Prove on metrization of uncountable product

Question

I am given the following problem:

Show that an uncountable product of unit intervals is not first countable, and thus not metrizable.

My answer would be that a), since the elements of the neighborhood basis have to contain open sets, only a finite amount of their elements can be unequal to the entire unit interval, which means that some elements will always be equal to the entire unit interval and thus the open sets not containing the unit interval in that element will also not contain any set in the supposed neighborhood basis, and b) that it is always possible to construct a countable neighborhood basis in a metric space using open balls with rational diameters.

Is this proof correct? The question is marked as more difficult than the other ones, and my answer seems to contradict that a bit.

Answer 1

Unfortunately your proof is not correct - you do not use the fact that we have an uncountable product.

Let $P = \prod_{\alpha \in A} I_\alpha$ where $A$ is uncountable and each $I_\alpha = I$. Let $\xi \in P$.

Assume that exists a countable neigborhood basis $\{U_n\}$ at $\xi$. For each $n$ there exists a finite subset $A_n \subset A$ such that $\xi \in \prod_{\alpha \in A} V^n_\alpha \subset U_n$ where the $V^n_\alpha$ are open in $I$ and $V^n_\alpha = I$ for $\alpha \notin A_n$. The set $B = \bigcup_n A_n$ is countable. Pick $\beta \in A \setminus B$. Let $\xi_\beta$ be the $\beta$-th coordinate of $\xi$ and $W_\beta$ be an open neigborhood of $\xi_\beta$ in $I$ such that $W_\beta \ne I$. Then $W = \prod_{\alpha \in A} W_\alpha$ with $W_\alpha = I$ for $\alpha \ne \beta$ is an open neigborhoood of $\xi$. It must contain some $U_n$, hence also some $\prod_{\alpha \in A} V^n_\alpha$. But $\beta \notin A_n$, thus $V^n_\beta = I$. Therefore $\prod_{\alpha \in A} V^n_\alpha$ contains points whose $\beta$-th coordinate is not in $W_\beta$ - but these points cannot be contained in $W$.

This contradiction proves that no countable neigborhood basis exists at $\xi$.