One of the most useful concepts for analysis that arise from topology and metric spaces is the concept of compactness; recall that a space is compact if every open cover of has a finite subcover, or equivalently if any collection of closed sets with the finite intersection property (i.e. every finite subcollection of these sets has non-empty intersection) has non-empty intersection. In these notes, we explore how compactness interacts with other key topological concepts: the Hausdorff property, bases and sub-bases, product spaces, and equicontinuity, in particular establishing the useful Tychonoff and Arzelá-Ascoli theorems that give criteria for compactness (or precompactness).

Exercise 1 (Basic properties of compact sets)

- Show that any finite set is compact.
- Show that any finite union of compact subsets of a topological space is still compact.
- Show that any image of a compact space under a continuous map is still compact.
Show that these three statements continue to hold if “compact” is replaced by “sequentially compact”.

** — 1. Compactness and the Hausdorff property — **

Recall from Notes 8 that a topological space is Hausdorff if every distinct pair of points can be separated by two disjoint open neighbourhoods of respectively; every metric space is Hausdorff, but not every topological space is.

At first glance, the Hausdorff property bears no resemblance to the compactness property. However, they are in some sense “dual” to each other, as the following two exercises show:

Exercise 2Let be a compact topological space.

- Show that every closed subset in is compact.
- Show that any weaker topology on also yields a compact topological space .
- Show that the trivial topology on is always compact.

Exercise 3Let be a Hausdorff topological space.

- Show that every compact subset of is closed.
- Show that any stronger topology on also yields a Hausdorff topological space .
- Show that the discrete topology on is always Hausdorff.

The first exercise asserts that compact topologies tend to be weak, while the second exercise asserts that Hausdorff topologies tend to be strong. The next lemma asserts that the two concepts only barely overlap:

Lemma 1Let be a weak and strong topology respectively on a space . If is compact and is Hausdorff, then . (In other words, a compact topology cannot be strictly stronger than a Hausdorff one, and a Hausdorff topology cannot be strictly weaker than a compact one.)

*Proof:* Since , every set which is closed in is closed in , and every set which is compact in is compact in . But from Exercises 2, 3, every set which is closed in is compact in , and every set which is compact in is closed in . Putting all this together, we see that and have exactly the same closed sets, and thus have exactly the same open sets; in other words, .

Corollary 2Any continuous bijection from a compact topological space to a Hausdorff topological space is a homeomorphism.

*Proof:* Consider the *pullback* of the topology on by ; this is a topology on . As is continuous, this topology is weaker than , and thus by Lemma 1 is equal to . As is a bijection, this implies that is continuous, and the claim follows.

One may wish to compare this corollary with Corollary 2 from Notes 9.

Remark 1Spaces which are both compact and Hausdorff (e.g. the unit interval with the usual topology) have many nice properties and are moderately common, so much so that the two properties are often concatenated as CH. Spaces that arelocallycompact and Hausdorff (e.g. manifolds) are much more common and have nearly as many nice properties, and so these two properties are often concatenated as LCH. One should caution that (somewhat confusingly) in some older literature (particularly those in the French tradition), “compact” is used for “compact Hausdorff”.

(Optional) Another way to contrast compactness and the Hausdorff property is via the machinery of ultrafilters. Define an *filter* on a space to be a collection of sets of which is closed under finite intersection, is also monotone (i.e. if and , then ), and does not contain the empty set. Define an *ultrafilter* to be a filter with the additional property that for any , exactly one of and lies in . (See also this blog post of mine for more discussion of ultrafilters.)

Exercise 4 (Ultrafilter lemma)Show that every filter is contained in at least one ultrafilter. (Hint: use Zorn’s lemma, see Notes 7.)

Exercise 5A collection of subsets of has the finite intersection property if every finite intersection of sets in the collection has non-empty intersection. Show that every filter has the finite intersection property, and that every collection of sets with the finite intersection property is contained in a filter (and hence contained in an ultrafilter, by the ultrafilter lemma).

Given a point and an ultrafilter on , we say that *converges* to if every neighbourhood of belongs to .

Exercise 6Show that a space is Hausdorff if and only if every ultrafilter has at most one limit. (Hint:For the “if” part, observe that if cannot be separated by disjoint neighbourhoods, then the neighbourhoods of and together enjoy the finite intersection property.)

Exercise 7Show that a space is compact if and only if every ultrafilter has at least one limit. (Hint:use the finite intersection property formulation of compactness and Exercise 5.)

** — 2. Compactness and bases — **

Compactness is the property that every open cover has a finite subcover. This property can be difficult to verify in practice, in part because the class of open sets is very large. However, in many cases one can replace the class of open sets with a much smaller class of sets. For instance, in metric spaces, a set is open if and only if it is the union of open balls (note that the union may be infinite or even uncountable). We can generalise this notion as follows:

Definition 3 (Base)Let be a topological space. A base for this space is a collection of open sets such that every open set in can be expressed as the union of sets in the base. The elements of are referred to asbasic open sets.

Example 1The collection of open balls in a metric space forms a base for the topology of that space. As another (rather trivial) example of a base: any topology is a base for itself.

This concept should be compared with that of a *basis* of a vector space: every vector in that space can be expressed as a linear combination of vectors in a basis. However, one difference between a base and a basis is that the representation of an open set as the union of basic open sets is almost certainly not unique.

