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$\def\bdy{\operatorname{bdy}}\def\interior{\operatorname{int}}$(1) Prove that a function from a metric space X into the metric space Y is continuous if and only if for each $A \subset X , f(\bar A) \subset \overline {f(A)} .$

(2) Prove that if $f$ is a one to one mapping from the metric space X into the metric space Y, show that $f$ is a homeomorphism if and only if, for each $A \subset X, f(\bar A)=\overline {f(A)}.$

Attempt at one. $\bar A=\interior(A) \cup \bdy A$. So x $\in A$ then $f(x) \in f(\bar A)$ also $f(x) \in \overline {f(A)}$. If $ x \in \bdy A$ then $f(x) \in f(\bar A)$ like wise $f(x) \in \overline {f(A)}$. But $f(\bar A)=f(A) \cup f(\bdy A) $ $f(\bar A) $maybe a open set where $\overline {f(A)}$ is a closed set containing the $\bdy f(A) \cup f(A).$ Now $\bdy f(\bar A) \not= \bdy\overline {f(A)}$ because $f(\bar A)$may be an open set in Y. Where as $\overline {f(A)}$ is closed so it contains all boundary points. Do not really have a clue how to do the converse.

Attempt at 2: Since $f$ is one to one and a homeomorphism we know that for every $f(a) \in f( A) $ are distinct so we need only to show that $\bdy f(A)=f(\bdy A).$ Not quite sure how do prove that and also not sure how to prove converse.

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    That $\bdy$ notation makes me feel weird. I'm used to $\partial A$ for the boundary of $A$.2012-11-26

2 Answers 2

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(1) I can't quite follow what you're trying to do here. I would suggest instead looking at the sequential characterization of continuity for metric spaces. That is a function $f: X \rightarrow Y$ is continuous if and only if for every sequence $x_n \rightarrow x$ we have $f(x_n) \rightarrow f(x)$.

(2) This question needs a bit of tweaking. Either the assumptions should include that $f$ is onto or the problem should be to show that $f$ is an embedding. Regardless the idea here is to use apply (1) to $f$ and $f^{-1}$ to get your desired result.

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  1. If $f$ is continuous then $f(\bar A)\subset\overline{f(A)}$ for all $A\subset X$: Remember that if $B$ is any set then $x\in\bar B$ if and only if for every open nbhd $U$ of $x$, $U\cap B\ne\emptyset$. So, let $y\in f(\bar A)$. To show that $y\in\overline{f(A)}$, let $V$ be an open nbhd of $y$. Now there exists some $x\in\bar A$ such that $y=f(x)$ and, since $f$ is continuous, $f^{-1}(V)$ is an open nbhd of $x$. Can you finish from here?
  2. If $f(\bar A)\subset\overline{f(A)}$ for every $A\subset X$ then $f$ is continuous: To show that $f$ is continuous, we only need to show that the preimage of every closed set is closed. So, let $F\subset Y$ be closed. Now it's enough to show that $\overline{f^{-1}(F)}\subset f^{-1}(F)$. From the given property of $f$ we get that $f\left(\overline{f^{-1}(F)}\right)\subset\overline{f(f^{-1}(F))}\subset\overline F=F.$ Therefore, since $B\subset f^{-1}(f(B))$ for all $B$, we have $\overline{f^{-1}(F)}\subset f^{-1}\left(f\left(\overline{f^{-1}(F)}\right)\right)\subset f^{-1}(F).$
  3. If $f$ is one to one and onto, and $f(\bar A)=\overline{f(A)}$ for every $A\subset X$, then $f$ is a homeomorphism: Now $f(\bar A)\subset\overline{f(A)}$ for every $A\subset X$ so, by (2), $f$ is continuous. On the other hand if $B\subset Y$ then using the fact that $f$ is one to one and onto, $\bar B=\overline{f(f^{-1}(B))}=f\left(\overline{f^{-1}(B)}\right)$ and hence $f^{-1}(\bar B)=\overline{f^{-1}(B)}.$ By (2) the function $f^{-1}$ is also continuous.
  4. If $f$ is a homeomorphism then $f(\bar A)=\overline{f(A)}$ for every $A\subset X$: The idea here is similar to (3) but this time use (1) instead of (2).

As a side note, this works for all topological spaces, not just metric ones.

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    Your (4) looks fine. Notice that you used the continuity of $f^{-1}$ to get $f^{-1}\left(\overline{f(A)}\right)\subset\bar A$.2012-11-27