## The dreaded product topology (Pt. III Tychonoff’s theorem and odds and ends)

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:**

**Lemma:**

**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.

**Proof:**

**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:**

**Lemma:** .

**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.

[…] little circle is a little square.” Given this normal topology one can prove that has the Heine-Borel Property that while in general compact subspaces of metric spaces are closed and bounded the converse is […]

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