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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?

Martin Sleziak
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Shuichi
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  • @DanielSchepler yes, I'm using box topology. – Shuichi Jan 08 '22 at 00:54
  • Hints: Wlog $x=0.$ Suppose there is a countable base at $x,$ say ${U_n}{n\in \mathbb N}.$ Then, $U_i=\prod{j=1}^\infty U_i^{(j)}$ Inductively. construct a basic set $V$ such that $U_n\nsubseteq V$ for any integer $n$. Start by taking $U_1^{(1)}$ and finding an open $V_1^{(1)}$ properly contained in $U_1^{(1)}.$ – Matematleta Jan 08 '22 at 01:09
  • I don't think your intermediate statement is true: for example, if $z_n = (\frac{1}{n}, 0, 0, \ldots)$ then $z_n$ converges to the zero sequence. – Daniel Schepler Jan 08 '22 at 01:13
  • For the overall statement in the title, see https://math.stackexchange.com/questions/2242615/metrizability-of-infinite-product-space-under-box-topology/2242627#2242627 . That could probably also serve as inspiration to come up with a proof that the zero sequence is not the limit of any sequence in $X^+$. – Daniel Schepler Jan 08 '22 at 01:14
  • hi Shuichi Saihara; I see that you have asked a number of questions on the site, and received many nice answers, but have not accepted any. are you familiar with the site's feature of "accepting" answers? if not, here is an explanation of how to do it. if your question has been resolved by someone, accepting their answer lets you both express your gratitude to them and takes your question off of the unanswered queue. – Atticus Stonestrom Jan 08 '22 at 02:05
  • An older post related to the same problem: $\mathbb {R}^{\omega}$ in the box topology is not first countable. – Martin Sleziak Jan 08 '22 at 09:42

2 Answers2

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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).

Henno Brandsma
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  • Hi professor, could I ask your assistance here, please? I am really curious to know if the argumentations I gave about uniform and pointwise convergence are correct. Forgive the bother. – Antonio Maria Di Mauro Jan 08 '22 at 09:59
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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).

orangeskid
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