฯ•:โ„•โ†’โ„•.

Lemma 3.46.

Let ฯ•:โ„•โ†’โ„• be strictly increasing. Then for all nโˆˆโ„• we have ฯ•โข(n)โ‰ฅn.

Proof.

By definition, ฯ•โข(0)โ‰ฅ0. Assume that for some nโˆˆโ„• we have ฯ•โข(n)โ‰ฅn. Then by assumption, ฯ•โข(n+1)>ฯ•โข(n)โ‰ฅn so ฯ•โข(n+1)โ‰ฅn+1. The lemma holds by the theorem of induction. โˆŽ

Proposition 3.47.

Let (E,๐’ฏ) be a topological space. If a sequence (xn)nโˆˆโ„• in E converges to l then all subsequences of (xn)nโˆˆโ„• converge to l.

Proof.

Assume that (xn)nโˆˆโ„• converges to l and let ฯ•:โ„•โ†’โ„• be a strictly increasing function. Let Vโˆˆ๐’ฑ๐’ฏโข(l). By assumption, there exists Nโˆˆโ„• such that for all nโ‰ฅN we have xnโˆˆV. Therefore, xฯ•โข(n)โˆˆV since ฯ•โข(n)โ‰ฅn. Hence the subsequence (xฯ•โข(n))nโˆˆโ„• converges to l as well. โˆŽ

Theorem 3.48.

Let (E,๐’ฏ) be a topological space. A sequence (xn)nโˆˆโ„• in E converges to l if and only if every subsequence of (xn)nโˆˆโ„• has a subsequence which converges to l.

Proof.

The condition is necessary by the previous result. Let us show it is sufficient. Assume that (xn)nโˆˆโ„• does not converge to l. Then there exists Vโˆˆ๐’ฑ๐’ฏโข(l) such that for all Nโˆˆโ„• there exists nโ‰ฅN such that xnโˆ‰V. Set ฯ•โข(0) to be the smallest nโˆˆโ„• such that xnโˆ‰V. Assume we have constructed ฯ•โข(0)<โ‹ฏ<ฯ•โข(N) for some Nโˆˆโ„• such that xฯ•โข(k)โˆ‰V for k=0,โ€ฆ,N. Then let AN={nโˆˆโ„•:nโ‰ฅฯ•โข(N)+1}. Then ANโ‰ โˆ…, so it has a smallest element ฯ•โข(N+1). Note that by construction, ฯ•โข(N+1)>ฯ•โข(N) and xฯ•โข(N+1)โˆ‰V. Hence we have constructed a subsequence (xฯ•โข(n))nโˆˆโ„• entirely contained in โˆEโขV. By construction, it has no subsequence which converges to l. โˆŽ

4 Limits and Continuity in Metric Spaces

In this section, the topology on any metric space is meant as the metric topology.

4.1 Convergence in metric spaces

Proposition 4.1.

Let (E,d) be a metric space. A sequence (xn)nโˆˆโ„• in E converges to l if and only if:

โˆ€ฮต>0โขโˆƒNโˆˆโ„•โขโˆ€nโ‰ฅNโขdโข(xn,l)<ฮตโข.
Proof.

Since open balls form a basis for the metric topology, the definition of limit for a sequence at infinity (along โ„•) is equivalent to:

โˆ€ฮต>0โขโˆƒNโˆˆโ„•โขโˆ€nโ‰ฅNโขxnโˆˆBโข(l,ฮต)

which is equivalent by definition to the statement in the proposition. โˆŽ

4.2 Squeeze Theorems

The following result is often useful.

Theorem 4.2.

Let (E,d) be a metric space. Let (xn)nโˆˆโ„• be a sequence in E and lโˆˆE. If there exists a sequence (rn)nโˆˆโ„• in [0,โˆž)โŠ†โ„ and Nโˆˆโ„• such that limnโ†’โˆžโกrn=0 and dโข(xn,l)โ‰คrn for all nโ‰ฅN, then lโขiโขmnโ†’โˆžโขxn=l.

Proof.

Let vโขaโขrโขeโขpโขsโขiโขlโขoโขn>0 be given. Since (rn)nโˆˆโ„• converges to 0 in [0,โˆž), there exists Nโˆˆโ„• such that for all nโ‰ฅN, rnโˆˆ[0,ฮต). Hence, for nโ‰ฅN we have dโข(xn,l)โ‰คฮต, as desired. โˆŽ

This theorem shows that a decent supply of sequences converging to 0 in โ„ is useful. The following results help in these constructions.

Proposition 4.3.

Let ฯ†:โ„•โ†’(0,โˆž) be increasing and not bounded. Let Mโˆˆ[0,โˆž). Then (Mฯ†โข(n))nโˆˆโ„• converges to 0.

Proof.

Let ฮต>0 be given. By assumption, there exists Nโˆˆโ„• such that ฯ†โข(n)>Mฮต. Since ฯ† is increasing, for all nโ‰ฅN we have ฯ†โข(n)โ‰ฅMฮต. Hence for all nโ‰ฅN we have Mฯ†โข(n)โ‰คฮตMโขM=ฮต, as desired. โˆŽ

The existence of unbounded functions ฯ† is a consequence of the fact that โ„ is Archimedean. Namely, the injection i:โ„•โ†’โ„ is unbounded precisely because of the Archimedean principle (in fact, the two statements are equivalent). Therefore, (1n)nโˆˆโ„•* converges to 0, as desired.

Theorem 4.4.

Let (xn)nโˆˆโ„•, (yn)nโˆˆโ„• and (zn)nโˆˆโ„• be three sequences in a linearly ordered set (E,โ‰ค), and Nโˆˆโ„• such that for all nโ‰ฅN we have xnโ‰คynโ‰คzn. If (xn)nโˆˆโ„• and (zn)nโˆˆโ„• have limit l then so does (yn)nโˆˆโ„•.

Proof.

โˆŽ

The above result is used in โ„ under many names: squeeze theorem, guardsmen theorem, pinching theorem, and more. It is often useful to prove limits.

Of course, โ„ is a field, and we will have to address the matter of the continuity of the operations. We shall do this later, however, as we do not need them at this moment.

4.3 Closures and Limits

Theorem 4.5.

Let (E,d) be a metric space. Let AโŠ†E. Then aโˆˆAยฏ if and only if there exists a sequence (xn)nโˆˆโ„• in A converging to a.

Proof.

If there exists a sequence in A converging to a then aโˆˆAยฏ.

Conversely, assume aโˆˆAยฏ. For nโˆˆโ„•*, there exists xnโˆˆB(a.1n)โˆฉA. By construction, dโข(xn,a)โ‰ค1n for all nโˆˆโ„•* so (xn)nโˆˆโ„•* converges to a, which completes our proof. โˆŽ

Limits of subsequences can be characterized in metric spaces as follows:

Theorem 4.6.

