If $X$ is disconnected, then $\prod_{f \in \mathcal{F}} I_f$ is disconnected

Question

I'm currently trying to prove a statement about the relationship between the connectedness of $X$ and the connectedness of $\beta(X)$. But the nature of this question regards a specific detail.

Let $X$ be a Tychonoff space and let $\mathcal{F} = \{f:X \to I_f\}$ be the set of continuous functions where $I_f$ is a compact interval of $\mathbb{R}$.

I was told that if $X$ is disconnected, then $\displaystyle\prod_{f \in \mathcal{F}} I_f$ is disconnected. But this result isn't so clear to me. After all, continuity preserves connectedness, but this isn't necessarily true for disconnectedness.

Can anyone help elucidate how this implication is true?

Answer 1

It's not true. Any product of connected spaces is connected.

You may be able to do something using $\displaystyle\prod_{f \in \mathcal{F}} f(X)$ instead. This will be disconnected if any $f(X)$ is, and of course many will be if $X$ is not connected.