Here is Prob. 1, Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
(a) A $G_\delta$ set in a space $X$ is a set $A$ that equals a countable intersection of open sets of $X$. Show that in a first-countable $T_1$ space, every one-point set is a $G_\delta$ set.
(b) There is a familiar space in which every one point set is a $G_\delta$ set, which nevertheless does not satisfy the first countability axiom. What is it?
Here and here are Math Stack Exchange posts on Prob. 1 (a).
So here I'll be attempting Prob. 1 (b), Sec. 30, in Munkres.
My Attempt:
Let $\mathbb{R}^\omega$ denote the set of all the (infinite) sequences of real numbers.
Here the set $\mathbb{R}$ of real numbers has the standard topology having as a basis all the (bounded) open intervals $(a, b)$ on the real line, where $a, b \in \mathbb{R}$ and $a < b$.
Let $X \colon= \mathbb{R}^\omega$ with the box topology having as a basis all the sets of the form
$$ \prod_{n=1}^\infty U_n, \tag{A} $$
where each $U_n$ is open in $\mathbb{R}$.
Let $\mathbf{x} \colon= \left( x_1, x_2, x_3, \ldots \right)$ be any point in $\mathbf{X}$.
Then we can write the one-point set $\{ \mathbf{x} \}$ as
$$ \{ \mathbf{x} \} = \bigcap_{m=1}^\infty \left( \prod_{n=1}^\infty \left( x_n – \frac{1}{m}, x_n + \frac{1}{m} \right) \right), $$
and all the sets $\prod_{n=1}^\infty \left( x_n – \frac{1}{m}, x_n + \frac{1}{m} \right)$ are open in $\mathbf{X}$. Therefore $\{ \mathbf{x} \}$ being the intersection of a countable collection of open sets of $\mathbf{X}$ is a $G_\delta$ set in $\mathbf{X}$.
Now we show that there exists no countable local basis at $\mathbf{x}$.
Let $\left\{ \mathbf{B}_n \colon n \in \mathbb{N} \right\}$ be any countable collection of open sets of $\mathbf{X}$ such that each set $\mathbf{B}_n$ contains $\mathbf{x}$.
Now for each $n \in \mathbb{N}$, as $\mathbf{B}_n$ is an open set in $X$ and as
$$\mathbf{x} = \left( x_1, x_2, x_3, \ldots \right) \in \mathbf{B}_n,$$
so there exist open sets $B_{n1}, B_{n2}, B_{n3}, \ldots$ in $\mathbb{R}$ such that
$$ \mathbf{x} \in \prod_{m=1}^\infty B_{nm} \subset \mathbf{B}_n. \tag{B} $$
Now as
$$\mathbf{x} = \left(x_1, x_2, x_3, \ldots \right) \in B_{n1}\times B_{n2} \times B_{n3} \times \cdots,$$
so we can conclude that $x_m \in B_{nm}$ for each $m \in \mathbb{N}$.
For each $n = 1, 2, 3, \ldots$ and for each $m = 1, 2, 3, \ldots$, as $x_m \in B_{nm}$ and as $B_{nm}$ is an open set in $\mathbb{R}$ (with the standard topology), so there exists a (bounded) open interval $\left( a_{nm}, b_{nm} \right)$ on the real line such that
$$ x_m \in \left( a_{nm}, b_{nm} \right) \subset B_{nm}, \tag{C} $$
where $a_{nm}$ and $b_{nm}$ are real numbers such that $a_{nm} < b_{nm}$, and let us choose real numbers $c_{nm}$ and $d_{nm}$ such that
$$ a_{nm} < c_{nm} < x_m < d_{nm} < b_{nm}. \tag{D} $$
Thus from (B), (C), and (D) we obtain
$$
\begin{align} \mathbf{x} &\in \prod_{m=1}^\infty \left( c_{nm}, d_{nm} \right) \\
&\subset \prod_{m=1}^\infty \left( a_{nm}, b_{nm} \right) \qquad \mbox{[ using (D) above ]} \\
&\subset \prod_{m=1}^\infty B_{nm} \qquad \mbox{[ using (C) above ]} \\
&\subset \mathbf{B}_n \qquad \mbox{[ using (B) above ]} \end{align} \tag{0} $$
for each $n \in \mathbb{N}$.
Now let us put
$$ \mathbb{U} \colon= \prod_{m=1}^\infty \left( c_{mm}, d_{mm} \right). \tag{Definition 0} $$
Then $\mathbf{U}$ is an open set in $X$ and from the preceding paragraph we can also conclude that
$\mathbf{x} \in \mathbf{U}$.
However, if there were a natural number $N$ for which $\mathbf{B}_N \subset \mathbf{U}$, then for that particular $N$ we would also obtain the inclusions
\begin{align}
& \ \ \prod_{m=1}^\infty \left( a_{Nm}, b_{Nm} \right) \\
&\subset \prod_{m=1}^\infty B_{Nm} \qquad [ \mbox{ by virtue of (0) above } ] \\
&\subset B_N \qquad [ \mbox{ by virtue of (B) above } ] \\
&\subset \mathbf{U} \\
&= \prod_{m=1}^\infty \left( c_{mm}, d_{mm} \right), \qquad [ \mbox{ refer to (Definition 0) above } ]
\end{align}
and hence for each $m \in \mathbb{N}$ we would obtain
$$ \left( a_{Nm}, b_{Nm} \right) \subset \left( c_{mm}, d_{mm} \right), $$
and so
$$ c_{mm} \leq a_{Nm} < b_{Nm} \leq d_{mm}. \tag{1} $$
And, for $m = N$ in particular from (1) we would obtain
$$ c_{NN} \leq a_{NN} < b_{NN} \leq d_{NN}, $$
which would contradict (D) above.
Thus although $\mathbf{U}$ is an open set in $\mathbf{X}$ containing $\mathbf{x}$, there exists no $n \in \mathbb{N}$ for which $\mathbf{B}_n \subset \mathbf{U}$. So no countable collection $\left\{ B_n \colon n \in \mathbb{N} \right\}$ of open sets of $\mathbb{X}$ containing $\mathbf{x}$ is a local basis at $\mathbf{x}$. Thus $\mathbf{X}$ is not first-countable.
Hence $\mathbf{X} = \mathbb{R}^\omega$ with the box topology is not first-countable although every one-point subset of $\mathbf{X}$ is a $G_\delta$ set in $\mathbf{X}$.
Is this proof correct? If so, is each and every step of it correct and clearly written? Or, are there problems in any detail thereof?
Best Answer
I think the proof is fine, I think. It's probably the example Munkres intends in the context of his book. He does mention somewhere the related example of $\Bbb R$ where $\Bbb Z$ is identified to a point (in the quotient topology). This would also have worked.
A minor critique: the proof of equality for $\{x\}$ is omitted, but this is not too hard: if $y \neq x$ is in $\Bbb R^\omega$, this means that for some $k$, $y_k \neq x_k$. But then for large enough $n$, $y_k \notin (x_k-\frac{1}{n}, x_k+\frac{1}{n})$ (pick $n$ such that $\frac{1}{n} < |y_k-x_k| (>0$) and then $y \notin \prod_k (x_k-\frac{1}{n}, x_k + \frac{1}{n})$, which is the $n$-th set in the intersection you (correctly) used. With your history of attention to detail, I would have expected that argument to have been included too. But in a paper or thesis, such things would be omitted of course.