# Abstract Nonsense

## The dreadful product topology (Pt. II)

In the previous post we gave the definitions of the product topology and some nice little tools about it. Here, we give some theorems regarding what properties of topological spaces are invariant under the formation of the product topology. Let us start out nice and easy

Theorem: Let $X_1,\cdots,X_n$ be a finite number of topological spaces with corresponding open bases $\mathfrak{B}_1,\cdots,\mathfrak{B}_n$. Then, $\mathfrak{M}=\left\{B_1\times\cdots\times B_n:B_k\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant m\right\}$ is an open base for $X_1\times\cdots\times X_n$ under the product topology.

Proof: Let $O$ be open in $X$, and let $o\in O$. There exists some $B\in\mathfrak{B}$ (the defining open base) such that $o\in B\subseteq O$. But, by theorem 1 in the last post we know that $\pi_k(B)$ is an open set containing $\pi_k(o)$ for $1\leqslant k\leqslant n$ and since $\mathfrak{B}_k$ is an open base for $X_k$ we know there exists some $J_k\in\mathfrak{B}_k$ such that $\pi_k(o)\in J_k\subseteq \pi_k(B)$. Clearly then $o\in J_1\times J_n\subseteq B$ and since $J_1\times J_n\in\mathfrak{M}$ and $o,O$ were arbitrary the conclusion follows. $\blacksquare$

We know use this to prove a nice little theorem.

Theorem: Let $X_1,\cdots,X_n$ be a finite number of second countable spaces, then $X=X_1\times\cdots\times X_n$ under the product topology is second countable.

Proof: Let $\mathfrak{M}$ be as above. Since $\mathfrak{M}$ is an open base for $X$ is remains to prove that it is countable. Let $\eta:\mathfrak{M}\mapsto\mathfrak{B}_1\times\cdots\times\mathfrak{B}_n$ be given by $B_1\times\cdots\times B_n\mapsto \left(B_1,\cdots,B_n\right)$. It is clear that this is a bijection. Recalling that the finite product of countable sets is countable finishes the argument. $\blacksquare$

It turns out that the above is a little more restrictive than is necessary as the following theorem illustrates.

Theorem: Let $\left\{X_n\right\}_{n\in\mathbb{N}}$ be a sequence of second countable spaces, then $\displaystyle X=\prod_{n=1}^{\infty}X_n$ is second countable under the product topology.

Proof: Let $\left\{\mathfrak{B}_n\right\}$ be the corresponding countable open bases and define $\displaystyle \mathcal{B}_m=\left\{E:E=\prod_{j=1}^{\infty}G_j\right\}$ where $\pi_k(G_j)\in\mathfrak{B}_k,\text{ }1\leqslant k\leqslant m$ and $\pi_k(G_j)=X_j,\text{ }m.

It is relatively easy to prove that $\eta:\mathcal{B}_m\mapsto\mathfrak{B}_1\times\cdots\times\mathfrak{B}_m$ given by $\displaystyle B_1\times\cdots\times B_m\times\prod_{j=m+1}^{\infty}X_{j+1}\mapsto \left(B_1,\cdots,B_m\right)$  is a bijection. It follows then, since the finite product of countable sets is countable, that $\mathcal{B}_m$ is countable.

So, now define $\displaystyle \mathfrak{M}=\bigcup_{j=1}^{\infty}\mathcal{B}_j$. It is clear that since $\mathfrak{M}$ is the countable union of countable sets that it is countable. So now, it remains to prove that $\mathfrak{M}$ is in fact an open base for $X$

So, let $O$ be open in $X$ and let $o\in O$. There exists some $B$ in the defining open base such that $o\in B\subseteq O$. Considering the form in which basic open sets in the product topology take we know that there is a finite set of indices $\left\{j_1,\cdots,j_n\right\}$ such that $\pi_{j_k}(B)\ne X_k$. So, let $m=\max\{j_1,\cdots,j_n\}$. Clearly we have that $\pi_\ell(O)$ is a neighborhood of $\pi_k(o)$ for $1\leqslant k\leqslant m$ and since $\mathfrak{B}_k$ is an open base there exists some $B_k$ such that $\pi_k(o)\in B_k\subseteq\pi_k(O)$. So, let $U=B_1\times\cdots\times B_m\times$$\displaystyle \prod_{j=m+1}^{\infty}X_j$. Then, it is evident that $o\in U\subseteq B\subseteq O$. Noting that $U\in\mathfrak{M}$ finishes the argument. $\blacksquare$.

Note the necessity for finiteness again.

Theorem: Let $X_1,\cdots,X_n$ be a finite number of toplogical spaces and let $\mathfrak{D}_1,\cdots,\mathfrak{D}_n$ be a number of corresponding dense sets. Then, $\mathfrak{D}_1\times\cdots\times\mathfrak{D}_n$ is dense in $X=X_1\times\cdots\times X_n$ under the product topology.

Proof: Let $x\in X$ be arbitrary and let $N$ be any neighborhood of $x$. We then have by theorem 1 in the last post that $\pi_k(N)$ is a neighborhood of $\pi_k(x)$ for $1\leqslant k\leqslant n$ and since $\mathfrak{D}_k$ is dense in $X_k$ there exists some $\mathfrak{d}_k\in\mathfrak{D}_k$ such that $\mathfrak{d}_k\in\pi_k(N)$. Clearly then $\left(\mathfrak{d}_1,\cdots,\mathfrak{d}_n\right)\in N$, and clearly $\left(\mathfrak{d}_1,\cdots,\mathfrak{d}_n\right)\in\mathfrak{D}_1\times\cdots\times\mathfrak{D}_n$. Noting the arbitrariness of $x, N$ finishes the argument. $\blacksquare$

From this we can derive the obvious consequence.

