$\Bbb R^{\omega}$ in the box topology is not metrizable

Question

Munkres Topology section 21 example 1 shows that $\Bbb R^{\omega}$ in the box topology is not metrizable.

(The hidden problems have been solved, but I have another new problem. Please see below.) I don't understand this sentence "the point $a_n$ cannot belong to $B'$ because its $n$th coordinate $x_{nn}$ does not belong to the interval $(-x_{nn},x_{nn})$."//We have an exercise in section 20 which gives a sequence of points in $\Bbb R^{\omega}$ converging to $0$. $z_1 = (1,1,0,0...), z_2 = (1/2,1/2,0,0...), z_3= (1/3,1/3,0,0...),...$. Could anyone explain it?//Let $a_n = (1/n, 1/n, ...)$ and $(a_n)$ be a sequence of points in $\Bbb R^{\omega}$. Does $(a_n)$ converge to $0$ in the box topology of $\Bbb R^{\omega}$?

To show the sequence lemma does not hold for $\Bbb R^{\omega}$, we need to show that there exists(?) $x \in \bar{A}$ s.t. for any sequence $(x_n)$ in $A$, $x_n \not \to x$. (I'm not sure if we only need to prove there exists such $x \in \bar{A}$ or we actually need to prove for all $x\in \bar{A}$?)

Pick $0\in \bar{A}$ and $a_n$ is arbitrary. $$a_1 = (x_{11}, x_{21}, x_{31}, \cdots, x_{i1}, \cdots),\\ a_2 = (x_{12}, x_{22}, x_{32}, \cdots, x_{i2}, \cdots),\\ a_3 = (x_{13}, x_{23}, x_{33}, \cdots, x_{i3}, \cdots),\\\cdots\\ a_n = (x_{1n}, x_{2n}, x_{3n}, \cdots, x_{in}, \cdots)\\ \cdots$$

$B' = (-x_{11},x_{11})\times (-x_{21}, x_{21})\times (-x_{31},x_{31})\times \cdots \times(-x_{i1},x_{i1}) \times \cdots$ .

$a_1 \not \in B'$ because $x_{11}\not \in (-x_{11},x_{11})$. Similarly, $x_n \not \in B'$ because $x_{nn}\not \in (-x_{nn},x_{nn})$ for any $n$.

---

---

Answer 1

The structure of the proof is as follows: So we have the set $A$ of the all positive sequences.

Then we observe that $0 \in \overline{A}$, as every box-open neighbourhood of $0$ intersects $A$.

So if $\mathbb{R}^\omega$ were metrisable, there is a sequence $x_n$, where all $x_n \in A$ such that $x_n \to x$. But then we show that no sequence from $A$ can converge to $0$ at all, contradiction.

The no-convergence part is a classical diagonalisation argument (literally, here, as well):

Suppose $x_n \in A$ for all $n$, so we write (as each point in $A$ is a sequence of (strictly positive) reals:

$$x_n = (x^{(n)}_1,x^{(n)}_2, x^{(n)}_3, \ldots)$$

where I choose a slightly different notation: the upper $n$ indicates it's the $n$-th point in the sequence from $A$, the lower index is just the coordinate index. As said, being in $A$ means $\forall n,m: x^{(n)}_m > 0$ so we can define the following box-open set:

$$O = \prod_{i=1}^{\infty} ( -x^{(i)}_i, x^{(i)}_i)$$

which contains the point $0 \in \mathbb{R}^\omega$, as each coordinate contains $0$ being a symmetric neighbourhood $(-a,a)$ around $0$.

Then for each $n$, $x_n \notin O$: at the $n$-th coordinate we have $x^{(n)}_n \notin (-x^{(n)}_n, x^{(n)}_n)$, so $x_n \notin O$.

So $0$ has a neighbourhood that contains no point of the sequence, while if $x_n \to 0$, by definition of convergence, every neighbourhood of $0$ contains all but finitely many points of the sequence. So indeed $x_n \not\to 0$ in the box topology. This final contradiction then shows that the box product fails the sequence lemma for a specific $A$ and $0$, and this is enough to refute the metrisability (or even first countability) of the box product.

BTW, $0$ is not a special point, the space is not first countable at any point. Taking this $A$ and $0$ makes the proof easier to follow, I think.

I hope this clarifies your questions?

Answer 2

Consider the members of $X=\Bbb R^{\omega}$ to be functions from $\omega$ to $\Bbb R.$ Let $X^+=\{g\in X:\forall n\in \omega\; (g(n)>0)\}.$

For $f\in X$ and $g\in X^+$ let $B(f,g)=\{h\in X: \forall n\in \omega\;(|f(n)-h(n)|<g(n)\}.$ Then $\{B(f,g): f\in X\land g\in X^+\}$ is a base (basis) for the box-product topology on $X.$

For $f\in X$ let $f^+=\{f+g:g\in X^+\}$ . Then $f\in \overline {f^+}.$ Because if $U$ is a nbhd of $f$ then $f\in B(f,g)\subset U$ for some $g\in X^+,$ so $f+\frac {1}{2}g\in f^+\cap B(f,g)\subset f^+\cap U.$

If $F=\{f+g_j:j\in \omega\}$ is a countable subset of $f^+,$ let $h(n)=\frac {1}{2}\min \{g_j(n): j\leq n\}$ for each $n\in \omega.$ For each $j\in \omega$ we have $|f(j)-(f+g_j)(j)|=g_j(j)>\frac {1}{2}g_j(j)\geq h(j),$ implying $f+g_j\not \in B(f,h).$ So $B(f,h)$ is a nbhd of $f$ that is disjoint from $F,$ so $f\not \in \overline F.$

So $f\in \overline {f^+},$ but no sequence in $f^+$ converges to $f .$