Given a base , define a *basic open neighbourhood* of a point to be a basic open set that contains . Observe that a set is open if and only if every point in has a basic open neighbourhood contained in .

Exercise 8Let be a collection of subsets of a set . Show that is a basis for some topology if and only if it it covers and has the following additional property: given any and any two basic open neighbourhoods of , there exists another basic open neighbourhood of that is contained in . Furthermore, the topology is uniquely determined by .

To verify the compactness property, it suffices to do so for basic open covers (i.e. coverings of the whole space by basic open sets):

Exercise 9Let be a topological space with a base . Then the following are equivalent:

- Every open cover has a finite subcover (i.e. is compact);
- Every basic open cover has a finite subcover.

A useful fact about compact metric spaces is that they are in some sense “countably generated”.

Lemma 4Let be a compact metric space.

- (i) is separable (i.e. it has an at most countably infinite dense subset).
- (ii) is second-countable (i.e. it has an at most countably infinite base).

*Proof:* By Theorem 1 of Notes 8, is totally bounded. In particular, for every , one can cover by a finite number of balls of radius . The set of points is then easily verified to be dense and at most countable, giving (i). Similarly, the set of balls can be easily verified to be a base which is at most countable, giving (ii).

Remark 2One can easily generalise compactness here to -compactness; thus for instance finite-dimensional vector spaces are separable and second-countable. The properties of separability and second-countability are much weaker than -compactness in general, but can still serve to provide some constraint as to the “size” or “complexity” of a metric space or topological space in many situations.

We now weaken the notion of a base to that of a sub-base.

Definition 5 (Sub-base)Let be a topological space. Asub-basefor this space is a collection of subsets of such that is the weakest topology that makes open (i.e. is generated by ). Elements of are referred to assub-basic open sets.

Observe for instance that every base is a sub-base. The converse is not true: for instance, the half-open intervals for form a sub-base for the standard topology on , but not a base. In contrast to bases, which need to obey the property in Exercise 8, no property is required on a collection in order for it to be a sub-base; every collection of sets generates a unique topology with respect to which it is a sub-base.

The precise relationship between sub-bases and bases is given by the following exercise.

Exercise 10Let be a topological space, and let be a collection of subsets of . Then the following are equivalent:

- is a sub-base for .
- The space of finite intersections of (including the whole space , which corresponds to the case ) is a base for .

Thus a set is open iff it is the union of finite intersections of sub-basic open sets.

Many topological facts involving open sets can often be reduced to verifications on basic or sub-basic open sets, as the following exercise illustrates:

Exercise 11Let be a topological space, and be a sub-base of , and let be a base of .

- Show that a sequence converges to a limit if and only if every sub-basic open neighbourhood of contains for all sufficiently large . (Optional: show that an analogous statement is also true for nets.)
- Show that a point is adherent to a set if and only if every basic open neighbourhood of intersects . Give an example to show that the claim fails for sub-basic open sets.
- Show that a point is in the interior of a set if and only if contains a basic open neighbourhood of . Give an example to show that the claim fails for sub-basic open sets.
- If is another topological space, show that a map is continuous if and only if the inverse image of every sub-basic open set is open.

There is a useful strengthening of Exercise 9 in the spirit of the above exercise, namely the Alexander sub-base theorem:

Theorem 6 (Alexander sub-base theorem)Let be a topological space with a sub-base . Then the following are equivalent:

- Every open cover has a finite subcover (i.e. is compact);
- Every sub-basic open cover has a finite subcover.

*Proof:* Call an open cover *bad* if it had no finite subcover, and *good* otherwise. In view of Exercise 9, it suffices to show that if every sub-basic open cover is good, then every basic open cover is good also, where we use the basis coming from Exercise 10.

Suppose for contradiction that every sub-basic open cover was good, but at least one basic open cover was bad. If we order the bad basic open covers by set inclusion, observe that every chain of bad basic open covers has an upper bound that is also a bad basic open cover, namely the union of all the covers in the chain. Thus, by Zorn’s lemma, there exists a maximal bad basic open cover . Thus this cover has no finite subcover, but if one adds any new basic open set to this cover, then there must now be a finite subcover.

Pick a basic open set in this cover . Then we can write for some sub-basic open sets . We claim that at least one of the also lie in the cover . To see this, suppose for contradiction that none of the was in . Then adding any of the to enlarges the basic open cover and thus creates a finite subcover; thus together with finitely many sets from cover , or equivalently that one can cover with finitely many sets from . Thus one can also cover with finitely many sets from , and thus itself can be covered by finitely many sets from , a contradiction.

From the above discussion and the axiom of choice, we see that for each basic set in there exists a sub-basic set containing that also lies in . (Two different basic sets could lead to the same sub-basic set , but this will not concern us.) Since the cover , the do also. By hypothesis, a finite number of can cover , and so is good, which gives the desired a contradiction.

Exercise 12(Optional) Use Exercise 7 to give another proof of the Alexander sub-base theorem.

Exercise 13Use the Alexander sub-base theorem to show that the unit interval (with the usual topology) is compact, without recourse to the Heine-Borel or Bolzano-Weierstrass theorems.

