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I have to prove that these statements are equivalents:

(i) $f: X \rightarrow Y$ is continuous

(ii) $f(A') \subset \overline {f(A)} , \forall A\subset X$

(iii) $Fr(f^{-1} (B)) \subset f^ {-1} (Fr(B)) , \forall B \subset Y$

I could only show (i) implies (ii).

I don't know what I'm missing to show the rest.

$Fr$ is boundary. I didn't find anything that relates to boundary.

Hints are much appreciated

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What does $A'$ mean here? In part (ii), I would think $\bar{A}$ to be more appropriate –  Will Jagy Jun 30 '13 at 19:11
    
@Will: Probably the derived set: all the limit points of $A$. –  Asaf Karagila Jun 30 '13 at 19:11
    
@will yes, it is the set of all limit points. –  Charlie Jun 30 '13 at 19:17
    
@AsafKaragila, alright. I'll look it up. Yes, Munkres Topology A First Course, page 97, Theorem 6.6. Boundary is on page 101, Exercise 18. And we are not assuming Hausdorff. –  Will Jagy Jun 30 '13 at 19:21
    
@Cameron Buie, youtube.com/watch?v=qQkBeOisNM0 –  Will Jagy Jun 30 '13 at 19:24
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1 Answer 1

up vote 2 down vote accepted

To see that (3) implies (1), take a closed set $B$ in $Y$, and note that closedness is equivalent to $B=\overline B$ and that $\overline B=B\cup\partial B=B\cup B'$. Then $f^{-1}(B)=f^{-1}(B)\cup f^{-1}(\partial B)$. By (3) this contains $\partial f^{-1}(B)$. Can you go on from here?

For (2) implies (1), note that $f(A')\subseteq\overline{f(A)}$ is equivalent to $f(\overline{A})\subseteq\overline{f(A)}$. Now let $B=\text{cl}B$. Deduce that $B\supseteq\text{cl}f(f^{-1}(B))\supseteq f(\text{cl}f^{-1}(B))$. Conlude that $f^{-1}(B)$ is closed.

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Thank you, I believe I can proceed. –  Charlie Jun 30 '13 at 19:25
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