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Show $(A^o)^c=\overline{A^c}$.

($\rightarrow$) $(A^o)^c\subseteq\overline{A^c}$

I want to show that $(A^o)^c$ is closed and that $A^c\subseteq (A^o)^c$. Then ($\rightarrow$) follows. Since $A^o$ is open $(A^o)^c$ is closed. Since $A^o\subseteq A$, $A^c\subseteq (A^o)^c$. Done.

($\leftarrow$) $\overline{A^c}\subseteq (A^o)^c$

This one is trickier. $(A^o)^c$ is closed and since $A^o\subseteq A\implies$ $A^c\subseteq (A^o)^c$. So $\overline{A^c}\subseteq (A^o)^c$ since $\overline{A^c}$ is the smallest closed set which $A^c$ fits into?

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Yes: one characterization of $\overline{A}$ is that it is the intersection of all closed sets containing $A$. Thus, if $C$ is any closed set containing $A$, we must have $\overline{A}\subseteq C$. – Brian M. Scott Feb 19 '12 at 8:36
However, you’ve a problem with the first inclusion: $A^c\subseteq \varnothing^c$, and $\varnothing^c$ is closed, but that doesn’t guarantee that $\varnothing^c\subseteq\overline{A^c}$. – Brian M. Scott Feb 19 '12 at 8:45
I see your objection to that part. $(A^o)^c$ could as very well be a larger closed set containing $A^c$ than $\bar{A^c}$. I'm a little stumped here -- I think I need to reformulate that entire part. – Emir Feb 19 '12 at 8:56
up vote 2 down vote accepted

Your second inclusion is fine. $A^o\subseteq A$, so $A^c\subseteq (A^o)^c$, and $A^o$ is open, so $(A^o)^c$ is closed. $\overline{A^o}$ is the intersection of all closed sets containing $A^o$, and $(A^o)^c$ is one of those closed sets, so $\overline{A^c}\subseteq (A^o)^c$.

The argument for the first inclusion, however, is incorrect. The easiest way to show that $(A^o)^c\subseteq\overline{A^c}$ is to chase points: if $x\in(A^o)^c$, then $x\notin A^o$, so every $\epsilon$-ball centred at $x$ intersects $A^c$, and therefore $x\in\overline{A^c}$.

In fact, you really need only that one argument, since each of the implications goes both ways: for any $x$,

$$\begin{align*}x\in(A^o)^c&\iff x\notin A^o\\ &\iff \text{ every }\epsilon\text{-ball about }x\text{ meets }A^c\\ &\iff x\in\overline{A^c}\;, \end{align*}$$

so of course $(A^o)^c=\overline{A^c}$.

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Many thanks! One question -- how does "$\text{ every }\epsilon\text{-ball about }x\text{ meets }A^c$" imply $x\in\bar{A^c}$? – Emir Feb 19 '12 at 9:01
@Emir: That’s one definition of the closure: $\overline{A}=\{x:\text{every }\epsilon\text{-ball about }x\text{ meets }A\}$. If it’s not the definition that you’re using, you should try proving that it’s equivalent to your definition. – Brian M. Scott Feb 19 '12 at 9:06
@Emir: By the way, you can get the bar to extend better if you use \overline{...} instead of \bar{...}. – Brian M. Scott Feb 19 '12 at 9:07

In your antepenultimate question you proved that $$A^{\circ} = A\setminus\partial A.$$ Therefore, $$\begin{align*} A^{\circ} &= A\setminus\partial A\\ A^{\circ} &= A\cap (X-\partial A)\\ (A^{\circ})^c &= (A\cap (X-\partial A))^c\\ (A^{\circ})^c &= (X-A)\cup \partial A\\ (A^{\circ})^c &= A^c \cup \partial(A^c) &&\text{(since }\partial A=\partial A^c\text{)}\\ (A^{\circ})^c &= \overline{(A^c)} &&\text{(since for any }B,\ \overline{B}=B\cup\partial B\text{)} \end{align*}$$

Again the same Moral: Use previous results when possible!

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