Is $\mathbb{R}^{\omega}$ under the box topology Lindelöf? And is it locally metrizable?

Question

As stated in the question, is $\mathbb{R}^{\omega}$ under the box topology Lindelöf? And is it locally metrizable?

Answer 1

If $\Box_{n \in \omega} \mathbb R$ were Lindelöf, then so, too, would be any closed subspace. However $\Box_{n \in \omega} \{ 0,1 \}$ is easily seen to be a closed discrete subspace of $\Box_{n \in \omega} \mathbb R$ of cardinality $2^{\aleph_0}$, and so is not Lindelöf. Thus $\Box_{n \in \omega} \mathbb R$ cannot be Lindelöf.

As Henno Brandsma mentions in a comment, $\Box_{n \in \omega} \mathbb R$ is nowhere first-countable, and so it cannot be locally metrizable. The proof of nowhere first-countablity is a straightforward diagonal argument.

• Suppose that $\mathbf x = ( x_n )_{n \in \omega} \in \Box_{n \in \omega} \mathbb R$ and $\{ U_i : i \in \omega \}$ is any family of open neighborhood of $\mathbf x$ in $\Box_{n \in \omega} \mathbb R$. For each $i$ there is a sequence $( \delta_{i,n} )_{n \in \omega}$ of positive reals such that $\mathbf x \in \prod_{n \in \omega} ( x_n - \delta_{i,n} , x_n + \delta_{i,n} ) \subseteq U_i$. Consider $V = \prod_{n \in \omega} ( x_n - \frac{\delta_{n,n}}2 , x_n + \frac{\delta_{n,n}}2 )$. Then $V$ is an open neighborhood of $\mathbf x$ in $\Box_{n \in \omega} \mathbb R$, but $U_n \nsubseteq V$ for each $n$.