Exercise 14Let be a well-ordered set, endowed with the order topology (Exercise 9 of Notes 8); such a space is known as an ordinal space. Show that is Hausdorff, and that is compact if and only if has a maximal element.

One of the major applications of the sub-base theorem is to prove Tychonoff’s theorem, which we turn to next.

** — 3. Compactness and product spaces — **

Given two topological spaces and , we can form the product space , using the cylinder sets as a sub-base, or equivalently using the open boxes as a base (cf. Example 15 from Notes 8). One easily verifies that the obvious projection maps , are continuous, and that these maps also provide homeomorphisms between and , or between and , for every . Also observe that a sequence (or net ) converges to a limit in if and only if and (or and ) converge in and to and respectively.

This operation preserves a number of useful topological properties, for instance

Exercise 15Prove that the product of two Hausdorff spaces is still Hausdorff.

Exercise 16Prove that the product of two sequentially compact spaces is still sequentially compact.

Proposition 7The product of two compact spaces is compact.

*Proof:* By Exercise 9 it suffices to show that any basic open cover of by boxes has a finite subcover. For any , this open cover covers ; by the compactness of , we can thus cover by a finite number of open boxes . Intersecting the together, we obtain a neighbourhood of such that is covered by a finite number of these boxes. But by compactness of , we can cover by a finite number of . Thus all of can be covered by a finite number of boxes in the cover, and the claim follows.

Exercise 17(Optional) Obtain an alternate proof of this proposition using Exercise 14 from Notes 8.

The above theory for products of two spaces extends without difficulty to products of finitely many spaces. Now we consider infinite products.

Definition 8 (Product spaces)Given a family of topological spaces, let be the Cartesian product, i.e. the space of tuples with for all . For each , we have the obvious projection map that maps to .

- We define the product topology on to be the topology generated by the cylinder sets for and as a sub-base, or equivalently the weakest topology that makes all of the continuous.
- We define the box topology on to be the topology generated by all the boxes , where for all .
Unless otherwise specified, we assume the product space to be endowed with the product topology rather than the box topology.

When is finite, the product topology and the box topology coincide. When is infinite, the two topologies are usually different (as we shall see), but the box topology is always at least as strong as the product topology. Actually, in practice the box topology is too strong to be of much use – there are not enough convergent sequences in it. For instance, in the space of real-valued sequences , even sequences such as do not converge to the zero sequence as (why?), despite converging in just about every other sense.

Exercise 18Show that the arbitrary product of Hausdorff spaces remains Hausdorff in either the product or the box topology.

Exercise 19Let be a sequence of metric spaces. Show that the the function on the product space defined byis a metric on which generates the product topology on .

Exercise 20Let be a product space with the product topology. Show that a sequence in that space converges to a limit if and only if converges in to for every . (The same statement also holds for nets.) Thus convergence in the product topology is essentially the same concept as pointwise convergence (cf. Example 14 of Notes 8).

The box topology usually does not preserve compactness. For instance, one easily checks that the product of any number of discrete spaces is still discrete in the box topology. On the other hand, a discrete space is compact (or sequentially compact) if and only if it is finite. Thus the infinite product of any number of non-trivial (i.e. having at least two elements) compact discrete spaces will be non-compact, and similarly for sequential compactness.

The situation improves significantly with the product topology, however (which is weaker, and thus more likely to be compact). We begin with the situation for sequential compactness.

Proposition 9 (Sequential Tychonoff theorem)Any at most countable product of sequentially compact topological spaces is sequentially compact.

*Proof:* We will use the “Arzelá-Ascoli diagonalisation argument”. The finite case is already handled by Exercise 16 (and can in any event be easily deduced from the countable case), so suppose we have a countably infinite sequence of sequentially compact spaces, and consider the product space with the product topology. Let be a sequence in , thus each is itself a sequence with for all . Our objective is to find a subsequence which converges to some limit in the product topology, which by Exercise 20 is the same as pointwise convergence (i.e. as for each ).

Consider the first coordinates of the sequence . As is sequentially compact, we can find a subsequence in such that converges in to some limit .

Now, in this subsequence, consider the second coordinates . As is sequentially compact, we can find a further subsequence in such that converges in to some limit . Also, we inherit from the preceding subsequence that converges in to .

We continue in this vein, creating nested subsequences for whose first components converge to respectively.

None of these subsequences, by themselves are sufficient to finish the problem. But now we use the diagonalisation trick: we consider the diagonal sequence . One easily verifies that converges in to as for every , and so we have extracted a sequence that is convergent in the product topology.

Remark 3In the converse direction, if a product of spaces is sequentially compact, then each of the factor spaces must also be sequentially compact, since they are continuous images of the product space and one can apply Exercise 1.

The sequential Tychonoff theorem breaks down for uncountable products. Consider for instance the product space of functions . As (with the discrete topology) is sequentially compact, this is an (uncountable) product of sequentially compact spaces. On the other hand, for each we can define the evaluation function by . This is a sequence in ; we claim that it has no convergent subsequence. Indeed, given any , we can find such that does not converge to a limit as , and so does not converge pointwise (i.e. does not converge in the product topology).

However, we can recover the result for uncountable products as long as we work with topological compactness rather than sequential compactness, leading to Tychonoff’s theorem:

Theorem 10 (Tychonoff theorem)Any product of compact topological spaces is compact.

