Why is Hausdorff needed here?

Question

Let $\{X_i, i \in I\}$ be a family of Hausdorff spaces, where $I$ is an infinite index set and every $X_i$ is Hausdorff and contains at least two different points. Prove that the product space $X=\prod_{i\in I}X_i$ does not contain isolated points.

Attempt: Let $x = (x_i)_{i\in I} \in X$. Consider a basic open neighborhood $U$ of $x$. Then $U = \prod_{i \in F}U_i \times \prod_{i\in F^c}X_i$ where $F\subset I$ is finite and every $U_i$ is open in $X_i$. Define $y \in X$ as follows: $$y_i = \begin{cases}x_i, \text{ if }i\in F \\ w_i, \text{otherwise}\end{cases}$$ where $w_i$ is a point in $X_i$ different from $x_i$. Now clearly $y \in U \cap (X\setminus \{x\})$ so $x$ cannot be an isolated point of $X$. Since $x$ was arbitrary $X$ contains no isolated points.

What's the problem with this proof? If it's correct then the assumption that every $X_i$ is Hausdorff is redundant!

Thanks.

Answer 1

You are right: in full generality, a product space contains an isolated point if and only if all its factors contain an isolated point and all but finitely many of them are one-point spaces. If this is the case, then $x\in\prod\limits_{i\in I}X_i$ is isolated if and only if $x_i$ is isolated for all $i\in I$ (such that $X_i$ is not a singleton).

Answer 2

Another way to convince yourself that this statement does not require Hausdorffness assumption, without actually proving either version: fix any $(X_i)_i$ such that each $X_i$ contains at least two distinct points. For each $i$, let $X_i'$ have the same underlying set as $X_i$, endowed with the discrete topology.

Then it is easy to check the identity mapping $X'=\prod_i X_i'\to X$ is continuous and by the Hausdorff version of your statement, $X'$ has no isolated point. Since the preimage of an isolated point by a continuous bijection is obviously an isolated point, it easily follows that $X$ has no isolated points, either.