# Abstract Nonsense

## 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 $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty class of non-empty compact topological spaces, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ 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 $\mathcal{S}=\left\{S_k\right\}_{k\in\mathcal{K}}$ be such a class. Consider some arbitrary $j\in\mathcal{j}$, since $S_k$ is subbasic closed for ever $S_k\in\mathcal{S}$ we know that $\pi_j(S_k)$ is closed, and so $\left\{\pi_j(S_k)\right\}_{j\in\mathcal{J}}$ is a class of closed subsets of $X_j$, and noticing that $\pi_{j}(S_{k_1})\cap\cdots\cap\pi_j(S_{k_n})\supseteq\pi_j\left(S_{k_1}\cap\cdots\cap S_{k_n}\right)\supseteq\varnothing$ and so by assumption we have that $\displaystyle \bigcap_{k\in\mathcal{K}}\pi_j(S_k)\ne \varnothing$. In other words, we have there exists some $\displaystyle x_j\in\bigcap_{k\in\mathcal{K}}\pi_j(S_k)$ and so every element of $\mathcal{S}$ contains an element which has $x$ as it’s $j$th coordinate. Doing this for each $j\in\mathcal{J}$ produces a set $\left\{x_j\right\}_{j\in\mathcal{J}}$ such that $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\bigcap_{k\in\mathcal{K}}S_k$. In light of our initial comments the conclusion follows. $\blacksquare$

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 $n$-cell is a multi-dimensional generalization of   a closed interval in $\mathbb{R}^n$ . In other words, we call $E$ an $n$-cell if $\pi_k(E)=[a,b]$ for every $1\leqslant k\leqslant n$. In other words, $E$ is an $n$-cell if there exists some $[a,b]\subseteq\mathbb{R}$ such that $E=\underbrace{[a,b]\times\cdots\times[a,b]}_{n\text{ times}}=[a,b]^n$.

Theorem (Generalized Heine-Borel): Every closed and bounded subspace of $\mathbb{R}^n$ is compact.

Proof: Let $E$ be such a set. Since $E$ is bounded we know that $E$ is contained in some $n$-cell $[a,b]^n$. So, if we prove that every $n$-cell is compact we are done (since a closed subspace of a compact space is compact).

But, we know that $[a,b]^n=[a,b]\times\cdots\times[a,b]$ and since $[a,b]\subseteq\mathbb{R}$ is compact we have by Tychonoff’s theorem that the product of compact spaces is compact. The conclusion follows. $\blacksquare$

Be careful about taking this and attempting to try to make a generalization of the extreme value theorem, namely “\$If $f:X\mapsto\mathbb{R}^n$ is continuous and $X$ compact then $\sup\text{ }f\left(X\right),\inf\text{ }f\left(X\right)\in f\left(X\right)$“. It fails for an easily over-looked and decidedly non-topological property of $\mathbb{R}^n,text{ }n\geqslant 2$. Particularly, we have that $\mathbb{R}$ has a canonical ordering (the one we know and love), but what is the ordering on $\mathbb{R}^n$? 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 $T_2$) 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 $X_1,\cdots,X_n$ be a finite number of discrete spaces, then the product set $X=X_1\times\cdots\times X_n$ is the same topological space under both the product and discrete topology.

Proof: Let $\mathcal{J}$ be the product topology on $X$ and, of course, let $\wp\left(X\right)$ be the discrete topology (I use the $\wp$ for power set). It is clear that $\mathcal{J}\subseteq\wp\left(X\right)$ and so we must merely show the reverse inclusion to finish.

So, let $E\in\wp\left(X\right)$ then $E=U_1\times\cdots\times U_n$ for some subsets $U_k\subseteq X_k$, but since $X_k$ is discrete we know that each $U_k$ is open. Thus, $E$ is the finite product of open sets and thus open in the product (box) topology. The conclusion follows. $\blacksquare$

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 $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of discrete spaces, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ under the box topology coincides with $X$ 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 $\displaystyle X=\prod_{j=1}^{\infty}X_j$ where $X_j=[0,1]$ under the discrete topology. Clearly then $(0,1)\times\cdots\times(0,1)\times\cdots$ is open in $X$ under the discrete topology but NOT the product topology. $\blacksquare$

The last thing we will talk about is the product of maps.

Product of maps: Let $f:X\mapsto Y$ and $f':X'\mapsto Y'$ be two maps, then define the product of the maps to be $f\times f':X\times X'\mapsto Y\times Y'$ by $(x,x')\mapsto (f(x),f'(x'))$.

We extend this definition to an arbitrary number of topological spaces as follows: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ and $\left\{Y_j\right\}_{j\in\mathcal{J}}$ be two sets of topological spaces and let $\left\{f_j\right\}_{j\in\mathcal{J}}$ be a class of mappings such that $f_j:X_j\mapsto Y_j$ (this can obviously be formulated in the guise of ordered triples as well), we then define $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{j}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$  by $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\mapsto\prod_{j\in\mathcal{j}}\{f_j(x_j)\}$

The reader may recall that this concept was used to prove that if $f:X\mapsto Y$ is continuous and $Y$ Hausdorff then $\Gamma_f$ is a closed subset of $X\times Y$.

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 $\left\{X_j\right\}_{j\in\mathcal{J}}$ and $\left\{Y_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of topological spaces and let $\left\{f_j\right\}_{j\in\mathcal{J}}$ be a corresponding set of continuous mappings such that $f_j:X_j\mapsto Y_j$. Then, $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is continuous, where $\displaystyle \prod_{j\in\mathcal{J}}X_j,\prod_{j\in\mathcal{J}}Y_j$ are under the product topology.