*Proof:* Write for this product of compact topological spaces. By Theorem 6, it suffices to show that any open cover of by sub-basic open sets has a finite sub-cover, where is some index set, and for each , and is open in .

For each , consider the sub-basic open sets that are associated to those with . If the open sets here cover , then by compactness of , a finite number of the already suffice to cover , and so a finite number of the cover , and we are done. So we may assume that the do not cover , thus there exists that avoids all the with . One then sees that the point in avoids all of the , a contradiction. The claim follows.

Remark 4The axiom of choice was used in several places in the proof (in particular, via the Alexander sub-base theorem). This turns out to be necessary, because one can use Tychonoff’s theorem to establish the axiom of choice. This was first observed by Kelley, and can be sketched as follows. It suffices to show that the product of non-empty sets is again non-empty. We can make each compact (e.g. by using the trivial topology). We then adjoin an isolated element to each to obtain another compact space , with closed in . By Tychonoff’s theorem, the product is compact, and thus every collection of closed sets with finite intersection property has non-empty intersection. But observe that the sets in , where is the obvious projection, are closed and has the finite intersection property; thus the intersection of all of these sets is non-empty, and the claim follows.

Remark 5From the above discussion, we see that the space is compact but not sequentially compact; thus compactness does not necessarily imply sequential compactness.

Exercise 21Let us call a topological space first-countable if, for every , there exists a countable family of open neighbourhoods of such that every neighbourhood of contains at least one of the .

- Show that every metric space is first-countable.
- Show that every second-countable space is first-countable (see Lemma 4).
- Show that every separable metric space is second-countable.
- Show that every space which is second-countable, is separable.
- Show that any compact space which is first-countable, is also sequentially compact. (The converse is not true: Exercise 9 from Notes 8 provides a counterexample.)

(Optional) There is an alternate proof of the Tychonoff theorem that uses the machinery of *universal nets*. We sketch this approach in a series of exercises.

Definition 11A net in a set isuniversalif for every function , the net converges to either or .

Exercise 22Show that a universal net in a compact topological space is necessarily convergent. (Hint: show that the collection of closed sets which contain for sufficiently large enjoys the finite intersection property.)

Exercise 23 (Kelley’s theorem)Every net in a set has a universal subnet . (Hint: First use Exercise 5 to find an ultrafilter on that contains the upsets for all . Now let be the space of all pairs , where , ordered by requiring when and , and let be the map .)

Exercise 24Use the previous two exercises, together with Exercise 20, to establish an alternate proof of Tychonoff’s theorem.

Exercise 25Establish yet another proof of Tychonoff’s theorem using Exercise 7 directly (rather than proceeding via Exercise 12).

** — 4. Compactness and equicontinuity — **

We now pause to give an important application of the (sequential) Tychonoff theorem. We begin with some definitions. If is a topological space and is a metric space, let be the space of bounded continuous functions from to . (If is compact, this is the same space as , the space of continuous functions from to .) We can give this space the uniform metric

Exercise 26If is complete, show that is a complete metric space. (Note that this implies Exercise 2 from Notes 6.)

Note that if is continuous if and only if, for every and , there exists a neighbourhood of such that for all . We now generalise this concept to families.

Definition 12Let be a topological space, let be a metric space, and Let be a family of functions .

- We say that this family is
pointwise boundedif for every , the set is bounded in .- We say that this family is
pointwise precompactif for every , the set is precompact in .- We say that this family is equicontinuous if for every and , there exists a neighbourhood of such that for all and .
- If is also a metric space, we say that the family is
uniformly equicontinuousif for every there exists a such that for all and with .

Remark 6From the Heine-Borel theorem, the pointwise boundedness and pointwise precompactness properties are equivalent if is a subset of for some . Any finite collection of continuous functions is automatically an equicontinuous family (why?), and any finite collection of uniformly continuous functions is automatically a uniformly equicontinuous family; the concept only acquires additional meaning once one considers infinite families of continuous functions.

Example 2With and , the family of functions for are pointwise bounded (and thus pointwise precompact), but not equicontinuous. The family of functions for , on the other hand, are equicontinuous, but not pointwise bounded or pointwise precompact. The family of functions for are pointwise bounded (even uniformly bounded), but not equicontinuous.

Example 3With and , the functions are pointwise bounded (even uniformly bounded), are equicontinuous, and are eachindividuallyuniformly continuous, but are not uniformly equicontinuous.

Exercise 27Show that the uniform boundedness principle (Theorem 2 from Notes 9) can be restated as the assertion that any family of bounded linear operators from the unit ball of a Banach space to a normed vector space is pointwise bounded if and only if it is equicontinuous.

Example 4A function between two metric spaces is said to be Lipschitz (orLipschitz continuous) if there exists a constant such that for all ; the smallest constant one can take here is known as theLipschitz constantof . Observe that Lipschitz functions are automatically continuous, hence the name. Also observe that a family of Lipschitz functions with uniformly bounded Lipschitz constant is equicontinuous.

One nice consequence of equicontinuity is that it equates uniform convergence with pointwise convergence, or even pointwise convergence on a dense subset.

