Countable product of first countable Spaces is first countable

Solution 1:

Pretty much any non-trivial product of uncountably many first countable spaces fails to be first countable:

Let $A$ be an uncountable index set, and for each $\alpha\in A$ let $X_\alpha$ be a first countable space with points $p_\alpha$ and $q_\alpha$ such that $p_\alpha$ has an open nbhd $U_\alpha$ such that $q_\alpha\notin U_\alpha$. Let $X=\prod_{\alpha\in A}X_\alpha$, and let $p=\langle p_\alpha:\alpha\in A\rangle\in X$; then $X$ is not first countable at $p$.

Let $\{V_n:n\in\Bbb N\}$ be any countable local base at $p$ in $X$. For each $n\in\Bbb N$ there are a finite $F_n\subseteq A$ and open sets $U(n,\alpha)$ in $X_\alpha$ for each $\alpha\in F_n$ such that the basic open set

$$B_n=\{x\in X:x_\alpha\in U(n,\alpha)\text{ for each }\alpha\in F_n\}$$

in the product $X$ is a nbhd of $p$ contained in $V_n$. Clearly $\{B_n:n\in\Bbb N\}$ is a local base at $p$.

Now let $C=\bigcup_{n\in\Bbb N}F_n$; each $F_n$ is finite, so $C$ is countable. $A$ is uncountable, so there is an $\alpha_0\in A\setminus C$. Let

$$W=\{x\in X:x_{\alpha_0}\in U_{\alpha_0}\}\;,$$

and note that $W$ is an open nbhd of $p$. Let $z=\langle z_\alpha:\alpha\in A\rangle\in X$ be defined by

$$z_\alpha=\begin{cases} p_\alpha,&\text{if }\alpha\in C\\ q_\alpha,&\text{if }\alpha\in A\setminus C\;; \end{cases}$$

then $z\in B_n\setminus W$ for each $n\in\Bbb N$, so for all $n\in\Bbb N$ we have $B_n\nsubseteq W$, and $\{B_n:n\in\Bbb N\}$ therefore cannot be a local base at $p$ after all.

One can prove a somewhat similar result for box products of countably infinitely many factors.

For each $n\in\Bbb N$ let $X_n$ be a first countable space with a non-isolated point $p_n$. Let $X$ be the box product of the spaces $X_n$, and let $p=\langle p_n:n\in\Bbb N\rangle\in X$; then $X$ is not first countable at $p$.

Suppose that $\mathscr{B}=\{B_n:n\in\Bbb N\}$ is a countable local base at $p$. Without loss of generality we may assume that for each $n,k\in\Bbb N$ there are open nbhds $U(n,k)$ of $p_k$ in $X_k$ such that

$$B_n=\prod_{k\in\Bbb N}U(n,k)\;.$$

For each $k\in\Bbb N$ let $V_k$ be an open nbhd of $p_k$ such that $V_k\subsetneqq U(k,k)$; this is possible because $p_k$ is not isolated in $X_k$.

Let $V=\prod_{k\in\Bbb N}V_k$, and suppose that $B_n\subseteq V$ for some $n\in\Bbb N$. Then $U(n,k)\subseteq V_k$ for each $k\in\Bbb N$, and in particular $U(n,n)\subseteq V_n\subsetneqq U(n,n)$, which is absurd. Thus, $V$ is an open nbhd of $p$ that does not contain any member of $\mathscr{B}$, contradicting the assumption that $\mathscr{B}$ was a local base at $p$. It follows that $X$ cannot be first countable at $p$.