Let (E,d) be a metric space. Let (xn)nโˆˆโ„• be a sequence in E. Then lโˆˆE is a limit for a subsequence of (xn)nโˆˆโ„• if and only if:

lโˆˆโ‹‚nโˆˆโ„•{xk:kโ‰ฅn}ยฏโข.
Proof.

Assume l is the limit of some subsequence (xฯ•โข(n))nโˆˆโ„• of (xn)nโˆˆโ„•. Let nโˆˆโ„•. Then l is the limit of the truncated sequence {kโˆˆโ„•:kโ‰ฅn}โ†ฆxฯ•โข(k) at โˆž, so lโˆˆ{xk:kโ‰ฅn}ยฏ since ฯ•โข(n)โ‰ฅn. Hence the condition is necessary.

Assume now lโˆˆโ‹‚nโˆˆโ„•{xk:kโ‰ฅn}ยฏ. To ease notations, write Xn={xk:kโ‰ฅn} for all nโˆˆโ„•. Since lโˆˆX0ยฏ, the set {kโˆˆโ„•:dโข(xk,l)<1} is not empty. Let ฯ•โข(0) be its smallest element.

Assume now we have constructed ฯ•โข(0)<โ‹ฏ<ฯ•โข(n) for some nโˆˆโ„• such that dโข(xฯ•โข(k),l)<1k+1 for k=0,โ€ฆ,n. Since lโˆˆXฯ•โข(n)+1ยฏ, the set {kโˆˆโ„•:k>ฯ•โข(n)โˆงdโข(xฯ•โข(k),l)<1n+2} is not empty. Let ฯ•โข(n+1) be its smallest element.

One checks that l=limnโ†’โˆžโกxฯ•โข(n) as desired. โˆŽ

4.4 Limits at a point

Theorem 4.7.

Let (E,dE), (F,dF) be metric spaces. Let f:Eโ†’F be a map. Let xโˆˆE, AโŠ†E with xโˆˆAยฏ, and lโˆˆF. Then f is has limit l at x along A if and only if for any sequence (xn)nโˆˆN in A which converges to x, we have limnโ†’โˆžโกfโข(xn)=l.

Proof.

Note that since xโˆˆAยฏ and (E,d) is metric, there is a sequence of elements of A converging to x. The theorem on composition of limits shows that the condition is necessary. Let us show that it is sufficient by contraposition. Assume f does not converge to l at x. Hence, there exists ฯต>0 such that for all ฮด>0, there exists zโˆˆE such that dEโข(x,z)<ฮด yet dFโข(fโข(z),l)>ฯต. For each nโˆˆโ„•*, pick xnโˆˆA such that dEโข(xn,z)<1n and dFโข(fโข(xn),l)>ฯต. Note that since xโˆˆAยฏ, we can always choose such an xn. By the squeeze theorem, we conclude that lโขiโขmnโ†’โˆžโขxn=x. On the other hand, (fโข(xn))nโˆˆโ„•* does not converge to l. This proves our result. โˆŽ

4.5 Continuity at a Point

Theorem 4.8.

Let (E,dE), (F,dF) be metric spaces. Let f:Eโ†’F be a map. Let xโˆˆE. Then f is continuous at x if and only if for any sequence (xn)nโˆˆN which converges to x in E, we have limnโˆˆโ„•โกfโข(xn)=fโข(a).

Proof.

This is a direct application of the result on limits in a metric space. โˆŽ

5 Filters

This chapter introduces filters as a mean to analyze topological spaces. In a sense to be illustrated below, filters provide a generalization of sequences, albeit in a dual manner.

5.1 Filters on sets

Definition 5.1.

Let E be a set. A filter base on E is a subset โ„ฌ of ๐ŸE such that:

  1. 1.

    โ„ฌ is not empty and,

  2. 2.

    โˆ…โˆ‰โ„ฌ

  3. 3.

    If A,Bโˆˆโ„ฌ then there exists Cโˆˆโ„ฌ such that CโŠ†AโˆฉB.

Definition 5.2.

Let E be a set. A filter on E is a subset โ„ฑ of ๐ŸE such that:

  1. 1.

    โˆ…โˆ‰โ„ฑ,

  2. 2.

    Eโˆˆโ„ฑ,

  3. 3.

    If Aโˆˆโ„ฑ and CโŠ†E with AโŠ†C then Cโˆˆโ„ฑ,

  4. 4.

    If A,Bโˆˆโ„ฑ then AโˆฉBโˆˆโ„ฑ.

Hence a filter is a proper nonempty upset closed under finite intersections. Note that the empty set is never in a filter. Note that a filter is a filter basis.

Proposition 5.3.

Let โ„ฌ be a filter basis on a set E. Let:

โ„ฑโ„ฌ={AโŠ†E:โˆƒBโˆˆโ„ฌโขBโŠ†A}โข.

Then โ„ฑโ„ฌ is a filter on E, and it is the smallest filter on E containing โ„ฌ. We call it the filter generated by โ„ฌ.

Proof.

By definition, โ„ฌโŠ†โ„ฑโ„ฌ so โ„ฑ is not empty. Moreover, if Aโˆˆโ„ฑโ„ฌ and CโŠ†E such that AโŠ†C then, by construction, there exists Bโˆˆโ„ฌ such that BโŠ†A and thus BโŠ†C so Cโˆˆโ„ฑโ„ฌ. If A1,A2โˆˆโ„ฑโ„ฌ then there exists B1,B2โˆˆโ„ฌ such that BiโŠ†Ai for i=1,2. Thus B1โˆฉB2โŠ†A1โˆฉA2. Since โ„ฌ is a filter basis, there exists Cโˆˆโ„ฌ such that CโŠ†B1โˆฉB2 so CโŠ†A1โˆฉA2. Hence A1โˆฉA2โˆˆโ„ฑโ„ฌ. Last, if โˆ…โˆˆโ„ฑโ„ฌ then there must be Aโˆˆโ„ฌ such that AโŠ†โˆ…, i.e. โˆ…โˆˆโ„ฌ which is a contradiction. So โˆ…โˆ‰โ„ฑโ„ฌ. So โ„ฑโ„ฌ is a filter.


Let ๐’ข be some filter containing โ„ฌ. Let Aโˆˆโ„ฑโ„ฌ. Then by definition, there exists Bโˆˆโ„ฌ such that BโŠ†A. Sicne Bโˆˆ๐’ข and ๐’ข is a filter, we conclude Aโˆˆ๐’ข. Hence โ„ฑโ„ฌโŠ†๐’ข as desired. โˆŽ

Proposition 5.4.

Let E, F be sets and f:Eโ†’F be a function. If โ„ฑ is a filter basis on E, then fโข[โ„ฑ], defined as:

fโข[โ„ฑ]={fโข(A):Aโˆˆโ„ฑ}

is a filter basis.

Proof.