Exercise 28Let be a topological space, let be a complete metric space, let be an equicontinuous family of functions. Show that the following are equivalent:

- The sequence is pointwise convergent.
- The sequence is pointwise convergent on some dense subset of .
If is compact, show that the above two statements are also equivalent to

- The sequence is uniformly convergent.
(Compare with Corollary 1 from Notes 9.) Show that no two of the three statements remain equivalent if the hypothesis of equicontinuity is dropped.

We can now use Proposition 9 to give a useful characterisation of precompactness in when is compact, known as the Arzelá-Ascoli theorem:

Theorem 13 (Arzelá-Ascoli theorem)Let be a metric space, be a compact metric space, and let be a family of functions . Then the following are equivalent:

- (i) is a precompact subset of .
- (ii) is pointwise precompact and equicontinuous.
- (iii) is pointwise precompact and uniformly equicontinuous.

*Proof:* We first show that (i) implies (ii). For any , the evaluation map is a continuous map from to , and thus maps precompact sets to precompact sets. As a consequence, any precompact family in is pointwise precompact. To show equicontinuity, suppose for contradiction that equicontinuity failed at some point , thus there exists , a sequence , and points such that for every . One then verifies that no subsequence of can converge uniformly to a continuous limit, contradicting precompactness. (Note that in the metric space , precompactness is equivalent to sequential precompactness.)

Now we show that (ii) implies (iii). It suffices to show that equicontinuity implies uniform equicontinuity. This is a straightforward generalisation of the more familiar argument that continuity implies uniform continuity on a compact domain, and we repeat it here. Namely, fix . For every , equicontinuity provides a such that whenever and . The balls cover , thus by compactness some finite subcollection , of these balls cover . One then easily verifies that whenever with .

Finally, we show that (iii) implies (i). It suffices to show that any sequence , , which is pointwise precompact and uniformly equicontinuous, has a convergent subsequence. By embedding in its metric completion , we may assume without loss of generality that is complete. (Note that for every , the set is precompact in , hence the closure in is complete and thus closed in also. Thus any pointwise limit of the in will take values in .) By Lemma 4, we can find a countable dense subset of . For each , we can use pointwise precompactness to find a compact set such that takes values in . For each , the tuple can then be viewed as a point in the product space . By Proposition 9, this product space is sequentially compact, hence we may find a subsequence such that is convergent in the product topology, or equivalently that pointwise converges on the countable dense set . The claim now follows from Exercise 28.

Remark 7The above theorem characterises precompact subsets of when is a compact metric space. One can also characterise compact subsets by observing that a subset of a metric space is compact if and only if it is both precompact and closed.

There are many variants of the Arzelá-Ascoli theorem with stronger or weaker hypotheses or conclusions; for instance, we have

Corollary 14 (Arzelá-Ascoli theorem, special case)Let be a sequence of functions from a compact metric space to a finite-dimensional vector space which are equicontinuous and pointwise bounded. Then there is a subsequence of which converges uniformly to a limit (which is necessarily bounded and continuous).

Thus, for instance, any sequence of uniformly bounded and uniformly Lipschitz functions will have a uniformly convergent subsequence. This claim fails without the uniform Lipschitz assumption (consider, for instance, the functions ). Thus one needs a “little bit extra” uniform regularity in addition to uniform boundedness in order to force the existence of uniformly convergent subsequences. This is a general phenomenon in infinite-dimensional function spaces: compactness in a strong topology tends to require some sort of uniform control on regularity or decay in addition to uniform bounds on the norm.

Exercise 29Show that the equivalence of (i) and (ii) continues to hold if is assumed to be just a compact Hausdorff space rather than a compact metric space (the statement (iii) no longer makes sense in this setting).Hint:need not be separable any more, however one can still adapt the diagonalisation argument used to prove Proposition 9. The starting point is the observation that for every and every , one can find a neighbourhood of and some subsequence which only oscillates by at most (or maybe ) on .

Exercise 30 (Locally compact Hausdorff version of Arzelá-Ascoli)Let be a locally compact Hausdorff space which is also -compact, and let be an equicontinuous, pointwise bounded sequence of functions. Then there exists a subsequence which converges uniformly on compact subsets of to a limit . (Hint:Express as a countable union of compact sets , each one contained in the interior of the next. Apply the compact Hausdorff Arzelá-Ascoli theorem on each compact set (Exercise 29). Then apply the Arzelá-Ascoli argument one last time.)

Remark 8The Arzelá-Ascoli theorem (and other compactness theorems of this type) are often used in partial differential equations, to demonstrate existence of solutions to various equations or variational problems. For instance, one may wish to solve some equation , for some function . One way to do this is to first construct a sequence of approximate solutions, so that as in some suitable sense. If one can also arrange these to be equicontinuous and pointwise bounded, then the Arzelá-Ascoli theorem allows one to pass to a subsequence that converges to a limit . Given enough continuity (or semi-continuity) properties on , one can then show that as required.More generally, the use of compactness theorems to demonstrate existence of solutions in PDE is known as the

compactness method. It is applicable in a remarkably broad range of PDE problems, but often has the drawback that it is difficult to establish uniqueness of the solutions created by this method (compactness guarantees existence of a limit point, but not uniqueness). Also, in many cases one can only hope for compactness in rather weak topologies, and as a consequence it is often difficult to establish regularity of the solutions obtained via compactness methods.

