# Show that $f(\bar A) \subset \overline{f(A)}$.

Let $X$ be a metric space, and $Y\subset X$ a subset. A point $x\in X$ is adherent to $Y$ if $B(x;r) \cap Y \neq \emptyset$ $\forall r > 0.$ The closure of $Y$ is then defined as $\bar Y := \{x\in X \mid x \text{ is adherent to } Y\}$.

I am trying to prove the next statement. Is there any mistake or error?

$f$ is a homeomorphism. Show that $f(\bar A) \subset \overline{f(A)}$.

Since $f$ is continuous $x\in \bar A \implies f(x)\in f(\bar A)$. (Is this right? or trivial?) I want to show $f(x)\in f(\bar A) \implies f(x)\in \overline{f(A)}$. Consider any open ball $V = B(f(x);r)$. Since $f$ is continuous, $f^{-1}(V)$ is an open subset of $X$ and $x\in f^{-1}(V)$.

Because $x\in \bar A$, there exists an element in the intersection of $A$ and any open set. i.e. $f^{-1}(V)\cap A \neq \emptyset$. Take $y\in f^{-1}(V)\cap A$, then $$f(y)\in f(f^{-1}(V)\cap A) \subset V\cap f(A).$$ From this, we see that $V\cap f(A)\neq \emptyset$ for any open ball $V=B(f(x);r)$. By the definition of closure, $f(x)\in \overline{f(A)}$.

• "$f(\bar A) \implies \overline{f(A)}$"? – Pedro Tamaroff Apr 17 '14 at 1:08
• The line $f(\overline{A})\Rightarrow \overline{f(A)}$ does not make mathematical sense. Are you trying to show that, for a continuous map $f$ of metric spaces, that $f(\overline{A})\subset \overline{f(A)}$? Nobody can confirm if your solution is correct or not until you tell us what exactly the problem is that you are solving! – roo Apr 17 '14 at 1:08
• Yes I omit that mistakely, Sorry. – jakeoung Apr 17 '14 at 2:50
• – Martin Sleziak Apr 17 '14 at 6:48

The proof is fine. I just want to point a few things:

• "Since $f$ is continuous $x\in \bar A \implies f(x)\in f(\bar A)$. (Is this right? or trivial?)" Recall the definition $f(\overline{A}) =\{f(x) \in Y \mid x \in \overline{A}\}$, so this step is actually trivial.

• It's overkill asking that $f$ is a homeomorphism. We only need $f$ to be continuous.

• Assuming $f$ continuous, the reciprocal is actually true, too.

• Do notice that you didn't really used the fact that $V$ was an open ball. This, in fact, is not important - the argument holds with $V$ being any open set containing $x$.

$\bar A$=A+A' Where A' is derived set. f($\bar A$)=f($A\cup A'$)=f(A)$\cup f(A')$......f(A$\cup B$)=f(A)$\cup$f(B)
f(A) $\subset$ $\overline {f(A)}$
We have to show that f(A')$\subset$ $\overline {f(A)}$
consider x is limit point of A i.e x$\in A'$ So there exist $x_n$ in A which conveges to x .Now by continuty f($x_n$) converges to f(x),That means f(x)is limit point of f(A).i.e.f(x)$\in$$\overline {f(A)}. f(A')\subset$$\overline {f(A)}$ that means f($\bar A$)$\subset\overline {f(A)}$