Proof:

Lemma: $\displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f_j^{-1}\left(O_j\right)$

Proof: Let $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\left( \prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)$ then, $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}\in\prod_{j\in\mathcal{J}}O_j$ which clearly means

that $f_j(x_j)\in O_j,\text{ }\forall j\in\mathcal{J}$ and so $x_j \in f^{-1}(O_j)$ and thus

$\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in\prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)$

Conversely, let $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in \prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)$. Then, $x_j\in f^{-1}\left(O_j\right)\implies f(x_j)\in O_j,\text{ }\forall j\in\mathcal{J}$ and thus $\displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)\in \prod_{j\in\mathcal{J}}O_j$ and thus $\displaystyle \prod_{j\in\mathcal{J}}\{x_j\}\in \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)$. The conclusion follows. $\blacksquare$

Now, as was previously proven we need only prove that the inverse image of each open basic set is open. So, let $B\in\mathfrak{B}$ (the defining open base for $\displaystyle \prod_{j\in\mathcal{J}}Y_j$). Then, $\displaystyle B=\prod_{j\in\mathcal{J}}O_j$ where $O_j=Y_j$for all but finitely many $j$, and thus $\displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(B\right)=\left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f^{-1}\left(O_j\right)$. But, we see that since $O_j$ is an open set which equals $Y_j$ for all but finitely many $j$‘s that $f^{-1}(O_j)$ is going to be an open set (since each $f_j:X_j\mapsto Y_j$ is continuous) which equals $X_j$ for all but finitely many $j$‘s. The conclusion follows. $\blacksquare$

Theorem: Let everything be as in the previous theorem, except instead of each $f_j:X_j\mapsto Y_j$ being continuous, let it be bijective. Then $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is bijective.

Proof:

Injectivity: Suppose that $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{y_j\}\right)$. Then, by definition $\displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{f_j(y_j)\}$ 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 $f_j(x_j)=f_j(y_j),\text{ }\forall j\in\mathcal{J}$ and since each $f_j$ is bijective (and thus injective) it follows that $x_j=y_j,\text{ }\forall j\in\mathcal{J}$, from where injectivity follows.

Surjectivity: Let $\displaystyle \prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}Y_j$ be arbitrary. For each $j\in\mathcal{J}$ we have that there exists some $x_j\in X_j$ such that $f_j(x_j)=y_j$ and thus $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{y_j\}$ from where surjectivity immediately follows. $\blacksquare$

Theorem: Let everything be as before, except now instead of continuous or bijective assume that each $f_j:X_j\mapsto Y_j$ is surjective and open. Then, $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is open.

Proof:

Lemma: $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f_j(O_j)$.

Proof: Let $\displaystyle \prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)$ then $\displaystyle \prod_{j\in\mathcal{J}}\{y_j\}=\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}x_j\right)=\prod_{j\in\mathcal{J}}\{f_j(x_j)\}$ for some $\displaystyle \left(\prod_{j\in\mathcal{J}}f_j\right)^{-1}\left(\prod_{j\in\mathcal{J}}O_j\right)=\prod_{j\in\mathcal{J}}f^{-1}(O_J)$. Clearly then we have that $x_j\in O_j$ and so $f_j(x_j)\in O_j$ and so $\displaystyle \prod_{j\in\mathcal{J}}\{f_j(x_j)\}=\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j(O_j)$.

Conversely, let $\displaystyle\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}\{f_j(O_j)\}$ then $y_j=f_j(x_j),\text{ }\forall j\in\mathcal{J}$ for some $x_j\in O_j$ and thus $\displaystyle \prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}\{x_j\}\right)=\prod_{j\in\mathcal{J}}\{y_j\}\in\prod_{j\in\mathcal{J}}f_j\left(\prod_{j\in\mathcal{J}}O_j\right)$

The conclusion follows. $\blacksquare$.

Now, let $E$ be open in $\displaystyle \prod_{j\in\mathcal{J}}X_j$ then $\displaystyle E=\prod_{j\in\mathcal{J}}O_j$ where $O_j=X_j$ for all but finitely many $j$ it follows then that $f_j(O_j)$ is an open set which equals $Y_j$ for all but finitely many $j$ (since each $f_j:X_j\mapsto Y_j$ is surjective and open). Thus $\displaystyle \prod_{j\in\mathcal{J}}f_j(E)=\prod_{j\in\mathcal{J}}f_J\left(\prod_{j\in\mathcal{J}}\right)=\prod_{j\in\mathcal{J}}f_j(O_j)$ which by prior comment is the product of open sets in $\displaystyle \prod_{j\in\mathcal{J}}Y_j$ which equals the full space for all but finitely many $j$. It follows that $f(E)$ is open in $\displaystyle \prod_{j\in\mathcal{J}}Y_j$ under the product topology. $\blacksquare$
Theorem: Combining these three we see that if $X_j\approx Y_j$ for all $j\in\mathcal{J}$, then there exists some homeomorphism (a bijective continuous open map)  $f_j:X_j\mapsto Y_j$ and so $\displaystyle \prod_{j\in\mathcal{J}}f_j:\prod_{j\in\mathcal{J}}X_j\mapsto\prod_{j\in\mathcal{J}}Y_j$ is a bijective continuous open map and thus a homeomorphism. It follows that $\displaystyle \prod_{j\in\mathcal{J}}X_j\approx\prod_{j\in\mathcal{J}}Y_j$ where each are under the product topology.