[*Update*, Feb 20: some corrections, new exercise added (note renumbering).]

## 53 comments

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9 February, 2009 at 11:15 am

David SpeyerIn the proof of Remark 4, you write “The axiom of choice was used in the proof (via the Alexander sub-base theorem).” Wasn’t it also used directly, to select the element in ? (Note that this selection takes place infinitely many times, once for each .)

9 February, 2009 at 11:28 am

Terence TaoTrue enough… though I find this invocation of the axiom of choice to be “shallower” than that in the sub-base theorem (because I consider Zorn’s lemma to be “deeper” somehow than the axiom of choice, despite being logically equivalent); it doesn’t seem to lead as easily to such “unconstructible” objects as non-principal ultrafilters, etc., as the Zorn lemma applications do. I don’t know how to make this distinction rigorous, though.

9 February, 2009 at 11:39 am

K. P. HartExample 2: the functions fn are not pointwise bounded; did you mean xn?

9 February, 2009 at 12:35 pm

Terence TaoOops, thanks for the correction!

9 February, 2009 at 4:51 pm

whuhDear Prof. Tao (and Prof. Trevisan),

An aesthetic comment – congratulations, this looks fantastic! The serif fonts in the boxes are far superior and sharper to the standard fare outside. Would it be possible to use serif everywhere? Presumably this is now far easier to change with your script.

Does the script support equation numbering? And would it be possible to slightly increase the contrast or the weight of the mathematics font?

There are consistency issues with the italics in the boxes – it sometimes seems to stop for no reason – e.g. in Example 4 and Remark 8. By the way there is a unclosed end{proof} tag after Theorem 10.

It is also a little disconcerting to have clearer structural elements for theorems, remarks and examples than for sections. Presumably this, too can be incorporated?

But these are all nitpicks, it looks amazing. Will it be possible for other WordPress bloggers to use it?

9 February, 2009 at 5:07 pm

Terence TaoThanks for the feedback (and for the corrections)! Luca can of course say more, but I believe he plans to release his code to the public once some of the kinks (such as those you pointed out) are worked out.

9 February, 2009 at 8:52 pm

Nate ChandlerIn the introduction, you wrote “In these notes, we explore how compactness interacts with other key topological concepts: the Hausdorff property, bases and sub-bases, product spaces, and equicontinuity, in particular establishing the useful and the Arzelá-Ascoli theorem that give criteria for compactness (or precompactness).” I think that you left out the noun which “useful” is modifying and that it should read like “the useful [theorem X] and the Arzelá-Ascoli theorem”.

[Corrected, thanks – T.]9 February, 2009 at 10:09 pm

lucaI think you used \label{bascov} twice, once in

Exercise 9 and once in Theorem 6.

(The effect is that \ref{bascov} points into Exercise 9, but it

reads as a reference to Exercise 6)

10 February, 2009 at 5:58 am

EricRe the first two comments: The use of AC that David points out, while seemingly less deep, is in fact stronger than the other uses of AC. Indeed, by Exercise 11, the sub-base lemma, which is the only other place AC is used in an essential way, requires only the ultrafilter lemma, while Tychonoff’s theorem is equivalent to the full power of AC. To get a more precise statement, consider the ultrafilter proof of Tychonoff’s theorem: given an ultrafilter on a product of compact spaces, it has a limit on each coordinate, and these give the coordinates of a limit . Here again, AC is used at two points: the ultrafilter lemma is used to be able to prove compactness using only ultrafilters, and AC is used to choose each . However, with this proof, we can say more: if each is Hausdorff, then the are all unique, so the second use of AC is unnecessary. Thus Tychonoff’s theorem for Hausdorff spaces requires only the ultrafilter lemma (and is in fact equivalent to it), whereas the full Tychonoff’s theorem is strictly stronger. This shows that the use of the strange non-Hausdorff spaces in Remark 4 was unavoidable.

10 February, 2009 at 10:15 am

RRProfessor Tao:

I enjoy a lot your postings. They keep my love for mathematics alive. Anyway, my browser (Mozilla Firefox) has problems with the following two lines of code:

my box (running scientific linux) only shows: “Formula does not parse”

Thanks ,

RR

10 February, 2009 at 12:56 pm

EricI also get a few ‘formula does not parse’ errors in:

1. Exercise 18 after between “Let” and “be a sequence”

2. Paragraph before Remark 3 after “diagonal sequence.”

3. Example 3 after “the functions”

4. Two paragraphs before Remark 7, after “finite subcollection B(x_i,\delta_i /2),”

10 February, 2009 at 8:26 pm

Wikipedia-like blog entries? « Shell Beach[…] of academic bloggers posting their lecture notes on specific topics online (some even making the lecture notes themselves posts rather than linking to PDFs), so parts of me feel like we’re already seeing this kind of […]

11 February, 2009 at 4:50 pm

Terence TaoI’ve recompiled the HTML. Let me know if there are still display problems.

11 February, 2009 at 6:12 pm

EricLooks like the FDNP persists in Exercise 18 between “Let” and “be a sequence.”

13 February, 2009 at 7:37 pm

DimaThere is something funny with the RSS feed for this post: most, albeit not all, of the formulae in this post look broken (“formula does not parse” yellow boxes all over) when looked in an RSS feed reader (e.g. reader.google.com).

