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Let $C(X)$ denote the ring of all continuous real-valued functions on a completely regular Hausdorff space $X$.

Let $e:X\longrightarrow \mathbb R$ be an idempotent function in $C(X)$ (i.e $e^2=e$)

Suppose that $Z(e)=\{x\in X| e(x)=0\}$. I need to to show $Z(e)$ is clopen.

Clear it is a closed set. But why is it open?

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    $Z(e) = \{x\in X \mid e(x)\neq 1 \}$. – 2017-01-25
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    @Slade I don't get . Sorry – 2017-01-25
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    @J.Doe X can be any space. – 2017-01-25
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    $Z$ is the preimage of an open set. – 2017-01-26

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For all $x\in X$, $e(x)^2 = e(x)\rightarrow e(x)(e(x) -1) = 0$, so $e(x) = 0$ or $e(x) = 1$.

So $X = \{x: e(x) = 0\} \cup \{x: e(x) =1\} = e^{-1}[\{0\}] \cup e^{-1}[\{1\}]$.

Both these sets are closed (as inverse images of closed sets under a continuous $e$) and disjoint, so each other's complement. So both are open as well. The first set is just $Z(e)$.