0
$\begingroup$

How to prove that space $\mathbb{R}_w$, the countably infinite product of $\mathbb{R}$ in the box topology, is not metrizable? I have tried finding a solution to this problem, but failed. Kindly help me find this answer.

$\endgroup$
2
$\begingroup$

At no point of this product in the box topology, the space is first countable: e.g. for $p=(p_1, p_2,p_3,\ldots)$: suppose $\{U_n: n \in \mathbb{N}\}$ is a local base at $p$. Every $U_n$ contains an open box around $p$ and as all open sets in $\mathbb{R}$ are unions of open intervals, for each $m$:

$$\exists r^{(m)}_1,r^{(m)}_2, \ldots,r^{(m)}_i,\ldots: p \in \prod_i (p_i - r^{(m)}_i, p_i + r^{(m)}_i) \subseteq U_m$$.

Then define $O = \prod_i (p_i - \frac{1}{2} r^{(i)}_i, p_i + \frac{1}{2}r^{(i)}_i)$ which is box-open, and contains $p$. But no $U_n$ can be a subset of $O$ ($U_n$ fails at the $n$-th coordinate), and so the $U_n$ cannot form a local base at $p$.

All metric spaces do have local countable bases everywhere: $\{B(p, \frac{1}{n}):n \in \mathbb{N}\}$ will do.

So this box product is not metrisable.

$\endgroup$
0
$\begingroup$

Hint: Do it by contradiction method. You must have learnt the sequence lemma: It says " If A a subset of a topological space X and there is a sequence in A converging to a point x then $x\in cl (A) $. The converse holds if X is metrizable". So assuming $\mathbb {R} ^{\omega}$ is metrizable try to find a contradiction to this theorem(i.e., prove the converse of the above Lemma isn't true.)
Try it with the set $$A=\{(x_1,x_2,\cdots)| x_i>0 for\, all \, i\in \mathbb {N}\} $$.
$\mathbf {0} $ is a limit point of this set but there is no sequence of points in A that converges to $\mathbf {0} $ in box topology. Hope it helps?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.