In this post we will discuss some points that may be put into one of the other categories, and it is just happenstance that I “forgot” to put it there. Topics that are indirectly related to compactness will also be included here. I also feel as though I should mention that a lot of things aren’t going to be discussed. It just isn’t feasible. Ascoli’s theorem for example, local compactness (at least not yet…maybe when we deal with the one-point compactification).
We begin with a few definitions.
Perfect set: We call a subset of a topological space perfect if where is the derived set or the set of all limit points of .
Condensation point: If is a topological space is a subset of then is a condensation point of if for every neighborhood of there are uncountably many points of .
Theorem: Let be a second countable topological space and latex an uncountable subset of . a) Then the set of all condensation points of , denoted , is perfect and b) all but countably many points of lie in
Proof: a) First suppose that but . Then, there exists some neighborhood latex of such that . In particular, for each there exists some such that is countable. We therefore see that is an open cover of and since is second countable we see that Lindelof’s theorem we are guaranteed a countable subcover . Thus, . But, by assumption we have that is countable and thus is the countable union of countable sets thus countable. This contradicts that is a condensation point of , and so
Conversely, let then then for every neighborhood of we have that there exists some such that . But, since is open there exists some neighborhood of such that , but by assumption every neighborhood of contains uncountably many points of and thus so does , but since it follows that contains uncountably many points of . Since was arbitrary we see that . The conclusion follows.
b) Let the guaranteed countable base of . Define and let . We claim that .
Let then for there exists some of such that is countable, but since is an open base for there exists some such that and clearly then since we see that
Conversely, let then there exists some such that and is countable. Thus, is a neighborhood of which has only countably many points of and so
It follows that . Thus, and since the right hand side is the countable union of countable sets the conclusion follows.
From this we can prove a nice little theorem.
Theorem: Let be a separable metric space, then every closed subset of may be written as where is perfect and is countable.
Proof: Since is closed we have that but clearly and so . Thus, we have that . But from the above we see that is perfect and countable. The conclusion follows. .
Corollary: Every compact subspace of a separable space mEay be written as the union of a perfect set and a closed set.
A theorem which has a cute little solution is the following.
Theorem: Let be a compact metric space and closed subsets of . Then for some and .
Proof: Notice that and since are closed subsets of a compact space we know that they are compact. And, by Tychonoff’s theorem we know that is compact and so given by is trivially continuous and so is continuous and by previous comments assumes its minimum on . The conclusion follows.
We can weaken the prerequisites of the sets involved slightly when dealing with
Theorem: Let be closed subsets of such that is compact. Then for some and some .
Proof: The idea here is that we could revert back to the previous theorem if was also compact. But, since every closed and bounded subset of is compact and since is already closed we must merely find someone to bound it. But, this is pretty obvious in the sense that since is bounded that a lot of contributes nothing to the distance between the two sets. More formally, let . Since is bounded we know that for some . So, consider the set . Clearly this is a closed and bounded subset of . We pause for a lemma:
Lemma: Let be a topological space, and a class of closed subspaces of such that at least one of them, call it , is compact. Then, is a compact subspace of .
Proof: Clearly we have that and since they are all closed we see they are a closed subspace of a compact space. Thus, compact. Since is the same as a subspace of as it is as subspace of the conclusion follows. .
With this in mind we see that is a compact subspace of . Now, the conclusion follows from the last theorem if we can prove that but this is a relatively simple task.
Before we get to the main part of the post we prove one nice little theorem:
Theorem: A metric space is compact if and only if it’s complete and totally bounded.
: This is relatively simple. We know that since every compact metric space is sequentially compact and every sequentially compact metric space is totally bounded we already have the second part. But, for didactic purposes we can prove the second part without the use of sequential compactness (in fact it’s easier).
Lemma: Every compact metric space is totally bounded.
Proof: Let be arbitrary and consider . This is clearly an open cover for and by assumption there exists a finite subcover . Clearly then will serve as a perfectly fine -net. The conclusion follows.
So it remains to show that is complete. But, we know that given any in which is Cauchy that it has some convergent subsequence (since every compact metric space is sequentially compact). So, consider the following lemma
Lemma: Let be a Cauchy sequence in a metric space then is convergent if and only if it has a convergent subsequence.
