Box topology does not preserve first countable

[R136a1]

So, in the topology text I'm reading is mentioned that if each $X_n$ is first countable, then $\prod_{n\in \mathbb{N}} X_n$ is first countable as well under the product topology. And then it says that this does not need to be true for the box topology. But there is no justification at all. Does anybody know a good counterexample?

 

[krome]

Let $X_n=\mathbb{R}$ and $X= \prod_{n \in \mathbb{N}} X_n = \mathbb{R}^{\mathbb{N}}$. Suppose $U_{n \in \mathbb{N}}$ is a countable neighborhood basis for the point $\vec{0} = (0,0, \ldots ) \in X$.

See if you can think of an open neighborhood of $\vec{0}$ that does not contain anyone of the $U_n$'s. That would contradict the assumption that the $U_n$'s form a countable neighborhood basis of $\vec{0}$ in the first place and would show that $X$ is not first countable.

 

[WannabeNewton]

As an added hint, recall Cantor's diagonal argument.

 

[R136a1]

Thanks a lot guys!

Here's what I came up with. We can of course assume that $U_n = \prod_{k\in \mathbb{N}} (a_k^n,b_k^n)$, and that all intervals $(a_k^n,b_k^n)$ are bounded. Now we consider the neighborhood $\prod_{k\in \mathbb{N}} (a_k^k/2, b_k^k/2)$. This is not included in any $U_n$. That seems to do it.

So it's not first countable. But maybe it is sequential? Is there something known about that?

 

[blue_raver22]

The statement that the box topology does not preserve first countable is indeed true. This means that if each $X_n$ is first countable, it does not necessarily follow that $\prod_{n\in \mathbb{N}} X_n$ is first countable under the box topology. This can be seen by considering the following counterexample:

Let $X_n$ be the set of real numbers with the standard topology for all $n\in \mathbb{N}$. This means that each $X_n$ is first countable. Now, let $X$ be the set of all real sequences, i.e. $X = \prod_{n\in \mathbb{N}} X_n$. Under the box topology, the open sets are generated by sets of the form $U = \prod_{n\in \mathbb{N}} U_n$, where each $U_n$ is an open set in $X_n$. However, for any sequence $x = (x_1, x_2, \ldots )\in X$, the collection of all open sets containing $x$ under the box topology is given by $\{U\subseteq X : x\in U\} = \{U \subseteq X : x_n\in U_n \text{ for all } n\in \mathbb{N}\}$. This collection is uncountable, since there are uncountably many choices for each $U_n$. Therefore, $X$ is not first countable under the box topology, since there is no countable basis for the topology.

In contrast, under the product topology, the open sets are generated by sets of the form $V = \prod_{n\in \mathbb{N}} V_n$, where each $V_n$ is an open set in $X_n$ and only finitely many of the $V_n$ are different from $X_n$. In this case, the collection of all open sets containing $x$ is given by $\{V\subseteq X : x\in V\} = \{V \subseteq X : x_n\in V_n \text{ for all but finitely many } n\in \mathbb{N}\}$. This collection is countable, since there are only countably many choices for the finitely many ##V