Theorem: Let $X_1,\cdots,X_n$ be finite number of separable spaces and let $X=X_1\times\cdots\times X_n$ under the product topology, then $X$ is separable.

Proof: Since each $X_k$ is separable it has a countable dense subset $\mathfrak{D}_k$ and by the above theorem we see that $\mathfrak{D}_1\times\cdots\times\mathfrak{D}_n$ is dense in $X$, but since the finite product of countable sets is countable the conclusion follows. $\blacksquare$

As in the case of second countable spaces we may extend this to the countable case

Theorem: Let $\left\{X_n\right\}_{n\in\mathbb{N}}$ be sequence of separable spaces, then $\displaystyle X=\prod_{j=1}^{\infty}X_j$ is separable.

Proof: Since each $X_n$ is separable it has a countable dense subset $\mathfrak{D}_n$ we may extract an arbitrary but fixed sequence from them as $\left\{d_n\right\}_{n\in\mathbb{N}}$ where $d_n\in\mathfrak{D}_n$. We now define $\displaystyle \mathcal{D}_m=\prod_{j=1}^{\infty}G_j$ where $G_j=\begin{cases} \mathfrak{D}_j & \mbox{if} \quad 1\leqslant j\leqslant m \\ \{d_j\} & \mbox{if} \quad j>m\end{cases}$

Clearly we have that $\eta:\mathcal{D}_m\mapsto\mathfrak{D}_1\times\cdots\times\mathfrak{D}_m$ by $D\mapsto \left\{\pi_1(D)\right\}\times\cdots\times \left\{\pi_m(D)\right\}$ is a bijection and since the finite product of countable sets is countable it follows that $\mathcal{D}_m$ is countable. So, let $\displaystyle \mathcal{D}=\bigcup_{m=1}^{\infty}\mathcal{D}_m$ being the countable union of countable sets is countable.

From the previous paragraph it is clear that we must merely show that $\mathcal{D}$ is dense in $X$. To do this let $x\in X$ be arbitrary and $N$ any neighborhood of $X$. There exists some $B\in\mathfrak{B}$ (the defining open base) such that $x\in B\subseteq N$. But, we know that $\pi_{k}(B)\ne X_k$ for finitely man indices, say $k_1,\cdots,k_m$. Let $r=\max\{k_1,\cdots,k_m\}$. We know then that $\pi_\ell(B)$ is a neighborhood of $\pi_\ell(x)$ for every $1\leqslant \ell\leqslant r$ and since $\mathfrak{D}_\ell$ is dense in $X_\ell$ there exists some $\mathfrak{d}_\ell$ such that $\mathfrak{d}_\ell\in\pi_\ell(B)$. Clearly then, we have that $\displaystyle D=\prod_{j=1}^{\infty}\{g_j\}$ where $g_j=\begin{cases} \mathfrak{d}_j & \mbox{if} \quad 1\leqslant j\leqslant r\\ d_j & \mbox{if} \quad j>r\end{cases}$ is in $B$ and thus $N$. But, $D\in\mathcal{D}$ and the conclusion follows. $\blacksquare$

We now discuss the product of Hausdorff spaces.

Theorem: $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of Hausdorff spaces, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ under the product topology is Hausdorff.

Proof: Let $\bold{x},\bold{y}\in X$ such that $\bold{x}\ne \bold{y}$ then $\pi_k(\bold{x})\ne\pi_k(\bold{x})$ for some $k\in\mathcal{J}$ and since $X_k$ is Hausdorff we know that there exists open sets $U,V$ such latex $\pi_k(\bold{x})\in U,\pi_k(\bold{y})\in V$ and $U\cap V$. So define $\displaystyle E=\prod_{j\in\mathcal{J}}G_j$ where $G_j=\begin{cases} U & \mbox{if}\quad j=k \\ X_j & \mbox{if}\quad k\ne j\end{cases}$ and define $\displaystyle F=\prod_{j\in\mathcal{J}}H_j$ where $H_j=\begin{cases} V & \mbox{if}\quad j=k \\ X_j & \mbox{if} \quad x\ne j\end{cases}$ clearly we have that $\bold{x}\in E$ and $\bold{y}\in F$ and since $\pi_k(F)\cap\pi_k(E)=\varnothing$ we must have $F\cap E=\varnothing$. The conclusion follows. $\blacksquare$

February 20, 2010 -

1. We can extend the last theorem even to c (continuum) many factors, but not to more. For a proof, see the Ask a Topologist forum of this year..

Comment by Henno | March 27, 2010 | Reply

2. Thank you very much Henno. I was aware of this, but I only post proofs that I am capable of doing by myself. Thus, while I have never tried the proof for the product of $\mathfrak{c}$ copies I’m not sure I could manage.

That aside, I have seen you on Ask a Topologist. Great work!

Comment by drexel28 | March 28, 2010 | Reply

3. It’s not that hard, really. But there is one “trick” to it that most wouldn’t think of by themselves.
Nice idea, this series! I’ll link to it on the forum!

Comment by Henno | March 29, 2010 | Reply

• Really? That would be great! I would love to hear feed-back/suggestions from more people!

Comment by drexel28 | March 29, 2010 | Reply