Let f be a real valued function defined and continuous on a closed set S in R. Prove that A is a closed set if
A = { x ∈ S : f(x) = 0}
What happens if S is not a closed set?
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1Consider $f : (0, 1) \to \mathbb{R}$ such that $f$ is constantly $0$. Then $A = (0, 1)$, which is open in $\mathbb{R}$. Or, if we replace $(0, 1)$ with $[0, 1)$, then $A = [0, 1)$ which is neither open nor closed. We could even replace $(0, 1)$ with $\mathbb{Q}$... you get the idea. – Theo Bendit Oct 29 '18 at 06:57
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In short, if $S$ is not a closed set, then a set which is closed in $S$ is not necessarily closed in $\mathbb R$. – Ivan Neretin Oct 29 '18 at 08:05