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 base 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 the take 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).]

## 61 comments

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22 January, 2023 at 6:44 am

AnonymousBelow Remark 3:

we can find

–>

we can find

At the end of Definition 12:

and with .

–>

and with .

[Corrected, thanks – T.]22 January, 2023 at 6:48 am

AnonymousRegarding Definition 12, can one still define the notion of “uniform equicontinuity” when is merely a topological space?

Does the following definition make sense?

We say that this family is “uniformly equicontinuous” (or “equi-uniformly-continuous”) if for every , there exists an open set such that for all and .

26 January, 2023 at 5:22 pm

Terence TaoThis is not an interesting notion (one could take to be the empty set, for instance). In order to have a good notion of uniform continuity (or equicontinuity), one needs the structure of a uniform space.

22 January, 2023 at 12:11 pm

AnonymousSome pedantic comments:

1. In the proof of “(iii) implies (i)” in Theorem 13,

is used both as a fixed variable and the dummy variable:

For each , the tuple can then be viewed as a point in the product space .

… is clearer

2. Possible typos in

hence we may find a subsequence such that is convergent in the product topology, or equivalently that pointwise

which seems to be:

hence we may find a subsequence such that is convergent in the product topology, or equivalently that pointwise

3.

The subscripts and are used in the proof. While it is clear that a sequence “” can be clearly understood as , the following in the proof:

… such that takes values in .

means

… such that , take values in .

not

… such that , take values in .

23 January, 2023 at 6:25 am

AnonymousIn the proof of “(iii) implies (i)” in Theorem 13:

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 .

1. I may miss something obvious. Why “any pointwise limit of the in will take values in “? One only knows from the quoted argument that the closure in is closed in , which means the limit should take values in ?

2. “hence the closure in is complete and thus closed in also” What is the use of completeness here? Is the closure in not automatically closed in ?

26 January, 2023 at 6:08 pm

Terence Tao1. Yes. If is a limit of the in , then it will lie in the Y-closure of the since this is already closed in .

2. No. For instance, is closed in , but is not closed in . However, any complete subset of will be closed in ; this is a good exercise for you to prove.