By definition, fโข[โ„ฑ] is not empty and does not contain the emptyset. Now, let A,Bโˆˆfโข[โ„ฑ]. By definition, there exists C,Dโˆˆโ„ฑ such that A=fโข(C) and B=fโข(D). Now, since โ„ฑ is a filter basis, there exists Hโˆˆโ„ฑ such that HโŠ†CโˆฉD. Hence fโข(H)โŠ†AโˆฉB, and by definition, fโข(H)โˆˆfโข[โ„ฑ]. โˆŽ

Example 5.5.

Let f:Eโ†’F be constant, equal to tโˆˆF, while E,F have more than one element. Let xโˆˆE. Then โ„ฑ={AโŠ†E:xโˆˆA} is a filter on E. Now, fโข[โ„ฑ]={{t}}. One checks easily that this is a filter basis, but not a filter (as it would contain F which is assumed to contain a set of cardinal 2). Hence, images of filters may not be filters. We could define fโข[[โ„ฑ]] to be the filter generated by the filter basis fโข[โ„ฑ], thus defining a map from filters to filters for each map from E to F.

5.2 Limits of Filters basis

Definition 5.6.

Let (E,๐’ฏ) be a topological space. Let xโˆˆE. Let โ„ฑ be a filter basis on E. We say that โ„ฑ converges to x when:

โˆ€Uโˆˆ๐’ฏ(xโˆˆU)โŸนโˆƒBโˆˆโ„ฑBโŠ†U.
Remark 5.7.

Equivalently, a filter basis โ„ฑ converges to x if the set ๐’ฑ๐’ฏโข(x) of open neighborhoods of x is contained in the filter generated by โ„ฑ.

Example 5.8.

The set ๐’ฑ๐’ฏโข(x) is a filter basis which converges to x. The elements of the filter generated by ๐’ฑ๐’ฏโข(x) are called neighborhoods of x.

The following theorem illustrates the connection between filters and sequences.

Theorem 5.9.

Let (E,๐’ฏ) be a topological space. Let (xn)nโˆˆโ„• be a sequence in E. For nโˆˆโ„• we define Xn={xk:kโ‰ฅn}. Then ๐’ณ={Xn:nโˆˆโ„•} is a filter basis. Moreover, (xn)nโˆˆโ„• converges to x if and only if ๐’ณ converges to x.

Proof.

Since Xn+1โŠ†Xn for all nโˆˆโ„•, it is immediate that ๐’ณ is a filter basis. Assume that (xn)nโˆˆโ„• converges to x. Let Uโˆˆ๐’ฏ such that xโˆˆU. Then by definition, there exists Nโˆˆโ„• such that for all nโ‰ฅN we have xnโˆˆU. Hence XNโŠ†U. Thus, ๐’ณ converges to x (since U was arbitrary).

Conversely, assume that ๐’ณ converges to x. Let Uโˆˆ๐’ฏ such that xโˆˆU. By definition, there exists XNโˆˆ๐’ณ such that XNโŠ†U. Hence for all nโ‰ฅN we have xnโˆˆU. So (xn)nโˆˆโ„• converges to x (since U was arbitrary). โˆŽ

Remark 5.10.

Let (xn)nโˆˆโ„• be a sequence. Let (xฯ†โข(n))nโˆˆโ„• be a subsequence. The filter generated by the filter basis {{xk:kโ‰ฅn}:nโˆˆโ„•} is a subset of the filter generated by the filter basis {{xฯ•โข(k):kโ‰ฅn}:nโˆˆโ„•}. So filters โ€œgrowโ€ as we take subsequences. From this observation, we note that by definition, it is now immediate to see that if a sequence converge to some x then all of its subsequences do too. The following proposition generalizes this observation.

Proposition 5.11.

Let (E,๐’ฏ) be a topological space. Let โ„ฑ and ๐’ข be two filter bases on E such that for all Fโˆˆโ„ฑ there exists Gโˆˆ๐’ข such that GโŠ†F (we say that ๐’ข is finer than โ„ฑ). If โ„ฑ converges to x, then ๐’ข converges to x.

Proof.

Let x be a limit of โ„ฑ. Let Uโˆˆ๐’ฏ such that xโˆˆ๐’ฏ. There exists Fโˆˆโ„ฑ such that FโŠ†U. By assumption, there exists Gโˆˆ๐’ข such that GโŠ†F. Hence ๐’ข converges to x. โˆŽ

Corollary 5.12.

Let (E,๐’ฏ) be a topological space. If โ„ฑโŠ†๐’ข are two filters on E and if โ„ฑ converges to x, then ๐’ข converges to x.

Limits of filters may not be unique. We have:

Theorem 5.13.

Let (E,๐’ฏ) be a topological space. The following are equivalent:

  1. 1.

    The space (E,๐’ฏ) is Hausdorff,

  2. 2.

    Every convergent filter basis on E has a unique limit.

Proof.

Assume that (E,๐’ฏ) is Hausdorff. Let โ„ฑ be a filter basis on E. Assume it converges to x and y in E. If xโ‰ y then, since ๐’ฏ is Hausdorff, there exists Ux,Uyโˆˆ๐’ฏ such that xโˆˆUx, yโˆˆUy and UxโˆฉUy=โˆ…. By definition of convergence, there exist Bx,Byโˆˆโ„ฑ such that BxโŠ†Ux and ByโŠ†Uy. By definition of filter basis, there exists Cโˆˆโ„ฑ such that CโŠ†BxโˆฉBy=โˆ…. This contradicts the fact โ„ฑ, as a filter basis, does not contain the empty set. Hence limits are unique if they exist.


Assume now that all filter bases have unique limits in (E,๐’ฏ). In particular, for any xโˆˆE, the filter basis ๐’ฑ๐’ฏโข(x) of open neighborhoods of x in ๐’ฏ, has x for unique limit. Let yโ‰ x. Suppose that for all Uโˆˆ๐’ฑ๐’ฏโข(y), for all Vโˆˆ๐’ฑ๐’ฏโข(x), we have UโˆฉVโ‰ โˆ…. Let:

โ„ฑ={UโˆฉV:Uโˆˆ๐’ฑ๐’ฏโข(x),Vโˆˆ๐’ฑ๐’ฏโข(y)}.

By assumption, โ„ฑ is a nonempty set of elements in ๐’ฏ not containing the emptyset. Moreover, it is closed by finite intersection (as ๐’ฏ is). Hence, it is a filter basis. Yet, for any Uโˆˆ๐’ฑ๐’ฏโข(x) and any Vโˆˆ๐’ฑ๐’ฏโข(y), we have U,Vโˆˆโ„ฑ so โ„ฑ converges to x and y by definition. This is a contradiction. Hence, for all Uโˆˆ๐’ฑ๐’ฏโข(y) there exists Vโˆˆ๐’ฑ๐’ฏโข(x) such that UโˆฉV=โˆ…, i.e. (E,๐’ฏ) is Hausdorff. โˆŽ

Notation 5.14.

When (E,๐’ฏ) is Hausdorff and โ„ฑ is a filter basis converging to x then we write: x=limโกโ„ฑ.

5.3 Some Applications of Filters

As a rule, filters can be used to replace sequences in general topology to obtain results limited to metric spaces. For instance:

Lemma 5.15.

