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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 $n$.

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user546106
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  • The sequence in the exercise you suggest isn't in $A$, since it has some zero coordinates. What don't you understand about that sentence? – B. Mehta Apr 15 '18 at 01:09
  • @B.Mehta Thank you, you are right. How about the sequence $(a_n)$, where $a_n = (1/n, 1/n, ...)$? Does it fail to converge to $0$? Why? – user546106 Apr 15 '18 at 01:13
  • What are your thoughts on this question? Specifically, does the proof given give you some hint? – B. Mehta Apr 15 '18 at 01:13
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    @B.Mehta $a_n$ converges to $a$ in $X$ iff for any nbhd. $U$ of $a$, there exists $N\in \Bbb Z_+$ s.t. for all $n>N$, $a_n\in U$. So we first pick a neighborhood of $0$ in the infinite product space, then we want to show that there is no element in this space converges to $0$. For the $a_n$ that I defined, I think we can pick a nbhd. $U= \prod(-1/n,1/n)$? Then $(a_n)$ does not converge. – user546106 Apr 15 '18 at 01:23
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    Yup, for no $n$ does the sequence $(a_n)$ lie in that $U$ so the sequence fails to converge. – B. Mehta Apr 15 '18 at 01:26
  • Yes $(1/n,1/n,\ldots)$ fails to converge to $0$ in the box topology. If you take a product of small intervals that are shrinking, this open set will fail to contain almost all of the sequence. – Cheerful Parsnip Apr 15 '18 at 02:28
  • You just need to find only one $x\in\overline{A}$ such that any sequence ${x_n}$ in $A$ does not converge to $x$. – ChoF Apr 15 '18 at 02:50
  • @ChoF Thank you! – user546106 Apr 15 '18 at 03:15

2 Answers2

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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?

Henno Brandsma
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  • (1).It is easy to show that this space is a $T_n$ space for $n\leq 3\frac {1}{2}.$ Is it a normal space? (2). Is it consistent with ZFC that the tightness of this space is a singular cardinal? – DanielWainfleet Apr 15 '18 at 10:05
  • @DanielWainfleet (1) is still an open problem, I believe. IIRC it is normal under CH. on (2) I have no thoughts. – Henno Brandsma Apr 15 '18 at 10:26
  • @DanielWainfleet I looked it up in "Open Problems in Topology" (p191, M.E. Rudin' s chapter) which states: Kunen and Rudin proved that under CH every box product of countably many locally compact separable metric spaces is paracompact. van Douwen showed that no non-trivial box product of metric spaces is normal if one factor is the irrationals. Lawrence showed that if $\mathfrak{b}=\mathfrak{d}$ or $\mathfrak{c} = \mathfrak{d}$ every countable box product of locally compact metric spaces is paracompact. – Henno Brandsma Apr 15 '18 at 10:36
  • @DanielWainfleet We don't even know whether a countable box product of copies of $\omega+1$ is normal or paracompact in ZFC. – Henno Brandsma Apr 15 '18 at 10:37
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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 $f .$

DanielWainfleet
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  • The tightness $\tau (X)$ of a topological space is the least infinite cardinal $k$ such that whever $Y\subset X$ and $p\in \overline Y,$ there exists $Z\subset Y$ where the cardinal of $Z$ is at most $k$, and $p\in \overline Z$. If $X$ is a metrizable space then $\tau (X)=\aleph_0. $ The tightness of the box-product topology on $\Bbb R^{\omega}$ is uncountable. – DanielWainfleet Apr 15 '18 at 09:59
  • At least it is known that $\mathbb{R}^\omega$ in the box topology is discretely generated, so there is a discrete subset $D$ of $f^+$ such that $f \in \overline{D}$. For compact Hausdorff spaces this is equivalent to countable tightness (so maybe in general spaces we might see it as a second best alternative to countable tightness, also any first countable $T_2$ space is discretely generated). – Henno Brandsma Apr 15 '18 at 10:48