Countable product of first countable Spaces is first countable 

Suppose $X_i$ are first countable Space , $X = \prod_{i=n}^{\infty} X_i $, Then $X$ is first countable  Space in product topology. Is it first countable in Box topology. Is uncountable product of first countable space is first countable


I have tried :
Let  $ x \in X$, we shall show that $X$ has a local countable basis at $x$.
we can assume that $x = \prod_{i =1}^{\infty} x_i$, where $x_i \in X_i$. Since $X_i$ is first countable space , then there is a local countable basis at $x_i$ says $B_{x_i} = \{ B_{x_i}(j) | j \in \mathbb N \} $
Define $W_n = \{ \prod_{i=1}^n B_{x_i}(j) | j \in \mathbb N \}  = \{ B_{x_1}(j_1) \times B_{x_2}(j_2 \times \dots  \times B_{x_n}(j_n) \ \  | \ \ j_i \in \mathbb N\} $ , Clearly $W_n$ is countable.
Let $W$ be the set of all subsets of the product space $X$ of the folowing form :
$\prod _{j=1}^\infty V_j$ , where there is some $n \in \mathbb N$ such that $\prod_{j=1}^n V_j \in W_n$ and for all $j > n , V_j = X_j$
we shall show that $W$ is a locall countable basis at $X$
Clearly $W$ is countable
Let $G$ be an open set containg $x$. So we can assume that $G = \prod_{i=1}^n G_i$, there exist some $ n \in \mathbb N$ such that $G_i$ is open in $X_i$ for all $i \leq n$ and $G_i = X_i$ for all $i >1$.
Since $B_{x_i} $ is a local countable basis at $x_i$. So for $x_i \in G_i$ for all $ i \leq n$, there is some $B_{x_i}(j_i)$ such that $ x_i \in B_{x_i}(j_i) \subset G_i$ for all $ i \leq n$.
let $V = \prod_{i=1}^\infty V_i$ such that $\prod_{i=1}^n V_i = \prod_{i=1}^n B_{x_i}(j_i)$ and $V_i = X_i $ for all $i>n$
Thus $V \in W_n$ and $V \subset G$, Thus $X$ is first countable.
Please see my solution and tell me $X$ is first countable in box topology and give me a counter example if collection $\{ X_{\alpha} \}$ is uncountable, then $\prod X_{\alpha} $ is not first countable in  product toplogy.
Thank you
 A: 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 point $p_n$ that has no smallest nbhd. (For example, $p_n$ might be a non-isolated point in a $T_1$ space $X_n$.) 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 $p$.
A: Your solution seems to be correct.
For box topology, you may consider the countably many product of real line. If $B$ is a countable collection of basis element
$$B = \left\{\prod_{k=1}^\infty (a_{nk},b_{nk}) : -\infty \le a_{nk}<b_{nk} \le \infty\right\}$$
we can consider
$$U = \prod_{n=1}^\infty (c_{n} , d_n).$$
for $a_{nn}<c_n < d_n<b_{nn}$. You can check that $U\subsetneq V_n$ for all $n$.
In the case of uncountable many product of first-countable space, you may consider the product of $2^{2^{\aleph_0}}$ many real line. For any given countable collection of open neighborhoods $\{V_n\}_{n\in \Bbb{N}}$ of the origin you can find a basic open set does not contains any $V_n$ for $n\in \Bbb{N}$, since $2^{\aleph_0}<2^{2^{\aleph_0}}$.
