Product of totally disconnected space is totally disconnected?

Question

I read that the cartesian product with the product topology of a family of totally disconnected topological spaces is totally disconnected, too. Is that true?

How are the connected components in the cartesian product with the product topology defined?

Didn't find it yet.

Thanks.

Answer 1

Yes, it’s true.

Added: It’s true, but the following argument proves a different result: it shows that the product of totally separated spaces is totally separated. If the spaces involved are compact Hausdorff, this is equivalent to their being totally disconnected, but the two properties are distinct in general. (I must have been thinking of compact: spaces when I wrote this answer.)

Let $X=\prod_{\alpha\in A}X_\alpha$, where each $X_\alpha$ is totally disconnected. To show that $X$ is totally disconnected, you just have to show that if $x$ and $y$ are distinct points of $X$, then there is a clopen set $U$ such that $x\in U$ and $y\notin U$: that’s exactly what it means to say that $x$ and $y$ are in different components of $X$.

Suppose, then, that $x=\langle x_\alpha:\alpha\in A\rangle,y=\langle y_\alpha:\alpha\in A\rangle\in X$, and $x\ne y$. Then there is some $\alpha\in A$ such that $x_\alpha\ne y_\alpha$. $X_\alpha$ is totally disconnected, so there is a clopen set $U$ in $X_\alpha$ such that $x_\alpha\in U$ and $y_\alpha\notin U$. Let $V=\pi_\alpha^{-1}[U]$, where $\pi_\alpha:X\to X_\alpha$ is the usual projection map; then $V$ is a clopen set in $X$, $x\in V$, and $y\notin V$.

Added: For a correct argument, suppose that $X$ is not totally disconnected, and let $C\subseteq X$ is connected. If $x=\langle x_\alpha:\alpha\in A\rangle$ and $y=\langle y_\alpha:\alpha\in A\rangle$ are distinct points of $C$. There is some $\alpha\in A$ such that $x_\alpha\ne y_\alpha$. But the canonical projection $\pi_\alpha:X\to X_\alpha$ is continuous, so $\pi_\alpha[C]$ is a connected subset of $X_\alpha$ containing the distinct points $x_\alpha$ and $y_\alpha$. Thus, $X_\alpha$ is not totally disconnected.