0
$\begingroup$

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 $Y = \mathbb{R}$, and $f$ is continuous, and $a \in \mathbb{R}$, what kind of set is $A = \{x \in X : f(x) \leq a\}$? 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?

$\endgroup$
  • $\begingroup$ Observe that $A=f^{-1}((-\infty,a])$ and that $(-\infty,a]$ is closed in $R$. Then you will need the converse of (a). $\endgroup$ – user647486 Apr 6 at 11:41
  • $\begingroup$ Your part B is good. Maybe state explicitly that $f^{-1}(B^c)^c = f^{-1}(B)$, but yes. $\endgroup$ – Kaj Hansen Apr 6 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.