$\prod_{n=1}^{\infty}{\mathbb{R}}$ endowed with the box topology is not first countable.

Question

What I'm trying to prove is that if $X^{+}\subseteq X:=\prod_{n=1}^{\infty}{\mathbb{R}}$ is the set of all positive sequences in $\mathbb{R}$, then no sequence of elements in $X^{+}$ converges to the zero sequence. This is my approach: I assumed that a sequence $(z_{n})_{n=1}^{\infty}$ in $X$ converges to $0$. I want to show that there is $n$ such that $z_{n}$ is eventually $0$. Let $U_{n}=(-1/n,1/n)$. Then $U:=\prod_{n=1}^{\infty}{U_{n}}$ is open in $X$, and so there exists $N$ such that $z_{n}\in U$ for all $n\geq N$. From there, we can deduce that such $z_{n}$ converges to $0$, but I'm stuck from this point. Is my approach correct? What can I possibly add from here?

Answer 1

So assume you have a sequence $(z_n)$ in $X^+$. Let $\textbf{0}$ be the all $0$ function $\Bbb N \to \Bbb R$. Each $z_n$ is also a function $\Bbb N \to \Bbb R$ (also known as a real sequence) such that $z_n(k)>0$ for each $k \in \Bbb N$ (that's because it is in $X^+$).

Now we can apply a "diagonal argument" by defining $O = \bigcap_n (-\frac12 z_n(n), \frac12 z_n(n))$ which is a well-defined box-open neighbourhood of $\textbf{0}$ (as all $z_n(n) >0$ and so each interval is open and contains $0$). Now it's clear that $z_n \notin O$ for all $O$ (why?) and so $z_n$ cannot converge to $\textbf{0}$ at all.

It's not good enough to work with the $(-\frac1n, \frac1n)$ interval: It's possible that all $z_n$ lie in your $U$ when they're smaller than that. The "problem" neighbourhood has to depend on the sequence you're considering. I just showed that whatever the sequence is, there is a neighbourhood of $\textbf{0}$ that misses all terms of the sequence, a sort of "anti-convergence".

Note that the sequence argument disproves first countability indirectly: it is true that $\textbf{0} \in \overline{X^+}$ but in a first countable space this would imply there is a sequence from $X^+$ converging to $\textbf{0}$, while we've shown the latter not to be the case. It follows that $X$ is not first countable (as a consequence of first countability fails).

Answer 2

HINT:

Let's show this: for every countable family $(U_n)$ of neighborhoods of $x=(x_m)$ there exists a neighborhood $V$ of $x=(x_m)$ that does not contain any of the $U_n$. Now, by the definition of the box topology, we may assume that
$$U_n= \prod_{m \in \mathbb{N}} U_{n m}$$ where $U_{n m}$ is a neighborhood of $x_m$ in $X_m$. Now, since $x_m$ is not isolated, there exists $V_m$ neighborhood of $x_m$, $ U_{m m}\not \subset V_m$. Now check that $V = \prod_{m \in \mathbb{N}} V_m$ works. (this is just the diagonal argument).