Let (E,๐’ฏ) be a topological space. Let AโŠ†E. Let โ„ฑ be a filter basis in A which converges to x in ๐’ฏ. Then xโˆˆAยฏ.

Proof.

Let Vโˆˆ๐’ฑ๐’ฏโข(x). By definition of convergence, there exists Bโˆˆโ„ฑ such that BโŠ†V. Now, BโŠ†A by assumption, so eโขmโขpโขtโขyโขsโขeโขtโ‰ BโŠ†VโˆฉA. So xโˆˆAยฏ since V is arbitrary. โˆŽ

Theorem 5.16.

Let (E,๐’ฏ) be a topological space. Let AโŠ†E. The closure of A in ๐’ฏ is the set of all limits in E of filters of A.

Proof.

All limits of filters of A are in Aยฏ by the previous lemma. Assume xโˆˆAยฏ. If Uโˆˆ๐’ฑ๐’ฏโข(x) then AโˆฉUโ‰ โˆ…. It is then obvious that:

{UโˆฉA:Uโˆˆ๐’ฑ๐’ฏโข(x)}

is a filter basis of A which converges to x. โˆŽ

Definition 5.17.

Let E be a set, (F,๐’ฏ) be a topological space, โ„ฑ a filter basis on E, and f:Eโ†’F. Then we say that f converges to xโˆˆF along โ„ฑ if fโข[โ„ฑ] converges to x.

When (F,๐’ฏF) is Hausdorff, we shall write: limโ„ฑโกf for this limit.

Theorem 5.18.

Let (E,๐’ฏE) and (F,๐’ฏF) be two topological spaces, and let f:Eโ†’F. Let AโŠ†E, aโˆˆAยฏ. Then f has limit l at a along A if and only if for all filter bases โ„ฑ of A converging to a, the filter basis fโข[โ„ฑ] converges to l in F.

Proof.

Assume first that f converges. Let โ„ฑ be a filter basis of A converging to a. Note that we have shown such a filter basis exists. Now, let Uโˆˆ๐’ฏF such that lโˆˆU. Since f converges to l at a along A, there exists Vโˆˆ๐’ฏE with aโˆˆV such that fโข(AโˆฉV)โŠ†U. Since โ„ฑ converges to a, there exists Cโˆˆโ„ฑ such that CโŠ†V. Since CโŠ†A by assumption, C=CโˆฉAโŠ†VโˆฉA so fโข(C)โŠ†U as desired. Since U was arbitrary, fโข[โ„ฑ] converges to l as desired.


Conversely, assume that fโข[โ„ฑ] converges to l for all filter bases โ„ฑ of A converging to a. Let Uโˆˆ๐’ฏV such that lโˆˆU. The filter basis ๐’ฑ={VโˆฉA:Vโˆˆ๐’ฏ,aโˆˆV} converges to a and is a subset of ๐ŸA so by assumption, fโข[๐’ฑ] converges to l. Thus, there exists Cโˆˆ๐’ฑ such that fโข(C)โŠ†U. Hence, by definition, there exists Vโˆˆ๐’ฏE such that fโข(VโˆฉA)โŠ†U. As U is arbitrary in ๐’ฑ๐’ฏFโข(l), we conclude that f converges to l at a along A. โˆŽ

Corollary 5.19.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces, xโˆˆE and f:Eโ†’F. Then f is continuous at x if and only if the image by f of all filter bases of E converging to x converge to fโข(x).

5.4 Ultrafilters

Definition 5.20.

Let E be a set. An ultrafilter on E is a filter โ„ฑ such that for all AโŠ†E, we have either Aโˆˆโ„ฑ or โˆEโขAโˆˆโ„ฑ.

Example 5.21.

Let xโˆˆE. Then {AโŠ†E:xโˆˆA} is an ultrafilter.

Since for no set A can we have A and โˆEโขA in a filter of E, this definition suggests the following:

Theorem 5.22.

A filter on a set E is an ultrafilter if and only if it is maximal among all filters on E (for the inclusion).

Proof.

Assume first that โ„ฑ is an ultrafilter on E. Let ๐’ข be a filter on E such that โ„ฑโŠ†๐’ข. Let Aโˆˆ๐’ข. Assume โˆEโขAโˆˆโ„ฑ. Then โˆEโขAโˆˆ๐’ข, and thus โˆ…=AโˆฉโˆEโขA is in ๐’ข, as filters are closed under finite intersections. This is a contradiction. Since โ„ฑ is an ultrafilter, Aโˆˆโ„ฑ. As A was arbitrary, ๐’ข=โ„ฑ.


Conversely, assume that โ„ฑ is a maximal filter for inclusion. Assume that it is not an ultrafilter. Let YโŠ†E such that Y,โˆEโขYโˆ‰โ„ฑ. Without loss of generality, assume Yโ‰ โˆ…. Let:

๐’ข={CโŠ†E:โˆƒAโˆˆโ„ฑโขAโˆฉYโŠ†C}.

Now, by construction, ๐’ข is not empty. Assume โˆ…โˆˆ๐’ข. Then there exists Aโˆˆโ„ฑ such that AโˆฉY=โˆ…. Thus AโŠ†โˆEโขY. As โ„ฑ is a filter, this would imply โˆEโขYโˆˆโ„ฑ which is a contradiction. So ๐’ข does not contain the empty set. By construction, if Aโˆˆ๐’ข and AโŠ†C then Cโˆˆ๐’ข. Moreover, if A1,A2โˆˆ๐’ข then there exist B1,B2โˆˆโ„ฑ such that BiโˆฉYโŠ†Ai for i=1,2. Thus B1โˆฉB2โˆฉYโŠ†A1โˆฉA2. Since B1โˆฉB2โˆˆโ„ฑ we conclude that A1โˆฉA2โˆˆ๐’ข. Thus ๐’ข is a filter. By definition, since AโˆฉYโŠ†A for all Aโˆˆโ„ฑ, we have โ„ฑโŠ†๐’ข. Since EโˆฉY=Y and Eโˆˆโ„ฑ, we have Yโˆˆ๐’ข, yet Yโˆ‰โ„ฑ. This contradicts the maximality of โ„ฑ. So โ„ฑ is an ultrafilter. โˆŽ

Maximality lends itself to the application of Zornโ€™s lemma to prove the existence of ultrafilters containing any given filer.

Theorem 5.23.

Let E be a set, and let โ„ฑ be a filter basis on E. Then there exists an ultrafilter ๐’ข containing โ„ฑ.

Proof.

Since every filter basis is contained in a filter, we will assume โ„ฑ is a filter. Let:

๐’ซ={โ„‹โขย filters onย โขE:โ„ฑโŠ†โ„‹}.

The set ๐’ซ is ordered by inclusion, and it is not empty (it contains โ„ฑ. Note that ๐’ซ is an upper set, and thus a maximal element of ๐’ซ is a maximal element of the set of all filters on E. Thus, it is enough to show that ๐’ซ has a maximal element.