Proof: Clearly if is convergent it will serve as it’s own convergent subsequence.
Conversely, let be the guaranteed convergent subsequence of with limit . Since there exists some such that and since is Cauchy it is relatively easy to see there exists some such that . Taking clearly gives that . The conclusion follows.
With this it’s clear that since any Cauchy sequence has a convergent subsequence (by sequential compactness) that every Cauchy sequence is convergent.
: Conversely, by previous argument we must only show that is sequentially compact. But, under the assumption of completeness it suffices to show that every sequence has a Cauchy subsequence. So let be sequence in it is clear that we could relabel this sequence as . Since is totally bounded, there exists a finite class of open balls with radius which cover . From this we see that has a subsequence all of which lie in some open ball of radius one-half. Applying the total boundedness of shows there exists a subsequence of which lies entirely in an open ball of radius . Continuing in this way we see that we may define which is Cauchy. The conclusion follows
Now, I’ve looked through a bunch of books and found what I believe to be the most useful/interesting results of compactness.
The following is actually a GREATLY generalized result that has a compactness related corollary.
Quasi-totally bounded: Call a metric space quasi-totally bounded if for every there exists some countable set such that is an open cover for .
Theorem: Every quasi-totally bounded metric space is separable.
Proof: Let be the guaranteed countable discusssed above with and let . Clearly we have is countable since it is the countable union of countable sets. We must merely prove that is dense in . So, let be arbitrary and be given. We must find some such that . But, by the Archimedean principle we have that there exists some such that . So considering we see that there exists some such that and so and so . The conclusion follows.
Corollary: Every compact metric space being totally bounded and thus quasi-totally bounded is separable.
Theorem: Let be a metric space and Let and be disjoint subsets such that is compact and closed, then .
Proof: Define by . We see that and similarly and thus and thus is Lipschitz and thus continuous. Since is compact we know then that for some . But, since and disjoint we have that . The conclusion follows.
Corollary: Let be disjoint compact subsets of then .
Corollary: Let be compact and latex then
We can now prove the following
Theorem: Let be compact metric space and let be continuous. Suppose that is expanding in the sense that . Then a) is injective and is continuous and b)
a) Injectivity follows immediately since if there existed such that but then .
Also, we see that is automatically uniformly continuous since it is Lipschitz with Lipschitz constant one since given any we have that
b) Suppose there exists some such that . Since is compact and continuous we know that is compact and thus by our previous theorem we know that .
So, consider the sequence where . Clearly this is a sequence in and by ‘s compactness we know that there exists some subsequence which converges, and is thus Cauchy. Consequently, there exists some such that . But
which is clearly a contradiction.
Corollary: Every isometry where is compact is a homeomorphism.
The next theorem is surprisingly powerful.
Theorem: Let be a metric space and a convergent sequence in with . Then, is compact.
Proof: Let be an open cover for , then there exists some such that and since is open there exists some . But, since we have that all but finitely many points of are in and consequently . Thus, taking and an open set containing each of the finitely many points not in procures the necessary finite subcover of . The conclusion follow.
Corollary: Let and be as above, then is closed.
Before we get to the next theorem we point something out that is true in any topological space.
Theorem: Let be a topological space and let and be a compact and a closed subspace of respectively. Then, is a compact subspace.
Proof: This follows since is closed subspace of .
Theorem: Let and be metric spaces and be continuous. Also, let be a decreasing sequence of non-empty compact subsets of . Then .
Prove: It is well-known that . So it remains to prove the reverse inclusion.
Firstly, it is easily proved that is a decreasing sequence of compact sets with the F.I.P, and thus we know that . Thus, let since every metric space is Hausdorff (thus ) we know that is closed and since is continuous we know that is closed as well. But, by the above theorem we see then that is compact for every . Also, and thus is decreasing sequence of non-empty compact sets. Thus, we have that and thus . The conclusion follows.
Theorem: Let be a compact metric space and . Prove that if for every which is continuous that attains a maximum then is compact.
Proof: Since is compact it suffices to show that is closed. If this is trivial, so assume otherwise and let . Define by . It is trivial to prove that this is a continuous mapping and so attains a maximum on , let’s say at . Thus, . But, since and so and the conclusion follows.
