I'm trying to show that $\mathbb {R}^{\omega}$ in the box topology is not first countable. But I cannot come up with a contradiction by assuming that for each $x \in \mathbb {R}^{\omega}$, there is a countable basis. My intuition is that I need to use the fact that a countable product of countable sets is uncountable, so I need to show that for any local basis, I can come up with a basis consisting of the product of countable open intervals in each coordinate. Can anyone help me through?
General Topology – Why R^? in the Box Topology Is Not First Countable
general-topology
Related Solutions
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 $p$.
The countable intersection V, of the open nhoods is an open nhood of x.
Remove from V a point other than x to obtain an open nhood of x.
There in is a contradiction.
Best Answer
Use a diagonal argument. Given a putative countable basis at x (wlog x = all zeros), construct a basic open set not containing any of the countable ones by making sure it's smaller than the $i^{\mathrm{th}}$ open set in the $i^{\mathrm{th}}$ coordinate.