So, I have the following problem :

Let $X$ be a topological space. Show that a bijection $f:X \to X$ is a homeomorphism iff $f(\bar A)=\overline{f(A)}$.

And I have got an online solution here. Now, I understand all the steps of the solution.

(There are typing errors in the last two statements. They should be $\forall H\subseteq S_2$ and not $S_1$ I think)

However I don't understand how the final conclusion proves the statement. Any kind of help will be highly appreciated. Thank you.


I think you are right about the typos. The last statement implies \begin{equation} f^{-1}(\overline H) = \overline{f^{-1}(H)} \end{equation} since you can take $f$ on the other side to get \begin{equation} \overline{f^{-1}(H)}\subseteq f^{-1}(\overline H) \end{equation} and you already had the other inclusion by continuity. Then you have shown the claim for $f^{-1}$, which is the same as showing it for $f$: just set $H=f(H')$, which is something you can do for every set $H'$, since $f$ is a bijection.

I don't know if it's also clear to you how the converse statement follows. Essentially, you should write the equality as a double inclusion and use this Continuity defined by closure the page refers to. If you want, I can elaborate on this point. :-)

  • $\begingroup$ Ok thanks, this part is clear. Now for the converse, aren't all the steps if and only if? As you said after breaking the equality into two inclusions, aren't all the steps the same as we did here in reverse? $\endgroup$ – User Not Found May 19 '16 at 9:38
  • $\begingroup$ yes! you go back through the same path $\endgroup$ – nelv May 19 '16 at 9:39
  • $\begingroup$ Ok Thanks a lot for your effort. $\endgroup$ – User Not Found May 19 '16 at 9:43

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