Up until this point we’ve talked about compactness mainly in the context of general topological spaces, with only brief allusions to metric spaces. Now, while topologically spaces may be a more fruitful study than metric spaces it is admittedly easier to apply facts about metric spaces. That said, we devote a few sections specifically to the study of compactness in metric spaces. Since metric spaces are in general much nicer than general topological spaces it isn’t surprising that there are other, more applicable, forms of compactness. So, we need to start with some definitions:
Sequential Compactness: A metric space is said to be sequentially compact if every sequence has a convergent subsequence.
Bolzano-Weierstrass Property (B.W.P): A metric space is said to have the Bolzano-Weierstrass Property (B.W.P) if every infinite subset has a limit point.
We begin by showing that these are equivalent statements.
Theorem: A metric space is sequentially compact if and only if it has the B.W.P.
: Suppose that is sequentially and let is an infinite subset of . It is clear that we can extract an infinite sequence of distinct points from and by assumption there exists some subsequence which is convergent. The conclusion follows by considering the following lemma
Lemma: Let where has an infinite number of points, then is a limit point of .
Proof: Let be arbitrary. Since we know that contains all but finitely many points of and in particular it contains more than two distinct points of . Since both of these can’t be equal to it follows that there exist some point of in distinct from . The conclusion follows
: Conversely, let be a sequence in . If has a value infinitely repeated then that constitutes a convergent subsequence. Otherwise, the set is infinite and by assumption has a limit point. It is easy to find a convergent subsequence from this. (consider taking to be any point in where is the limit point afforded by the B.W.P).
The conclusion follows.
Our ultimate goal is to prove that a metric space being sequentially compact, having the B.W.P. , and being compact are all equivalent. We have laid the ground-work for this above. The rest will actually uncover some equally fruitful concepts.
Theorem: A compact metric space has the B.W.P.
Proof: Suppose not. Then there exists some infinite subset which has no limit point in . Thus, for every there exists some such that . Clearly then is an open cover for and by assumption there exists a finite subcover. But, since is contained in the set of centers of these open balls it follows that is finite. This is clearly a contradiction, and the conclusion follows.
To complete our goal we must now prove that sequential compactness implies compactness. This, fortunately, is not as simple as the previous theorem. It is fortunate because the sequence of theorems needed to ultimately prove this fact ends up being equally, if not more so, useful than the theorem itself. We begin by discussing the concept of a Lebesgue number.
Lebesgue number: Let be a metric space and let be an open cover for . Then, we call a Lebesgue number for in if whenever there exists some such that .
In other words, an open cover has a Lebesgue number if whenever the diameter of a set gets small enough it is contained in at least one of the sets in the open cover.
We now prove that every sequentially compact set has a Lebesgue number.
Theorem: Let be a sequentially compact metric space. Then has a Lebesgue number for any open cover .
Proof: Call a subset of big if it is not contained in any and define . If there are no big sets, then and thus any real number will serve as a Lebesgue number. So, assume that there are big sets. Clearly we must have that . Once again, if any real number will do and if then will work. So, it is only problematic if . So, for a contradiction assume that it does.
Clearly we must have a sequence such that . So let be an arbitrary but fixed sequence such that . By assumption we have that has a convergent subsequence which converges to a point . Since covers there exists some such that and since is open we have that there exists some such that . Consider , since we have that contains all but finitely many points of , in particular there exists some such that . Since and it can be easily seen that which contradicts that is big. The conclusion follows.
To discuss the next topic we need to define an -net
-net: Let be a metric space. We see that a finite point set is an -net for if is an open cover for .
Totally bounded: A metric space is totally bounded if it has an -net for ever
Theorem: Every sequentially compact metric space is totally bounded.
Proof: Let and be arbitrary. If covers then is an -net and we’re done, so assume not. Let . If cover we are done, so assume not. Continuing on this way we get a sequence . If is infinite then it clearly has no convergent subsequence which contradicts the sequentially compactness of . It follows that is finite and thus an -net for . The conclusion follows.
Using this we can prove what we’re really after.
Theorem: If is sequentially compact then it is compact.
