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I want to prove that if I have two topological spaces $X$ and $Y$, with $A \subset X$, and a homeomorphism $f : X \to Y$, then $f(\partial A) = \partial \big(f(A)\big)$.

I saw a proof here: https://math.stackexchange.com/a/3502105/1053101 (the point (2)). But I don't understand. I mean.. why we can find $a, b \in X$ with $a \in A \cap f^{-1}(V)$ and $b \in A^c \cap f^{-1}(V)$ ? And why is $a,b \ne x$ if $f$ is bijective?

Or, if you could help me with another proof it will be nice. I searched in textbooks but I didn't find something useful. Thanks!

ProofSeeker
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  • if $A$ is closed then $\partial A=A\setminus A^{\circ }$. Then the result follows from the shown identity $f(A^\circ )=f(A)^\circ $ in the linked answer –  Apr 18 '24 at 15:04
  • But from where results that $A$ is closed? – ProofSeeker Apr 18 '24 at 15:10
  • And I still don't understand if it is possible to find $a,b \in X$. I think at $X = {1,2}$ with topology $\tau = {\emptyset, {1}, {2}, {1, 2}}$. How can I find 2 elements different from $x$? – ProofSeeker Apr 18 '24 at 15:18
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    forget $a,b\in X$. For any function it holds that $f(A\cup B)=f(A)\cup f(B)$. Now let $C$ closed, then $f(C)$ is closed (because $f$ have a continuous inverse) and if you already knows that $f(C^\circ )=f(C)^\circ$ then $\partial f(C)\cup f(C)^\circ=f(C)=f(\partial C\cup C^\circ)=f(\partial C)\cup f(C^\circ)=f(\partial C)\cup f(C)^\circ$, so we have that $f(\partial C)\subset \partial f(C)$ as $\partial A\cap A^\circ =\emptyset$ for any chosen set $A$. Now let $D:=f(C)$ and repeat the argument but using this time $f^{-1}$, then you find that $\partial f(C)\subset f(\partial C)$.∎ –  Apr 18 '24 at 15:32
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    (continue...) For general $A\subset X$, note that from the statement $f(A^\circ)=f(A)^\circ$ if follows easily taking complements that $\overline{f(A)}=f(\overline{A})$, from where it follows using the result from the previous comment that $\partial f(A)=f(\partial A)$. –  Apr 18 '24 at 15:48
  • @Masacroso thanks! I think I understand now! So a one more question regarding my initial proof. So that is wrong, correct? – ProofSeeker Apr 18 '24 at 16:02

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