$\mathbb{R} ^{\mathbb{N}}$ with box topology is not first countable

Question

I have $X:=\mathbb{R} ^{\mathbb{N}}$ equipped with the box topology. I want to show that it's not first countable: since every coordinate has a countable basis for its neighbourhoods, and not less (in $\mathbb{R}$ with usual topology) then a neighbourhood of $(a_n)_{n \in \mathbb{N}}$ must have a basis of cardinality at least $\aleph_0^{\aleph_0}=2^{\aleph_0}$. Is this correct? Moreover I have to show the following. Let $I$ be the set of the sequences converging to $0$ but never $0$: then there's no sequence in $I$ converging to it. Find a net in $I$ converging to it. I know that if my proof about the first countability is true then a sequence can't converge to $(0)_{i \in \mathbb{N}}$ because it would make a countable basis for the neighbourhoods of $0$.

Is everything correct? Can you fin a net in $I$ converging to $(0)_{i \in \mathbb{N}}$?

Answer 1

Sequences that are eventually constant are convergent, so in every space each point is the limit of a convergent sequence. What you mean, I suspect, is that no point is the limit of a sequence of other points. Showing this would indeed show that $X$ is not first countable, but in fact it’s not true here. For $n\in\Bbb Z^+$ let $x^{(n)}=\langle x_k^{(n)}:k\in\Bbb N\rangle$, where

$$x_k^{(n)}=\begin{cases} n,&\text{if }k=0\\ 0,&\text{otherwise}\;. \end{cases}$$

It’s easy to check that $\langle x^{(n)}:n\in\Bbb Z^+\rangle$ converges to the zero sequence $z=\langle 0,0,0,\ldots\rangle$ in $X$, even though each $x^{(n)}\in X\setminus\{z\}$.

However, it’s possible to show directly that $X$ is not first countable. Your cardinality argument is not sufficient as it stands: it shows that each point of $X$ has a local base of cardinality at most $\mathfrak{c}=2^\omega$, but you would need further argument to establish that no smaller local base exists.

Added: In fact it’s consistent with the usual axioms of set theory that $2^\omega=\omega_2$, and each point of $X$ has a local base of cardinality $\omega_1$: this is the case in models in which the dominating number $\mathfrak{d}=\omega_1$ and the continuum $\mathfrak{c}=\omega_2$.

In the second half of this answer I used this approach to show a more general result:

For each $n\in\Bbb N$ let $X_n$ be a first countable space with a non-isolated point $p_n$. Let $X$ be the box product of the spaces $X_n$, and let $p=\langle p_n:n\in\Bbb N\rangle\in X$; then $X$ is not first countable at $p$.

Since no point of $\Bbb R$ is isolated, this shows that $X$ is not first countable at any point. You might find it instructive to try to adapt that argument to the specific case of the space $X$.

If $Y$ is any space, and $p\in Y$ is not an isolated point, we can always find a net in $Y\setminus\{p\}$ that converges to $p$. I’ll do this in general and let you adapt that to the specific case in your problem.

Let $\mathscr{B}$ be any nbhd base at $p$, and let $\mathscr{D}=\{\langle B,x\rangle:x\in B\setminus\{p\}\in\mathscr{B}\}$. For $\langle B,x\rangle,\langle C,y\rangle\in\mathscr{D}$ let $\langle B,x\rangle\preceq\langle C,y\rangle$ if and only if $B\supseteq C$; then $\langle\mathscr{D},\preceq\rangle$ is a directed set. (Check this!) Let $\nu$ be the net

$$\nu:\mathscr{D}\to Y:\langle B,x\rangle\mapsto x\;;$$

I leave it to you to verify that $\nu$ converges to $p$.

Answer 2

Perhaps the simplest way to do this is to use a diagonalization argument.

Suppose $\mathbb R^{\omega}$ in the box topology is first countable and wlog $x=0$. Then if $\mathscr{N}=\left \{ (U^{(n)})_{m} \right \}_{m,n}$, and we list the elements of $\mathscr N$in the usual way, we may observe that $\frac{1}{2}U_{n}^{(n)}$ is open, contains $0$ but contains no member of $\mathscr N$, so $\mathscr N$ can not be a nbhd base about $0$.

Similarly, let $x_{m}$ be any sequence in $\mathbb R^{\omega}$ such that $x_m$ is $not$ eventually constant.

Then for $x_1$, choose $x^{n_1}_{1}\neq 0$ and in general, for $x_m\ $ an $x^{n_m}_{m}\neq 0$ such that $m>k\Rightarrow n_m>n_k$. This is possible since $x_m$ is not eventually constant. Now if we take the open set $U=\left ( -\frac{1}{2}x^{m_k}_{k}, \frac{1}{2}x^{m_k}_{k}\right )\cup (-1,1)_{n\neq m_k}$ we see that $U$ contains $no$ member of the sequence $x_m$ so $x_m\nrightarrow0$.

If $x_m$ is eventually constant then there is an integer $N$ such that if $n>N$ there are only finitely many non-zero coordinates $x^{(k)}_{n}$ in $x_n$ and it is easy to show that in this case $x_m\to 0$.

Finding a net converging to zero has already been covered in the other answer.