Proof: Let be an arbitrary open cover of . Since is sequentially compact it has a Lebesgue number . Let . Once again, by virtue of ‘s sequential compactness, we may find an -net of . Since we know that for some . Doing this for produces a finite subclass of such that . The conclusion follows.
We can summarize this by saying that
From this we can prove one of the most useful theorems in real analysis.
Theorem: Let be metric spaces and be continuous, then is uniformly continuous.
Proof: Let be arbitrary and consider , clearly then is an open cover for . And since is compact we know that this open cover has a Lebesgue number . In particular if and thus and thus . The conclusion follows
In this post we will discuss one of the most important in point-set topology, Tychonoff’s theorem, and prove up some miscellaneous facts that are either interesting or useful.
Theorem (Tychonoff): Let be a non-empty class of non-empty compact topological spaces, then is compact under the product topology.
Proof: It follows from an earlier theorem that we must merely show that every class of closed sets in the defining closed subbase with the F.I.P have non-empty intersection. So, let be such a class. Consider some arbitrary , since is subbasic closed for ever we know that is closed, and so is a class of closed subsets of , and noticing that and so by assumption we have that . In other words, we have there exists some and so every element of contains an element which has as it’s th coordinate. Doing this for each produces a set such that . In light of our initial comments the conclusion follows.
From this we can prove a nice little result that is used incessantly in multivariate analysis. But, before we continue we need a little terminology
n-cell: An -cell is a multi-dimensional generalization of a closed interval in . In other words, we call an -cell if for every . In other words, is an -cell if there exists some such that .
Theorem (Generalized Heine-Borel): Every closed and bounded subspace of is compact.
Proof: Let be such a set. Since is bounded we know that is contained in some -cell . So, if we prove that every -cell is compact we are done (since a closed subspace of a compact space is compact).
But, we know that and since is compact we have by Tychonoff’s theorem that the product of compact spaces is compact. The conclusion follows.
Be careful about taking this and attempting to try to make a generalization of the extreme value theorem, namely “$If is continuous and compact then “. It fails for an easily over-looked and decidedly non-topological property of . Particularly, we have that has a canonical ordering (the one we know and love), but what is the ordering on ? No ordering no concept of infimums and supremums.
We now take a look at some of the odds and ends that were previously mentioned. It should probably be mentioned for those that are reading this and already know topology that, for now, we omit mention of the majority (save Hausdorfness AKA ) of the separability axioms and things like local compacntess, connectedness, paracompactness, and path-connectedness. We shall mention them in due time, from where we will discuss which are invariant under product.
Theorem: Let be a finite number of discrete spaces, then the product set is the same topological space under both the product and discrete topology.
Proof: Let be the product topology on and, of course, let be the discrete topology (I use the for power set). It is clear that and so we must merely show the reverse inclusion to finish.
So, let then for some subsets , but since is discrete we know that each is open. Thus, is the finite product of open sets and thus open in the product (box) topology. The conclusion follows.
The fact that I noted that product topology and box topology were the same here was meaningful. We know that the two coincide for the product of finitely many spaces, but the next remark shows (once again) how they differ for infinite.
Theorem: Let be an arbitrary collection of discrete spaces, then under the box topology coincides with under the discrete topology. The same cannot be said about the product topology.
Proof: Follows the exact same reasoning as in the last theorem.
To see why the infinite product of discrete spaces need not be discrete under the product topology, one need only consider where under the discrete topology. Clearly then is open in under the discrete topology but NOT the product topology.
The last thing we will talk about is the product of maps.
Product of maps: Let and be two maps, then define the product of the maps to be by .
We extend this definition to an arbitrary number of topological spaces as follows: Let and be two sets of topological spaces and let be a class of mappings such that (this can obviously be formulated in the guise of ordered triples as well), we then define by
The reader may recall that this concept was used to prove that if is continuous and Hausdorff then is a closed subset of .
We begin by proving the product of continuous maps is continuous. This is the first of a sequence of three proofs which will lead to a very satisfying corollary.
Theorem: Let and be an arbitrary collection of topological spaces and let be a corresponding set of continuous mappings such that . Then, is continuous, where are under the product topology.
