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Let $f : X → Y$ be a given function, and suppose that $f^{-1}(C)$ is an open subset of $X$ whenever C is an open subset of $Y$ .

(a) Prove that f is continuous on $X$.

(b) Prove that $f^{-1}(B)$ is a closed subset of $X$ whenever B is a closed subset of $Y$

(c) If $Y = R$, and $f$ is continuous, and a $\epsilon$ $R$, what kind of set is A = {$x$ $\epsilon$ X : $f(x)$ <= a}? Justify your answer

I already solved part a, and my attempt for part b is:

$f^{−1}(B)$ = $(f^{−1}(B^c))^c$ ⋯ (1) ($E^c$ denoting the complement of $E$).

So if B is closed, then $B^c$ is open, $f^{−1}{(B^c)}$ is open and its complement is closed. This means $f^{−1}(B)$ is closed by (1).

But I'm finding trouble in solving part c. Any help please?

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    $\begingroup$ Hint: When $B$ is closed, then $Y\setminus B$ is open. $\endgroup$ – Severin Schraven Mar 28 at 9:12
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$f^{-1}(B)=(f^{-1}(B^{c}))^{c}$ $\cdots\, \,\,\, $ (1) ($E^{c}$ denoting the complement of $E$). So if $B$ is closed, then $B^{c}$ is open, $f^{-1}(B^{c})$ is open and its complement is closed. This means $f^{-1}(B)$ is closed by (1).

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  • $\begingroup$ then what? how can this help me? $\endgroup$ – souad bou chahine Mar 28 at 9:36
  • $\begingroup$ @souadbouchahine I have added more details. Remember that a set is closed iff its complement is open. $\endgroup$ – Kavi Rama Murthy Mar 28 at 9:39

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