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suppose I have a closed set $[a,b]$, and a continuous function $f$ defined on that set. How do I go about showing that the intervals on which $f$ is non-zero are countable? I think: since $f$ is continuous, around every single point x for which f(x) is non-zero, we can get an open interval (say the largest one possible), and since the $[a,b]$ is closed, there has to be a countable number (but I don't know why).

Is this right? Is there an easier way to do this?

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    Indeed, the specifics of your question aren't important here: Any set of disjoint intervals (of positive length) in $\mathbb{R}$ is countable for the same reason.2011-10-27

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Your approach is correct, and it’s probably the simplest one. To finish it off, let $\mathscr{J}$ be your collection of maximal open intervals. Every open interval contains a rational number, so for each $J\in\mathscr{J}$ let $q_J\in J$ be rational. Since distinct members of $\mathscr{J}$ are disjoint, the mapping $\mathscr{J}\to\mathbb{Q}:J\mapsto q_J$ is an injection (one-to-one), and therefore $|\mathscr{J}|\le |\mathbb{Q}|$. Since $\mathbb{Q}$ is countable, this implies that $\mathscr{J}$ is countable.

For completeness I’ll include a rigorous way to get the intervals in the first place. Define an equivalence relation $\sim$ on $[a,b]$ by setting $x\sim y$ iff either $x\le y$ and $f(z)\ne 0$ for all $z\in[x,y]$, or $y\le x$ and $f(z)\ne 0$ for all $z\in[y,x]$. It’s easy to check that this is an equivalence relation whose equivalence classes are intervals, and the continuity of $f$ implies that these intervals are open.