Let ๐’ฐ be a chain in ๐’ซ, i.e. a linearly ordered subset of ๐’ซ. Let ๐’ด=โ‹ƒ๐’ฐ. If A,Bโˆˆ๐’ด then there exists U,Vโˆˆ๐’ฐ such that AโˆˆU and BโˆˆV. Since ๐’ฐ is linearly ordered, we may assume UโŠ†V. Thus A,BโˆˆV, and V is a filter, so AโˆฉBโˆˆVโŠ†โ‹ƒ๐’ด. The set ๐’ด is not empty (filters never are), does not contain the empty set (filters never do), and is an upper set: if Aโˆˆ๐’ด and AโŠ†C for some C, then there exists Uโˆˆ๐’ฐ such that AโˆˆU, and since U is a filter, CโˆˆU so Cโˆˆ๐’ด. Therefore, ๐’ด is a filter. Trivially ๐’ดโˆˆ๐’ซ. Hence, ๐’ฐ admits an upper bound. Since ๐’ฐ is an arbitrary chain, Zornโ€™s lemma implies that ๐’ซ admits a maximal element ๐’ข. Thus ๐’ข is a maximal filter, and hence an ultrafilter, containing โ„ฑ. โˆŽ

Due to this result, it is possible to rephrase the theorems in the application section of this chapter in terms of ultrafilters only.

6 Compactness

6.1 Compact space

Definition 6.1.

A topoligical space (E,๐’ฏ) is compact if, given any ๐’ฐโŠ†๐’ฏ such that E=โ‹ƒ๐’ฐ, there exists a finite subset ๐’ฑ of ๐’ฐ such that E=โ‹ƒ๐’ฑ.

Remark 6.2.

A subset ๐’ฐ of ๐’ฏ such that E=โ‹ƒ๐’ฐ is called an open covering of E.

Theorem 6.3.

Let (E,๐’ฏ) be a topological space. Then the following are equivalent:

  1. 1.

    (E,๐’ฏ) is compact,

  2. 2.

    For any set โ„ฑ of closed subsets of E whose intersection is empty, there exists a finite subset ๐’ข of โ„ฑ whose intersection is empty.

Proof.

The theorem follows by taking complements. โˆŽ

Corollary 6.4.

Let (E,๐’ฏ) be a compact space. Let (Fn)nโˆˆโ„• be a decreasing sequence of nonempty closed subsets of E (where the order on sets in inclusion). Then:

โ‹‚nโˆˆโ„•Fnโ‰ โˆ…โข.
Proof.

If โ‹‚nโˆˆโ„•Fn=โˆ… then since (E,๐’ฏ) is compact, we can find Nโˆˆโ„• such that F1โˆฉF2โˆฉโ‹ฏโˆฉFN=โˆ…. Yet since the family is decreasing, F1โˆฉF2โˆฉโ‹ฏโˆฉFN=FN which is assumed not empty. Our result follows by contraposition. โˆŽ

Theorem 6.5.

Let (E,๐’ฏ) be a topological space. Let โ„ฑ be a set of closed subsets of E such that given any finite ๐’ขโŠ†โ„ฑ, we have โ‹‚๐’ขโ‰ โˆ… (we say that โ„ฑ has the finite intersection property). Then โ‹‚โ„ฑโ‰ โˆ… if and only if (E,๐’ฏ) is compact.

Proof.

Assume first that (E,๐’ฏ) is compact and consider a set โ„ฑ of closed subsets of E. Assume โ‹‚โ„ฑ=โˆ…. Since E is compact, there exists a finite set ๐’ขโŠ†โ„ฑ such that โ‹‚๐’ข=โˆ…. This proves the necessity of our theorem by contraposition.

Conversely, assume that for all sets โ„ฑ of closed subsets of E with the finite intersection property, โ‹‚โ„ฑโ‰ โˆ…. Let โ„ฑ be a set of closed subsets of E such that โ‹‚โ„ฑ=โˆ…. Then by our assumption, there exists at least one finite subset ๐’ขโŠ†โ„ฑ such that โ‹‚๐’ข=โˆ…. โˆŽ

Example 6.6.

The space (E,๐ŸE) is compact if and only if E is finite.

Example 6.7.

It is striaghtforward that if Un=(-n,โˆž) in โ„ for all nโˆˆโ„• then {Un:nโˆˆโ„•} is an open covering of โ„ which has no finite subcovering. So โ„ is not compact.

6.2 Compact Subspaces

Definition 6.8.

Let (E,๐’ฏ) be a topological space. Let AโŠ†E. Then A is a compact subspace of (E,๐’ฏ) if (A,๐’ฏA) is a compact space (where ๐’ฏA is the trace topology on A).

Remark 6.9.

We will say that A is compact in E when A is a compact subspace of E.

Theorem 6.10.

Let (E,๐’ฏ) be a topological space. Let AโŠ†E. Then A is compact if and only if for all ๐’ฐโŠ†๐’ฏ such that AโŠ†โ‹ƒ๐’ฐ, there exists nโˆˆโ„• and U1,โ€ฆ,Unโˆˆ๐’ฐ such that:

AโŠ†โ‹ƒk=1nUkโข.
Proof.

Assume that (A,๐’ฏA) is compact in (E,๐’ฏ). Let ๐’ฐโŠ†๐’ฏ such that AโŠ†โ‹ƒ๐’ฐ. Then A=โ‹ƒ{UโˆฉA:Uโˆˆ๐’ฐ}. By definition, UโˆฉAโˆˆ๐’ฏA for all Uโˆˆ๐’ฏ. Hence by compactness of A, there exists {U1,โ€ฆ,Un}โŠ†๐’ฐ such that A=โ‹ƒi=1n(AโˆฉUi). Hence AโŠ†โ‹ƒi=1nUi.


Assume now that any open covering of A in E admits a finite covering. Let ๐’ฑโŠ†๐’ฏA such that A=โ‹ƒ๐’ฑ. By definition of ๐’ฏA, for each Vโˆˆ๐’ฑ there exists UVโˆˆ๐’ฏ such that V=UVโˆฉA. Thus AโŠ†โ‹ƒ{UV:Vโˆˆ๐’ฑ}, so there exists UV1,โ€ฆ,UVnโˆˆ๐’ฏ such that AโŠ†โ‹ƒi=1nUVi. Hence A=โ‹ƒi=1nVi. โˆŽ

Theorem 6.11.

Let (E,๐’ฏ) be a topological space. Let AโŠ†E. Then A is compact if and only if for any family (Fi)iโˆˆI of closed sets in E such that:

Aโˆฉโ‹‚iโˆˆIFi=โˆ…

there exists a finite subset JโŠ†I such that:

Aโˆฉโ‹‚iโˆˆJFj=โˆ….
Proof.

Take complements in previous theorem. โˆŽ

Theorem 6.12.

Let (E,๐’ฏ) be a Hausdorff topological space. Then if AโŠ†E is compact, then A is closed.

