## Thoughts about compactness (Pt. III, compactness and mappings)

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 .

**Proof:**

**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:**

**Lemma:**

**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.

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