Let $X$ denote $\prod_{n=1}^\infty\mathbb{R}$, the Cartesian product of countably infinitely many copies of $\mathbb R$ (which is just the set of all infinite sequences of real numbers), endowed with the box topology. Now, let $X^+\subset X$ be the subset consisting of the sequences of strictly positive real numbers, and let $z$ denote the zero sequence, that is, the one whose terms are $z_i = 0$ for all $i$. Show that $z$ is in the closure of $X^+$, but there is no sequence of elements of $X^+$ converging to $z$.

I guess I did the first part. The closure of $X^+$ is $$\bigcap_{\substack{\text{closed }S\,\subseteq X,\\ X^+\subseteq S}}S.$$ But these subsets are of the form $\prod[-E,+\infty)$ for every $E\geq 0$, right? When $E = 0$, we have the required. Is it right? And what about the second part?

  • $\begingroup$ I assume "Produtorium" means product? $\endgroup$ – Zev Chonoles Apr 26 '12 at 1:56
  • $\begingroup$ Yes, sure! I'm sorry. $\endgroup$ – Br09 Apr 26 '12 at 2:15
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    $\begingroup$ You can look at this post for info on how to type math here; please feel free to edit your post to clarify what you mean. $\endgroup$ – Zev Chonoles Apr 26 '12 at 2:18
  • $\begingroup$ Identifying $X$ with the set of functions from $\mathbb N$ to $\mathbb R,$ a useful base for the box topology is $\{B(f,g)\;|\; f\in X \land g:\mathbb N\to \mathbb R^+\}$ where $B(f,g)=\{h\in X \;|\; \forall n \in \mathbb N\;(|f(n)-h(n)|<g(x))\}.$ $\endgroup$ – DanielWainfleet Apr 17 '17 at 6:40

Your argument for the first part is basically ok . Not all closed sets containing $X^+$ are as you describe; but given any closed set containing $X^+$, there is a closed set of your form contained in it.

For the first part, it may be easier to show that any open set containing $z$ must contain an element of $X^+$. Note that such a set must contain an open set of the form $\prod_{i=1}^\infty O_i$ where each $O_i$ is an open set in $\Bbb R$ containing $0$.

For the second part, suppose that $(x_i)$ is a sequence in $X^+$. In the $i^{\rm th}$ copy of $\Bbb R$ in the product, let $O_i$ be an open set centered at zero such that $x_i(i)\notin O_i$. What can you say about the open set $O=\prod_{i=1}^\infty O_i$?

  • $\begingroup$ The set O doesn't contain any term of the sequence, right? Because every x_i doesn't belong to O_i. This way, by the definition, we can't have a sequence converging to zero, is it? I think your argument for the second part still works if we have a finite product, doesn't it? $\endgroup$ – Br09 Apr 26 '12 at 2:40
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    $\begingroup$ @ItamarSales Yes to your first question. For your second question, this shows the given sequence doesn't converge to $z$. The answer to your third question is "no": if the product is finite, we will run out of copies of $\Bbb R$ before we run through the sequence (if that makes sense). For a finite product, in fact, there is a sequence in $X^+$ converging to $z$ (for example in $\Bbb R^3$, the sequence with terms $x_n=(1/n,1/n,1/n)$. $\endgroup$ – David Mitra Apr 26 '12 at 2:51
  • $\begingroup$ Hum, understood! Your construction clarified things. Thank you! $\endgroup$ – Br09 Apr 26 '12 at 3:13

Recall that: From [Wikipedia][1]

  1. One of the most important properties of first-countable spaces is that given a subset $A$, a point $x$ lies in the closure of $A$ if and only if there exists a sequence $\{x_n\}$ in $A$ which converges to $x$.

A nice consequence of the result proven above is that $X$ (which is $\mathbb{R}^\omega$ with the box topology) is not first countable and thus not metrizable (all metric spaces are first countable).


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