Proof.

Let xโˆˆโˆEโขA and yโˆˆA. Since ๐’ฏ is Hausdorff, there exists Ux,y,Vx,yโˆˆ๐’ฏ such that Ux,yโˆฉVx,y=โˆ…, with yโˆˆUx,y and xโˆˆVx,y. Let ๐’ฐx={Ux,y:yโˆˆA}. By construction:

AโŠ†โ‹ƒ๐’ฐx

and thus, since A is compact, there exists nโˆˆโ„• and y1,lโขdโขoโขtโขs,ynโˆˆA such that:

AโŠ†โ‹ƒk=1nUx,ykโข.

Let:

Wx=โ‹‚k=1nVx,ykโข.

As a finite intersection of open sets, Wx is open. Moreover:

A โŠ‚ โ‹ƒk=1nUx,yk
โŠ† โ‹ƒk=1nโˆEโขVx,yk
= โˆEโข(โ‹‚k=1nVx,yk)=โˆEโขWxโข.

Hence, WxโŠ†โˆEโขA. Therefore:

โˆEโขA=โ‹ƒxโˆˆAWx

and thus โˆEโขA is open, so A is closed as desired. โˆŽ

Theorem 6.13.

Let (E,๐’ฏ) be a compact space. Then if AโŠ†E is closed, then it is compact.

Proof.

Let ๐’ฐโŠ†๐’ฏ such that AโŠ†โ‹ƒ๐’ฐ and let A be closed. Then โˆEโขAโˆˆtโขaโขu so, if ๐’ฑ=๐’ฐโˆช{โˆEโขA}, then E=โ‹ƒ๐’ฑ. Since (E,๐’ฏ) is compact, there exists a finite set ๐’ฒโŠ†๐’ฑ such that E=โ‹ƒ๐’ฒ. Now, if ๐’ต=๐’ฒโˆ–{โˆEโขA}, one checks readily that AโŠ†โ‹ƒ๐’ต and by construction, ๐’ตโŠ†๐’ฐ with ๐’ต finite. โˆŽ

Corollary 6.14.

Let (E,๐’ฏ) be a compact Hausdorff space. Then AโŠ†E is compact if and only if A is closed.

6.3 Continuous image of a compact space

Theorem 6.15.

Let (E,๐’ฏE) be a compact space, (F,๐’ฏF) be a topological space, and f:Eโ†’F be a continuous function. Then fโข(E) is a compact subspace of F.

Proof.

Let ๐’ฐโŠ†๐’ฏF such that fโข(E)โŠ†โ‹ƒ๐’ฐ. Since f is continuous, for all Uโˆˆ๐’ฐ, we have f-1โข(U)โˆˆ๐’ฏE. Let ๐’ฑ={f-1โข(U):Uโˆˆ๐’ฐ}. Then ๐’ฑโŠ†๐’ฏE and by construction, โ‹ƒ๐’ฑ=E. Since (E,๐’ฏ) is compact, there exists nโˆˆN and U1,โ€ฆ,Unโˆˆ๐’ฐ such that E=โ‹ƒk=1nf-1โข(Uk). Hence fโข(E)โŠ†โ‹ƒk=1nUk as desired. โˆŽ

Corollary 6.16.

Let (E,๐’ฏE) and (F,๐’ฏF) be topological spaces. If f:Eโ†’F is continuous, and AโŠ†E is compact, then fโข(A) is compact.

Theorem 6.17.

Let (E,๐’ฏE) be a compact space, and (F,๐’ฏF) be a Hausdorff topological space. Let f:Eโ†’F be a continuous bijection. Then f is a homeomorphism.

Proof.

It suffices to show that f-1 is continuous. Let AโŠ†E be a closed set. Then A is compact since E is compact. So fโข(A) is compact in F. Since F is Hausdorff, fโข(A) is closed. Since f is a bijection, (f-1)-1โข(A)=fโข(A). So f-1 is continuous. โˆŽ

6.4 Compact Metric Spaces

Theorem 6.18.

Let (E,d) be a metric space. Then the following are equivalent:

  1. 1.

    E is compact in its metric topology.

  2. 2.

    Bolzanno-Weierstrass Every sequence in E admits a convergent subsequence.

Proof.

First, assume that E is compact. Let (xn)nโˆˆโ„• be a sequence in E. Let Xn={xk:kโˆˆโ„•,kโ‰ฅn}ยฏ for all nโˆˆโ„•. The sequence (Xn)nโˆˆโ„• is a decreasing sequence of nonempty closed subsets of E, which is compact, so โ‹‚nโˆˆโ„•Xnโ‰ โˆ…. By Theorem (LABEL:AccumulationSeqThm), this implies that (xn)nโˆˆโ„• admits a convergent subsequence.


Second, assume that the Bolzanno-Weierstass axiom holds for E. We first prove a few lemmas.

Lemma 6.19 (Lebesgues number).

Let ๐’ฐโŠ†๐’ฏd be an open covering of E. There exists ฯต>0 such that for all xโˆˆE, there exists Uโˆˆ๐’ฐ such that the open ball Bโข(x,ฯต) of center x and radius ฯต is contained in U.

Proof.

Assume that for all ฯต>0, there exists xโˆˆE such that for all Uโˆˆ๐’ฐ, we have Bโข(x,ฯต)โŠˆU. Let nโˆˆโ„•. There exists xnโˆˆE such that Bโข(xn,1n+1)โŠˆU for all Uโˆˆ๐’ฐ. The sequence (xn)nโˆˆโ„• admits a convergent subsequence (xฯ•โข(n))nโˆˆโ„• by the Bolzanno-Weierstrass axiom. Let lโˆˆE be its limit. Since ๐’ฐ covers E, there exists Uโˆˆ๐’ฐ such that lโˆˆU. Since U is open, there exists ฮด>0 such that Bโข(l,ฮด)โŠ†U. Since (xฯ•โข(n))nโˆˆโ„• converges to l, there exists N2โˆˆโ„• such that for all nโ‰ฅN2, we have xฯ•โข(n)โˆˆBโข(l,ฮด2). Let Nโˆˆโ„• such that 1N+1<ฮด2 and Nโ‰ฅN2. Let zโˆˆBโข(xฯ•โข(N),1ฯ•โข(N)+1). Then:

dโข(z,l)โ‰คdโข(z,xฯ•โข(N))+dโข(xฯ•โข(N),l)<1ฯ•โข(N)+1+ฮด2<ฮด

since ฯ•โข(N)โ‰ฅN. Hence, zโˆˆBโข(l,ฮด) so Bโข(xฯ•โข(N),1ฯ•โข(N)+1)โŠ†Bโข(l,ฮด)โŠ†U. This is a contradiction. โˆŽ

Lemma 6.20 (Precompact).

For any ฯต>0 there exists nโˆˆโ„• and x1,โ€ฆ,xnโˆˆE such that:

E=โ‹ƒj=1nBโข(xj,ฯต)โข.

.

Proof.

