Show that $\displaystyle\prod_{\Bbb{N}} \Bbb{R}$ with the box topology is Hausdorff but not metrizable.
$\Bbb{R}$ must be Hausdorff. For $x_1, x_2 \in \Bbb{R}$ (where $x_1 \not= x_2$), if $d$ denotes the distance between these two points, then we can choose an open ball $U_1$ with radius $d/2$ around $x_1$ and an open ball $U_2$ again of radius $d/2$ around $x_2$. Obviously, these two open balls will not intersect at any point.
So let $(x_1, x_2, …), (x'_1, x'_2, …) \in \Bbb{R^n}$. Then For each $x_i, x'_i \in \Bbb{R}$ (where $i \in \Bbb{N}$), there exist open balls $U_i$ and $U'_i$ in $\Bbb{R}$ around $x_i$ and $x'_i$, respectively, such that the open balls do not intersect. But then $(U_1 \times U_2 \times …) \cap (U'_1 \times U'_2 \times …) = \emptyset$.
We can show that $\displaystyle\prod_{\Bbb{N}} \Bbb{R}$ is not metrizable by proving that it is not first countable.
The first countability condition tells us that $\forall x \in X$ $\exists \{ U_n \}_{n\ \in \Bbb{N}} \subseteq P(X)$ such that
(i) $\forall n \in \Bbb{N}$, $U_n$ is a neighborhood of $x$.
(ii) For every neighborhood $V$ of $x$ $\exists n \in \Bbb{N}$ such that $U_n \subseteq V$.
So let's pick the point $(0,0,0,…) \in \displaystyle\prod_{\Bbb{N}} \Bbb{R}$. We know that, for all $n \in \Bbb{N}$ (in each $\Bbb{R}$) $(-1/n, 1/n)$ is a neighborhood of $0$.
So $(\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})…$ is a neighborhood of $(0,0,0,0,…)$.
But in order to have $U_n \in \{U_n\}_{n \in \Bbb{N}}$ such that $U_n \subseteq (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})…$ for all $n$, we know that the sequence $\{(\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})…\}_{n \in \Bbb{N}} = \{U_n\}_{n \in \Bbb{N}}$.
But a sequences approaches infinity, and as $n \rightarrow \infty$ $(\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})… = \{0\} \times \{0\} \times …$, which is a closed neighborhood of $(0,0,0,…)$, meaning that the 1st condition of first countability fails.
Is my answer correct? If not, can anybody give me a hint?
Thanks.
Best Answer
Your solution to Hausdorffness is not quite correct. Note what happens if your points are $\mathbf{x} = \langle 1 , 2, 3, \ldots \rangle$ and $\mathbf{x^\prime} = \langle 3 , 3 , 3 , \ldots \rangle$. (How do you choose the neighbourhoods $U_3$ and $U^\prime_3$?)
Recall that the usual product topology on $\mathbb{R}^\mathbb{N}$ is Hausdorff. Perhaps we can use this fact (or the proof of this fact.)
Your solution to the non-first-countability of the space is also somewhat off. You appear to be showing that one particular countable family of neighbourhoods of $\mathbf{0} = \langle 0 , 0 , \ldots \rangle$ is not a neighbourhood base. But you instead have to show that every countable family of neighbourhoods of $\mathbf{0}$ is not a neighbourhood base.
What you want to do is start with any old countably family $\{ U_n : n \in \mathbb{N} \}$ of open neighbourhoods of $\mathbf{0}$, and produce an open neighbourhood $V$ of $\mathbf{0}$ for which $U_n \not\subseteq V$ holds for all $n$. As a particularly strong hint, produce $V$ so that for each $n \in \mathbb{N}$ the projection of $V$ onto the $n$th coordinate is a proper subset of the projection of $U_n$ onto the $n$th coordinate.