An uncountable product of $\mathbb{R}$ with itself is not metrizable (product topology).

Question

An uncountable product of $\mathbb{R}$ with itself isn't metrizable in the product topology

I wrote what I believe is an incorrect proof of this statement since I feel like I ignored the fact that the product is uncountable. However I can't find where is my mistake.

Cannot this proof "work" for countably infinite product of $\mathbb{R}$ ? (But aren't these products of $\mathbb{R}$ metrizable with the product topology ?).

Proof:

Let $J$ be an uncountable index set.

Let $(\mathbb{R}^J,\tau)$ be a topological space under the product topology.

Let $\mathcal{N}(x)$ denote the set of every neighbourhood of $x$ in $\mathbb{R}^J$.

Let $A\subset\mathbb{R}^J$ such that $\forall x \in A$, $x_\alpha = 1\text{ except for finitely many }\alpha\in J$,

We want tot show that $0\in\overline{A}$ ?

Let $U\in\mathcal{N}(0)\subset\mathbb{R}$. \begin{gather} U\in\tau \Rightarrow U=\prod_{\alpha\in J}{U_\alpha}\text{ with finitely many } U_\alpha \neq \mathbb{R} \end{gather} Let $\Gamma = \left\lbrace \beta\in J \vert U_\beta \neq \mathbb{R} \right\rbrace$

Then for $\beta\in\Gamma$ we know $U_\beta$ is open in $\mathbb{R}$ and $0\in U_\beta$.

So, \begin{gather} U_\beta = ]a_\beta; b_\beta[\text{ with }a_\beta < 0 < b_\beta \end{gather}

Let $y\in\mathbb{R}^J$ s.t. for $\alpha\in J$, \begin{gather} y_\alpha = \frac{b_\alpha - a_\alpha}{2}\text{, }\alpha\in\Gamma\\ y_\alpha = 1\text{, }\alpha\notin\Gamma \end{gather} Then $y\in A$ therefore $\exists y\in A$ s.t. $y\in U \cap A$.

Since $U\in\mathcal{N}(0)$ we have $0\in A' \subset \overline{A}$.

Now we must show that all sequences $(x_n)$ of $A$ do not converge to $0$.

Suppose that there is a sequence $(x_n)$ of $A$ s.t. $x_n \rightarrow 0$ \begin{gather} \forall U\in\mathcal{N}(0):\exists N\in\mathbb{N}:n>N\Rightarrow x_n\in U \end{gather} We note $x_{n,\alpha}$ the $\alpha$th coordinate of $x_n$.

Let $D_n=\lbrace i\in J \vert x_{n,i}\neq 1\rbrace$ (obviously $D_n$ is finite $\forall n\in\mathbb{N}$).

Then, let \begin{gather} \Gamma_n = \prod_{\alpha\in J}U_\alpha \end{gather} such that \begin{gather} U_\alpha = ]-|x_{n,\alpha}|;|x_{n,\alpha}|[\text{, }\alpha\in D_n\\ U_\alpha = \mathbb{R}\text{, }\alpha\notin\ D_n \end{gather} Then $\forall n\in\mathbb{N}$ we have $\Gamma_n\in\mathcal{N}(0)$ and $x_n\notin\Gamma_n$.

So $x_n \nrightarrow 0$, which further imply that every sequence $(x_n)$ of $A$ cannot converge to $0$.

Therefore $\mathbb{R}^J$ isn't metrizable.

Answer 1

I just realised that I made an obvious mistake, and that I can easily transform my proof in order to show that in the box topology my claim is true.

It was false because it's clear that $x_i$ can be in $\Gamma_n$ if $i\neq n$.

However $\Gamma = \bigcap_{n\in\mathbb{N}}\Gamma_n$ is open in the box topology and every $x_n\notin\Gamma$ which would show (by deleting the parts saying that infinitely $U_\alpha=\mathbb{R}$) that the space isn't metrizable.