Assume that there exists ฯต>0 such E is not the union of finitely many open balls of radius ฯต. Let x0โˆˆE: since EโŠŠBโข(x0,ฯต), there exists x1โˆˆE such that dโข(x0,x1)โ‰ฅฯต.

Assume we have constructed x0,โ€ฆ,xnโˆˆE such that dโข(xi,xj)โ‰ฅฯต for iโ‰ jโˆˆ{0,โ€ฆ,n}. Since EโŠŠโ‹ƒj=0nBโข(xj,ฯต), there exists xn+1โˆˆEโˆ–(โ‹ƒj=0nBโข(xj,ฯต)), i.e. dโข(xn+1,xi)โ‰ฅฯต.

By induction, we have constructued a sequence (xn)nโˆˆโ„• which admits no convergent subseqence. This contradicts the Bolzanno-Weierstrass axiom. โˆŽ

We can now conclude our proof. Assume ๐’ฐ is an open covering of E. There exists ฯต>0 such that any open ball of radius ฯต or less is contained in some Uโˆˆ๐’ฐ. Now, there exists x1,โ€ฆ,xnโˆˆE such that E=โ‹ƒj=1nBโข(xj,ฯต). Let Ujโˆˆ๐’ฐ such that Bโข(xj,ฯต)โŠ†Uj. Then E=โ‹ƒj=1nUj. So E is compact, as desired. โˆŽ

Corollary 6.21.

A compact metric space is bounbed.

Proof.

The precompact lemma shows that a compact metric space can be covered with finitely many balls of radius 1. โˆŽ

Remark 6.22.

The definition of compacity by means of open covering is referred to the Heine-Borel axioms of compactity. Thus we have shown that in metric spaces, Heine-Borel and Bolzanno-Weierstrass are equivalent.

6.5 Compacts of โ„

We provide two proofs of the following result, both based on your real analysis class.

Theorem 6.23.

A subset A of โ„ is compact if and only if it is closed and bounded.

Necessity.

Since โ„ is Hausdorff, compact subsets are closed. Since โ„ is metric, compact subsets are bounded. โˆŽ

First proof of sufficiency: monotonicity.

Assume that AโŠ†โ„ is closed and bounded. Let (xn)nโˆˆโ„• be a sequence in A. By the monotone subsequence theorem, there exists a monotone subsequence of (xn)nโˆˆโ„•. Since A is bounded, this subsequence must converge. Hence A satisfies Bolzanno-Weierstrass axiom in a metric space and is thus compact. โˆŽ

Second proof of sufficiency: dichotomy.

Let a0<b0 be real numbers. Let (xn)nโˆˆโ„• be a sequence in [a0,b0]. Set ฯ•โข(0)=0.

Assume that for some nโˆˆโ„• we have constructed ฯ•โข(0)<โ€ฆ<ฯ•โข(n), a0โ‰คa1โ‰คโ€ฆโ‰คanโ‰คb0 and a0โ‰คbnโ‰คโ€ฆโ‰คb0 such that xฯ•โข(k)โˆˆ[ak,bk] and bk-ak=12kโข(b0-a0). Let mn=an+bn2. If {nโˆˆโ„•:nโ‰ฅฯ•โข(n)โˆงxnโˆˆ[an,mn]} is infinite, then let ฯ•โข(n+1) be its smallest element. We also let an+1=an, bn+1=mn. Otherwise, we set ฯ•โข(n+1) to be the smallest element of the (necessarily infinite) set {nโˆˆโ„•:nโ‰ฅฯ•โข(n)โˆงxnโˆˆ[mn,bn]}, and we let an+1=mn, bn+1=bn. One checks that by induction, we have constructed an subsequence (xฯ•โข(n))nโˆˆโ„• and monotone sequences (an)nโˆˆโ„• and (bn)nโˆˆโ„• such that:

  1. 1.

    (an)nโˆˆโ„• is increasing and bounded above by b0 so it converges.

  2. 2.

    (bn)nโˆˆโ„• is decreasing and bounded above by a0 so it converges.

  3. 3.

    We have limnโ†’โˆžโกan-bn=0 so limnโ†’โˆžโกan=limnโ†’โˆžโกbn.

  4. 4.

    Thus by the squeeze theorem, limnโ†’โˆžโกxฯ•โข(n)=limnโ†’โˆžโกan.

Hence again we have proven the Bolzanno-Weierstrass axiom for all closed intervals. Now, if AโŠ†โ„ is closed and bounded, then it is a closed subset of some closed interval, i.e. of a compact space. Hence it is compact, as needed. โˆŽ

Remark 6.24.

The second proof is longer but can be applied as is to prove that compact subsets of โ„n are closed bounded subsets of โ„n. Indeed, rather than cut an interval in two, one can repeat the argument, cutting a square in four, a cube in eight, and so forth. This process works well as long as we divide an hypercube into finitely many hypercubes at each stage. This proofs fails for infinite dimensional normed vector spaces, and in fact so does the result: one can show that locally compact normed vector spaces are all finite dimensional.

6.6 Uniform Continuity and Heine Theorem

Definition 6.25.

Let (E,dE) and (F,dF) be metric spaces. A function f:Eโ†’F is uniformly continuous when:

โˆ€ฯต>0โขโˆƒฮด>0โขโˆ€x,yโˆˆEโขdEโข(x,y)<ฮดโŸนdFโข(fโข(x),fโข(y))<ฯตโข.
Proposition 6.26.

A uniformly continuous function is continuous.

Proof.

Obvious. โˆŽ

Example 6.27.

A L-Lipshitz function is uniformly continuous: take ฮด=ฯตL if L>0 (i.e. nonconstant functions), and ฮด arbitrary for constants.

Example 6.28.

The function xโˆˆโ„โ†ฆx2 is not uniformly continuous in โ„. Indeed basic precalculus shows that x2<kโขx if and only if xโˆˆ(-k,k).

The following shows that continuity can be strengthened to uniform continuity when working on a compact metric space.

Theorem 6.29.

Let (E,dE) be a compact metric space. Let (F,dF) be a metric space. Let f:Eโ†’F be given. Then f is continuous if and only if f is uniformly continuous on E.

Proof.

If f is uniformly continuous then it is continuous. Let us prove the converse. Assume f is continuous on E. Let ฯต>0. Then for all xโˆˆE there exists ฮดx>0 such that if yโˆˆBโข(x,ฮดx) then dFโข(fโข(y),fโข(x))<ฯต2. Now, since E=โ‹ƒxโˆˆEBโข(x,ฮดx2), and E is compact, there exists x1,โ€ฆ,xn such that E=โ‹ƒj=1nBโข(xj,ฮดxj2). Let ฮด=12โขminโก{ฮดxj:j=1,โ€ฆ,n}>0. Let x,yโˆˆE with dEโข(x,y)<ฮด. There exists jโˆˆ{1,โ€ฆ,n} such that xโˆˆBโข(xj,ฮดxj). Then

dEโข(y,xj)โ‰คdEโข(y,x)+dEโข(x,xj)<ฮด+12โขฮดxjโ‰คฮดxjโข.