Proof: Let then, which clearly means
that and so and thus
Conversely, let . Then, and thus and thus . The conclusion follows.
Now, as was previously proven we need only prove that the inverse image of each open basic set is open. So, let (the defining open base for ). Then, where for all but finitely many , and thus . But, we see that since is an open set which equals for all but finitely many ‘s that is going to be an open set (since each is continuous) which equals for all but finitely many ‘s. The conclusion follows.
Theorem: Let everything be as in the previous theorem, except instead of each being continuous, let it be bijective. Then is bijective.
Injectivity: Suppose that . Then, by definition and since two functions (remember we think of arbitrary tuples as functions) are equal if and only if they take the same value at every point we see that and since each is bijective (and thus injective) it follows that , from where injectivity follows.
Surjectivity: Let be arbitrary. For each we have that there exists some such that and thus from where surjectivity immediately follows.
Theorem: Let everything be as before, except now instead of continuous or bijective assume that each is surjective and open. Then, is open.
Proof: Let then for some . Clearly then we have that and so and so .
Conversely, let then for some and thus
The conclusion follows. .
Now, let be open in then where for all but finitely many it follows then that is an open set which equals for all but finitely many (since each is surjective and open). Thus which by prior comment is the product of open sets in which equals the full space for all but finitely many . It follows that is open in under the product topology.
Theorem: Combining these three we see that if for all , then there exists some homeomorphism (a bijective continuous open map) and so is a bijective continuous open map and thus a homeomorphism. It follows that where each are under the product topology.
Up until now we’ve talked quite extensively about necessary and sufficient conditions for a topological space to be compact, but we’ve neglected to speak about how compactness interacts with mappings of topological spaces into another topological spaces. Probably the most widely used theorem is the following:
Theorem 1: The continuous image of a compact topological space is compact, or phrased differently if be continuous and compact, then is a compact subspace of .
Proof: Let be an open cover of then for some open set in . Clearly we have that covers and so and since is continuous and open in we see that is an open cover of and thus by assumption we have that there exists some which is also an open cover. Therefore, . Therefore, is an open cover for . The conclusion follows.
Corollary : If is a surjective continuous mapping and is compact, then is compact.
Corollary : If (that’s what I use for homeomorphic) then is compact if and only if is compact.
With this theorem we can prove one of the most highly used, and least thought about, theorems in elementary calculus.
Theorem 2: Let be continuous, where is a compact topological space. Then .
Lemma: A compact subspace of a metric space is bounded.
Proof: Let be compact. Consider the class . Clearly is an open cover for and so there exists some finite subcover . Clearly then, . The conclusion follows. .
Now, we know that is compact, and combining the above with the earlier proof that ever compact subspace of a Hausdorff space is closed we see that is closed and bounded. Thus, we have that exist and it’s a relatively easy exercise in the basic topology of the real line that if is closed then . It follows that .
Corollary 1: A continuous function assumes a maximum and minimum value on
Corollary 2: If is continuous and compact, then
Also, using corollary two you can prove the Heine-Borel theorem in more elementary, albeit more complex, terms.
Theorem (Heine-Borel): If is closed and bounded, then is compact.
Proof: Just as last time we note that since is bounded that it is contained in some interval . And so, if we can prove that is compact then ‘s compactness follows since it is a closed subspace of a compact space. So now we need to prove that is compact, but we even have to prove something simpler thanks to the above theorem and the following lemma.
Lemma: Let and then
Proof: If we can find a linear function we are done. So assume that one exists and is of the form . Then we need . Solving this gives which clearly satisfies the conditions of a homeomorphism.
With this lemma in mind it remains to prove that any closed interval is compact. We choose, of course, for convenience . So let be an open cover of , and assume that it admits no finite subcover. Clearly then either no finite subclass of can cover either . Let be one of these subintervals. Clearly has length . Now, using the same logic we see that no finite subclass of can cover either . Continuing in this manner we have a sequence of closed intervals . It is clear that . Clearly then each is an upper bound for the set and so let and . Clearly by prior comment but by definition and by the contrivance of our sequence implies that for all and so . Since covers and where exists some such that . Now, since is open there exists some such that .