Needless to say, the formulae are fine on the blog page.

This probably means that the converter does not work well with the RSS feed machinery. (I see no problem with other maths posts from this blog from within the RSS feed reader)

14 February, 2009 at 12:54 am

Emmanuel KowalskiThere is a standard terminological warning about compactness for readers who also read (or will read) French literature: the French tradition is that “compact” means “compact and Hausdorff”.

14 February, 2009 at 12:05 pm

Terence TaoThanks for the warning, Emmanuel!

Dima: I see the RSS quirk too, but am hoping that it will sort itself out with time. (The LaTeX code was recently changed, thanks to another post here on this blog; that may have something to do with it.)

16 February, 2009 at 7:28 am

实分析0-10 « Liu Xiaochuan’s Weblog[…] 第十节是关于拓扑空间中的紧致性质的探讨，无疑是非常重要的。分成了三个部分，而三个部分中常常会出现的关于ultrafilter的证明，我是最喜欢的。第一个部分是紧致性和Hausdorff性的关系，显然，以前我是没有注意到的。在某种意义上，他们有着对立的关系。最后的Alexander子基定理十分有用。第二部分讲述了乘积空间的紧致性的定理，即Tychonoff定理。不可避免的，选择公理由出现了。最后还在习题中介绍了另外一种证明方法，用到ultrafilter的方法。另外，box拓扑的概念是这么自然，怎么以前没自己研究研究呢。第三个部分是关于Arzelá-Ascoli定理，这定理重要的无以复加了。 […]

18 February, 2009 at 7:27 pm

liuxiaochuanDear Professor:

In Exercise 27, is the space X asked to be compact or not?

In the first two of Definition 12, the set seems to be , instead of

19 February, 2009 at 2:23 pm

AnonymousIn exercise 5, you say “A collection of subsets of {X} has the finite intersection property if every finite intersection of sets in the collection.”

Do you mean “A collection of subsets of {X} has the finite intersection property if every finite intersection of sets in the collection is nonempty.”?

19 February, 2009 at 10:17 pm

Terence TaoThanks for the correction!

21 February, 2009 at 1:59 pm

Converting LaTeX to WordPress « in theory[…] Tao has tested it on a couple of posts. Thanks to his feedback, the current version, while surely bug-filled and very limited, is […]

21 February, 2009 at 9:33 pm

245B, Notes 11: The strong and weak topologies « What’s new[…] has a subsequence which converges pointwise on , and thus converges pointwise on by Exercise 27 of Notes 10, and thus converges in the weak* topology. But as is closed in this topology, we conclude that is […]

24 February, 2009 at 1:24 pm

KeExercise 28 (about equicontinuity equating uniform convergence and pointwise convergence on a dense subset) requires completeness of Y. Example : X = [-1,1], Y = (-1,1), . The sequence is equicontinuous and pointwise convergent for but not convergent for .

24 February, 2009 at 3:26 pm

Terence TaoHmm, you’re right, thanks. Fortunately, for the purposes of Arzela-Ascoli I can pass from Y to its metric completion (using the pointwise precompactness to ensure the limit ends up back in Y rather than the completion).

27 February, 2009 at 12:26 pm

AnonymousSorry if this is trivial but I don’t understand exercise 11, part 3. The claim to be shown to fail is, as I understand it, “a point {x \in X} is in the interior of a set {U} if {U} contains a sub-basic open neighbourhood of {x}.”

A sub-basic open neighbourhood of x contained in U is an open set containing x and contained in U, so x is in the interior of U.

27 February, 2009 at 12:53 pm

Terence TaoOops; the “if” there should be “if and only if” (and similarly for part 2). I’ve amended the post accordingly.

27 February, 2009 at 8:47 pm

LaTeX to WordPress « Tranvinhduc’s Weblog[…] far LaTeX2WP has been used at what’s new and here, at in theory. You’ll notice that the typesetting styles for theorems and section names […]

2 March, 2009 at 8:17 am

245B, Notes 12: Continuous functions on locally compact Hausdorff spaces « What’s new[…] a countable union of closed sets. But as is a compact metric space, it is separable (Lemma 4 from Notes 10), and so has a countable dense subset . One then easily verifies that every point in the open set […]

11 March, 2009 at 12:23 am

SamirHi Professor Tao,

I think there may be a little typo in Exercise 19 above. In the sum on the right, it appears that the metrics should have subscripts and be written . If I’m correct, thanks for the correction!

[Corrected, thanks – T.]Samir

18 March, 2009 at 10:32 pm

245B, Notes 13: Compactification and metrisation (optional) « What’s new[…] that compact metric spaces are second countable (Lemma 4 of Notes 10), thus we have Corollary 4 A compact Hausdorff space is metrisable if and only if it is second […]

13 April, 2009 at 7:31 am

PDEbeginnerDear Prof. Tao,

There seems some typo in the third paragraph in the proof of Theorem 6. I guess ‘For’ there should be ‘For contradiction’.

I can understand the proof of Theorem 6, but can’t understand the following problem: In the usual topology of , every sub-basic set cover for has finite sub-cover (our sub-base is ), but there exist some basic set cover for does not has finite sub-cover.

I began to seriously learn topology from this blog, so my question is most probably trivial. Thanks in advance!

