If f is differentiable with a continuous derivative function, then the set of critical points of f is closed.
Is this a true statement? I'm kinda lost.
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Well, the critical points of $f$ are the set of $x$ such that $f'(x) = 0$. So this follows from the more general fact that if $g$ is a continuous function, $Z_g = \{x : g(x) = 0\}$ is closed. Let's use the limit point definition of closed sets: $D$ is closed if $x_n$ is a convergent sequence, its limit is also in $D$. So pick a sequence $(x_n) \subset Z_g$; then $g(x_n)$ = 0; and because $g$ is continuous, $g(\lim x_n) = \lim g(x_n) = 0$. So $\lim x_n \in Z_g$, and $Z_g$ is closed.
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Thank you very much! Your explanation is pretty clear. Just, shouldn't the $f$s in the last two lines be $g$? – user149131 Dec 17 '14 at 05:13
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Suppose $f'(x) \neq 0$. Since $f'$ is continuous, there is some open set $U$ containing $x$ such that $f'(y) \neq 0$ for $y \in U$. Hence the set $\{x | f'(x) \neq 0 \}$ is open, and so $\{x | f'(x) = 0 \} = \{x | f'(x) \neq 0 \}^c$ is closed.
copper.hat
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