Thus:

dFโข(fโข(x),fโข(y))โ‰คdFโข(fโข(x),fโข(xj))+dโข(fโข(xj),fโข(y))<ฯต

as desired.

โˆŽ

Alternate, use real analysis..

If f is uniformly continuous then it is continuous. Let us prove the converse. Assume f is continuous but not uniformly continuous. There exists ฯต>0 such that for all ฮด>0 there exists x,yโˆˆE such that dEโข(x,y)<ฮด and dFโข(fโข(x),fโข(y))โ‰ฅฯต. For nโˆˆโ„• let xn,ynโˆˆE such that dEโข(xn,yn)<1n+1 and dFโข(fโข(xn),fโข(yn))โ‰ฅฯต. Since E is compact, there exists a convergent subsequence (xฯ•โข(n))nโˆˆโ„•. Then there exists a convergent subsequence (yฯ•โข(ฯˆโข(n)))nโˆˆโ„• of (yฯ•โข(n))nโˆˆโ„•. Let lx=limnโ†’โˆžโกxฯ•โข(ฯˆโข(n)) and ly=limnโ†’โˆžโกyฯ•โข(ฯˆโข(n)). Now lx-ly=limnโ†’โˆžโกyฯ•(ฯˆ(n)))-xฯ•โข(ฯˆโข(n))=0. On the other hand, using the continuity of f, we have ฯตโ‰คdFโข(fโข(lx),fโข(ly)) which is a contradiction. โˆŽ

Remark 6.30.

The notion of completeness of metric spaces is not topological: there exist two homeomorphic metric spaces, one complete and one not. A possible choice of morphisms for the category of metric spaces are the uniformly continuous functions, in which case isomorphic metric spaces are either all complete or all not complete.

Now, if E and F are compact metric spaces, and if they are homeomorphic, then E is complete if and only if F is complete. This is a very remarkable fact: a purely topological notion (compactness) imposes a metric notion (completeness). This is seen by observing that a Cauchy sequence has at most one accumulation point, and we have seen that every sequence of a compact space has at least one. With the Heine theorem, we see a different reason: an homeomorphism between compact metric spaces is always bi-uniformly continuous, and hence preserve Cauchy sequences.

6.7 Filter characterization of compactness

Theorem 6.31.

Let (E,๐’ฏ) be a topological space. (E,๐’ฏ) is compact if and only if every ultrafilter in E converge.

Proof.

Assume that (E,๐’ฏ) is compact. Let โ„ฑ be an ultrafilter in E. Assume it does not converge. Then for all xโˆˆE there exists Uxโˆˆ๐’ฏ with xโˆˆUx such that Uxโˆ‰โ„ฑ. As โ„ฑ is an ultrafilter, โˆEโขUxโˆˆโ„ฑ. Now, E=โ‹ƒxโˆˆEUx, and as ๐’ฏ is compact, there exists a finite subset F of E such that E=โ‹ƒxโˆˆFUx. Now, โˆ…=โ‹‚xโˆˆFโˆEโขUxโˆˆโ„ฑ which is a contradiction. So โ„ฑ converges.


Assume now that all ultrafilters in E converge in ๐’ฏ. Assume ๐’ฏ is not compact. Let ๐’ฒโŠ†๐’ฏ such that E=โ‹ƒ๐’ฒ, with ๐’ฒ infinite, and such that no finite subset of ๐’ฒ covers E. Let:

โ„ฑ={โˆEโขโ‹ƒUโˆˆFU:FโŠ†๐’ฒ,finite}.

Since no finite subset of ๐’ฒ covers E, the emptyset is not in โ„ฑ. By construction, โ„ฑ is closed by finite intersections, and thus it is a filter basis. Assume it converges to x. Since ๐’ฒ covers E, there exists Wโˆˆ๐’ฒ such that xโˆˆW. By definition of convergence, there exists a finite set FโŠ†๐’ฒ such that โˆEโขโ‹ƒUโˆˆFUโŠ†W, i.e.:

E=โ‹ƒUโˆˆFโˆช{W}U

with Fโˆช{W} finite subset of ๐’ฒ. This is a contradiction. Hence โ„ฑ does not converge. Now, โ„ฑ is contained in some ultrafilter ๐’ข, which therefore does not converge at all. โˆŽ

6.8 Tychonoff theorem

Theorem 6.32.

Let (Ei,๐’ฏi)iโˆˆI be a family of compact topological spaces. Let ๐’ฏ be the product topology of (๐’ฏi)iโˆˆI. Then (โˆiโˆˆIEi,๐’ฏ) is compact.

Proof.

Let โ„ฑ be an ultrafilter in โˆiโˆˆIEi. Let iโˆˆI. Let pi:โˆjโˆˆIEjโ†’Ei be the canonical surjection. Let โ„ฑi=piโข[โ„ฑ]. Now, let Aโˆˆโ„ฑi and CโŠ†Ei such that CโŠ†Ai. Since โ„ฑ is a filter, and since AโŠ†pi-1โข(A)โŠ†pi-1โข(C), we conclude pi-1โข(A)โˆˆโ„ฑ and pi-1โข(C)โˆˆโ„ฑ. Since pi is a surjection, C=piโข(pi-1โข(C))โˆˆโ„ฑi by definition. It is also immediate that โ„ฑi is a nonempty subset of ๐ŸEi to which the empty set does not belong. Now, the same reasonning shows that if A,Bโˆˆโ„ฑi then AโˆฉBโˆˆโ„ฑi. Hence, โ„ฑi is a filter in Ei. Last, let CโŠ†Ei. Then either pi-1โข(C) or its complement are in โ„ฑ, since โ„ฑ is an ultrafilter. Since C=piโข(pi-1โข(C)) as pi is a surjection, either C or its complemenent belongs to โ„ฑi. Thus โ„ฑi is an ultrafilter on Ei.

Since Ei is compact, โ„ฑi converges to some xi.


Let us show that โ„ฑ converges to (xi)iโˆˆI in the product topology ๐’ฏ. Let Uโˆˆ๐’ฏ such that (xi)iโˆˆIโˆˆU. Let Wโˆˆ๐’ฏ and JโŠ†I finite such that piโข(W)=Ei for iโˆ‰J, and WโŠ†U, (xi)iโˆˆIโˆˆW. Write Wi=piโข(W) for all iโˆˆI and note that xiโˆˆWi, with Wiโˆˆ๐’ฏi for all iโˆˆI. Also note that W=โ‹‚jโˆˆFpj-1โข(Wj). By definition of convergence and (xi)iโˆˆI, we have Wiโˆˆโ„ฑi, so pj-1โข(Wj)โˆˆโ„ฑ. Since โ„ฑ is closed under finite intersections, we conclude Wโˆˆโ„ฑ, hence Uโˆˆโ„ฑ. So โ„ฑ converges to (xi)iโˆˆI as desired. โˆŽ