13 April, 2009 at 5:08 pm

Terence TaoDear PDEbeginner,

Thanks for the correction!

The collection of sub-basic sets covers , but has no finite subcover.

14 June, 2009 at 1:55 pm

Sequential compactness theorem @ unwanted capture[…] compact because it is a countable product of the sequentially compact . (See Proposition 9 here for instance.) So the sequence of theories in has a subsequence converging to some theory . […]

20 June, 2009 at 5:23 pm

AlanDear Prof Tao,

is the space of continuous functions from to

under uniform convergence topology Haussdorf?

thanks

24 May, 2020 at 6:16 am

Anonymousthat topology comes from a norm $||f||=\sup\{|f(x)|\,:\,x\in[0,1]\}.$

12 November, 2009 at 1:52 pm

AnonymousYes.

f(x) = a

g(x) = b

a ≠ b

r = |a-b|/2

|a-b| ≤ d(f,g)

The open balls of radius r centered on f and g are disjoint.

30 January, 2010 at 7:08 pm

Ultralimit analysis, and quantitative algebraic geometry « What’s new[…] So we see that compactness methods and ultrafilter methods are closely intertwined. (See also my earlier class notes for a related connection between ultrafilters and […]

11 September, 2010 at 7:13 am

Anonymoushint about exercise 14

23 September, 2010 at 2:33 pm

FatenDear Prof Tao,

Do not you have some papers or theorems about compactness in bi_topological spaces.

30 October, 2010 at 6:54 pm

245A, Notes 6: Outer measures, pre-measures, and product measures « What’s new[…] holds in this more general setting, but requires Tychonoff’s theorem, which we will cover in 245B Notes 10. However, some minimal regularity hypotheses of a topological nature are needed to make the […]

22 November, 2010 at 5:46 pm

Boolean rings, ultrafilters, and Stone’s representation theorem « Annoying Precision[…] on is stronger than or the same as the topology on . But since they are both compact Hausdorff, by Lemma 1 in the link the topologies must […]

1 February, 2012 at 8:14 pm

RexFor Example 3: I think that the family f_n(x) = arctan(nx) is not equicontinuous at the origin. The slope of the graph there keeps getting steeper and steeper.

[Corrected, thanks -T.]21 January, 2016 at 11:08 pm

nikitaPlease tell me the proof of following axiom that

every sequentially compact topological space is countably compact

as soon as possible i need the proof. Please send it to my email id

thanku

26 February, 2016 at 6:17 pm

AnonymousDear Prof. Tao

In Exercise 29, do we need Hausdorff condition? I think that the both implications do not use any separation axiom. Or is it for convenience( in the next exercise)?

Best

27 February, 2016 at 7:05 am

Terence TaoIt does indeed appear that the Hausdorff axiom is unnecessary here, though I don’t know of any application where one would wish to apply the Arzela-Ascoli theorem on a non-Hausdorff domain. (Generally, enough other things go wrong on non-Hausdorff spaces that it is a bit risky to invoke standard theorems in analysis on such spaces without carefully checking the hypotheses.)

20 November, 2016 at 9:52 am

ianmarqzReblogged this on ianmarqz and commented:

compact topology a la Terrence Tao

3 September, 2017 at 10:16 am

AnonymousA little bit of categorical language can shorten the exposition

in a couple of places. This is based on the observation that

the definition of compactness via ultrafilters can be expressed

using the notion of Quillen lifting property (orthogonality of morphisms):

a map is proper iff it is left-orthogonal (has right lifting property)

in the category of topological spaces to each map associated

with an ultrafilter in a certain natural way.

This gives a very short diagram chasing proof of Tychonoff theorem.

14 December, 2018 at 2:56 am

BernhardI have a funny question. Do you know any exercise where the equivalence of compactness and sequential compactness in metric spaces is needed? (It would be weird to consider this equivalence an important result (I do), when you could not point a single example where the equivalence is really needed)

14 December, 2018 at 2:59 am

BernhardOk, Alaoglu’s theorem in the non-separable case is one example. But are there ‘easy ones’ ?

15 December, 2018 at 11:22 am

Terence TaoWell, very few results in mathematics are truly “needed”, since one can usually always rearrange a proof to avoid invoking any specific result. But having several equivalent definitions of a basic concept such as compactness gives one a lot of flexibility when trying to prove facts about that concept. For instance, if one wants to prove that the product of compact sets is compact, or the continuous image of a compact set is compact, etc., one can choose (at least in the setting of metric spaces) to use either the topological definition or the sequential definition of compactness. In some cases one may be conceptually easier to use one notion, and in other cases it is easier to use the other, so it helps to have both available.

There are also variants of the compactness concept in which one formulation is easier to work with than the other; for instance, the theory of concentration compactness is generally described using the sequential formalism, whereas compactness theorems in logic typically use a formalism closer to the topological formalism, particularly the version involving the finite intersection property.

4 November, 2021 at 8:30 am

Anil Pedgaonkarbeautiful exposition

18 March, 2022 at 7:45 pm

ho boon suanexercise 21(v) doesn’t seem right; i learned from Eric Wofsey (https://math.stackexchange.com/a/4407568/) that a constant net , where is the first uncountable ordinal, has no subsequences, since every function has a strict upper bound .

[You are right, this part is in error and is now deleted -T.]