The Archimedean principle now furnishes us with some such that . Thus, . Now, . Therefore and . Consequently, . Therefore, we have that may be covered by a finite subclass of . Boom.
Another theorem which proves useful has to do with mappings from compact spaces into Hausdorff spaces, in particular:
Theorem: Let be compact and Hausdorff, then if is continuous and bijective, then it is a homeomorphism.
Proof: We must merely show that the mapping is open to conclude that it’s a homeomorphism, but since is bijective we know that given any that and so is open if and only if it’s closed. Thus, we must merely show that is closed. To do this we merely note that since is compact, that any closed must necessarily be compact. And by our first theorem then is a compact subspace of . But, by previous problem compact subspaces of Hausdorff spaces are closed and so is closed. The conclusion follows.
This theorem has some awfully amazing applications when trying to prove that two spaces are homeomorphic. It saves a lot of hassle. Our next theorem has to do with the graph of a continuous mapping from a compact space into a Hausdorff space, and although they’re are easier methods to prove this, I believe that a little elbow grease is worth the reward. That said, we need to prove some theorems not related to compactness.
Theorem: Let be continuous and define by . Then, this function is continuous.
Proof: Let , then and so and and so and and so . Conversely, let , then and . Therefore, and so . The conclusion follows.
Now, since we must only prove that the inverse image of any basic open set in is open in , but this follows immediately since any basic open set is of the form where both and are open in . And since is continuous we see that is open and so is the product of two open sets in and and thus basic open in and so clearly open. The conclusion follows.
Next, we prove a fact that is used incessantly as a counter example when dealing with Hausdorff spaces.
Theorem: Let (this is called the diagonal in ). Then is closed in .
Proof: We prove that is closed by proving that is open. So, let , then and since both and are in the Hausdorff space there exists open sets and such that and and . Clearly then $latext (x,y)\in O_x\times O_y$ which is open in and since we see that . The conclusion follows.
We prove our second to last theorem before we get to the relevant part.
Theorem: Let , , , and be as above, then where ().
Proof: Let , then but this clearly implies that . Therefore,
Conversely, let , then . The conclusion follows.
From this we quickly derive the following fact:
Theorem: Let be continuous and Hausdorff, then is a closed in .
Proof: Since is continuous we know that is continuous, and since is closed in it follows that is closed. The conclusion follows.
We now return to the actual theorem we wished to prove.
Theorem: If is continuous and is compact, then is a compact suspace of .
Proof: It follows from out last theorem that is a closed subspace of , but since is compact we know that is compact and since (see next post) the product of compact spaces is compact, we know that is compact. Thus, is a closed subspace of a compact space, and thus compact. The conclusion follows.
In the previous post we looked at the real number motivation for the concept of compactness and generalized it to topological spaces. We also gave some useful alternative formulations of compactness. We continue briefly here:
Basic open cover: If is a topological space and an open base for then is a basic open cover for if it is an open cover for and composed entirely of basic sets.
Theorem: A set is compact if and only if every basic open cover has a finite subcover.
: Let be an open cover for , clearly we have that each where is some class of open basic sets. Clearly then is a basic open cover for and by assumption there exists some set which covers clearly taking one of the ‘s that contain for finishes the argument.
: This is obvious since a basic open cover is an open cover.
Subbasic open cover: Similarly to before, a subbasic open cover is an open cover consisting entirely of sets from a given open subbase.
Theorem: A topological space is compact if and only if every subbasic open cover has a finite subcover.
Proof: It is clear from the last post that it is sufficient to show that is compact if and only if every class of closed subbasic sets with the F.I.P. have non-empty intersection.
So, let be the subbase in question. By the previous theorem it suffices to show that every set of closed basic sets with the F.I.P has non-empty intersection.
So, let be the open base generated by and let have the F.I.P. We first show that is contained in some maximal collection of open basic sets in the sense that if then does not have the F.I.P.
To see this we use Zorn’s lemma. Let be the collection of all superset of which have the F.I.P. It is clear that this class along with set inclusion induced a partially ordered set. Now, it is realtively easy to show that if is a chain, then is a class with the finite intersection property, and thus an upper bound. It follows by Zorn’s lemma that has a maximal element, call it and since it only remains to show that
Since each is a basic open set we know that where each is a subbasic closed set. And so if we can show that we are done because .
