Product of totally disconnected space is totally disconnected?

Solution 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.

Solution 2:

Suppose $X=\prod X_\alpha$ where each $X_\alpha$ is a totally disconnected space and $A\subseteq X$ is connected. For each $\alpha$, the image $\pi_\alpha(A)\subseteq X_\alpha$ must also be connected, where $\pi_\alpha:X\to X_\alpha$ is the projection. Since $X_\alpha$ is totally disconnected, $\pi_\alpha(A)$ must be a singleton, say $\{x_\alpha\}$. It follows that $A$ is a singleton, since the only possible element of $A$ is the element $x\in X$ whose $\alpha$th coordinate is $x_\alpha$ for each $\alpha$.

Thus any connected subset of $X$ is a singleton, so $X$ is totally disconnected.

More generally, if $X=\prod X_\alpha$ for arbitrary spaces $X_\alpha$, the connected components of $X$ are just the sets of the form $\prod A_\alpha$ where each $A_\alpha$ is a connected component of $X_\alpha$. Indeed, the argument above shows that any connected subset $A\subset X$ is contained in exactly one such set $\prod A_\alpha$ (namely, let $A_\alpha$ be the component of $X_\alpha$ containing the connected set $\pi_\alpha(A)$). On the other hand, a product of connected spaces is connected, so sets of the form $\prod A_\alpha$ are indeed connected. So these are exactly the maximal connected subsets of $X$.