It merely remains to show that at least one of is in for each . So now assume that none of the aforementioned sets are in . Since is a subbasic open set is is necessarily a basic closed set and so to assume that would mean that and by the maximality of we see that does not have the finite intersection property. This then implies that for some . Doing this for each of gives us a class whose union is disjoint from which contradicts having the F.I.P. Doing this for each finishes the proof.
These will be extremely useful later in the proof of some very powerful theorems. Here is one of them.
Theorem (Heine-Borel): A subset is compact if is closed and bounded.
Proof: Since is bounded we know that for some finite closed interval and if we can prove that is compact it our theorem will follow since will be a closed subspace of a compact space.
So, now if this is obvious, so assume . It is relatively easy to prove that the set is an open subbase for so the set is a closed subbase, we must merely show that any subclass of with the F.I.P. has non-empty intersection. So let be non-empty and posses the F.I.P. If contains only intervals of one of the two types then their intersection clearly contiains or , so assume that contains intervals of both types. Let . We finish the argument by showin that for ever . Suppose not, and there existed some . By definition then there exists some and so contradicting ‘s F.I.P. The conclusion follows.
One immediately corollary of this is the following:
Theorem: Let be sequence of decreasing closed intervals. Then, .
To continue from here we need to define the following:
Compact subspace: If and is compact with the relative topology we say that is a compact subspace of .
It is clear that not only is this definition necessary for many reasons one would be we might want to consider whether is compact implies that , but as one can see we run into the problem of under what topology? As we will see later, in a generalization, if are compact subspaces of some ambient space then is a compact subspace. But for now:
Theorem: If is a compact topological space and is a closed subset of then is a compact subspace.
Proof: Let be an open cover for , by definition we have that for some open in . Clearly covers and thus is an open cover for . By assumption then there exists some finite subcover of , if we may discard it and be left with some finite set which covers . Clearly then taking the corresponding in produces the necessary finite subcover.
Of course any student worth their weight in sawdust is asking “Is the converse necessarily true? Do compact subspaces have to be closed?”. The answer is, no. Consider for example the topological space with and . Clearly is a compact subspace since it’s finite but it is clearly not closed since . Fortunately, that’s not all she wrote. There is a partial converse:
Theorem: Let be a Hausdorff topological space and a compact subspace, then is closed in .
Proof: We prove that is closed by proving that is open. Let . For each let be the open sets containing respectively such that . Clearly, if we let then is clearly an open cover for and so there exists some such that . Clearly then we have that for the corresponding set . So let which is clearly an open set disjoint from . The conclusion follows.
We now proved what was promised earlier.
Theorem: Let be a topological space and let be a finite class of compact subspaces of . Then is a compact subspace of .
Proof: Let be an open cover for , then for some open set in . Clearly then covers and thus each . Clearly then are open covers for respectively which by assumption means that for each there exists some which is a finite subcover of . Let . Then is a class of open sets in that covers . Clearly then intersecting each element of will result in the desired subclass of . The conclusion follows.
Using this we can prove a nice little theorem. Suppose that you were a naive analyst and you encounter the following dilemma. You have a function where but is not compact. You wonder, could I possibly make compact by removing a few points? The answer is no.
Theorem: Let be a topological space and let be a non-compact subspace of , then
Proof: Suppose that was compact, then since each is compact (since they’re finite) it follows that is compact. Boom.
Also, one might ask “what about the infinite case?” The answer is…well no. It should be obvious intuitively that since the union of an infinite number of closed sets need not be closed and any compact subspace of a Hausdorff space is closed that this probably isn’t true. Just to sate your doubtlessly eager minds though, consider with the discrete topology and consider that is an infinite class of compact spaces who’s union is the full space and by previous discussion, this is not compact.
Now, what about the intersection of compact subspaces?
Theorem: Let be a non-empty class of compact closed subspaces of a topological space then if (we don’t ever want to talk about empty topological spaces) then is compact.
Proof: Clearly we have that given any that since is a closed subset of that is a closed subset of . And since is compact it follows that is a compact subspace of and thus a compact subspace of .