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Suppose $(X,d)$ is a metric space.

If $A$ is a closed subset of X then ${c(\bar A)} = \overline{c({A})}$ where $c$ is the complement of the set and $A$ is a subset of the metric space.

I think this is false since a counter example is a closed ball in $\mathbb{R}$ with the usual metric. Am I correct?

QUAN
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  • A is already closed? – Tsemo Aristide Mar 17 '16 at 20:07
  • $A \subseteq \overline A$ so $c(\overline A) \subset c(A) \subset \overline {c(A)}$ So... yeah. If A is closed/not open. c(A) is open/not closed a proper subset of its closure of which the compliment of the closure of A is a proper subset. So yes. $A = \overline A = [0,1]$; $c(\overline A) = (-\infty,0)\cup (1,\infty) \ne (-\infty,0]\cup[1,\infty) = \overline {c(A)}$. – fleablood Mar 17 '16 at 20:11

3 Answers3

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You're right.

Note that $A = \overline{A}$ since $A$ is closed. Hence $c(\overline{A}) = c(A).$ Moreover $c(A)$ is open and $\overline{c(A)}$ is closed. Thus, to be true, you need a closed-open subset.

Especially if $X$ is connected, the only $A$ possible are $X$ and $\emptyset$.

C. Dubussy
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Yes, you are correct.

It is noteworthy (and potentially worth proving) that $c(\bar A)$ will always be an open set, whereas $\overline{c(A)}$ will always be closed (regardless of whether $A$ is closed). In fact, we can say that $c(\bar A)$ is the interior of $c(A)$, while $\overline{c(A)}$ is the closure of $c(A)$.

Ben Grossmann
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  • "In fact, we can say that c(A¯) is the interior of c(A), while c(A)¯¯¯¯¯¯¯¯¯¯ is the closure of c(A)." That's a really nice way of putting it. Wish I had thought to express it as such. – fleablood Mar 17 '16 at 20:52
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$A \subseteq \overline A$ (equality holding iff A is closed)

So $\overline A^c \subseteq A^c$ (equality holding iff A is closed)

So $\overline A^c \subseteq A^c \subseteq \overline {A^c}$. The first equality holds iff A is closed. The second equality holds iff $A^c$ is closed iff $A$ is open.

So, unless I did something very wrong, this is only true when A is both open and closed.

Four examples:

i) A = (0,1). $\overline A^c = \overline{(0,1)}^c = [0,1]^c = (-\infty, 0)\cup (1,\infty)$

$\overline {A^c} = \overline {(-\infty, 0]\cup [1,\infty)} = (-\infty, 0]\cup [1,\infty)$

Not equal.

ii) A = [0,1]. $\overline A^c = \overline{[0,1]}^c = [0,1]^c = (-\infty, 0)\cup (1,\infty)$

$\overline {A^c} = \overline {(-\infty, 0)\cup (1,\infty)} = (-\infty, 0]\cup [1,\infty)$

Not equal.

iii) A = (0,1]. $\overline A^c = \overline{(0,1]}^c = [0,1]^c = (-\infty, 0)\cup (1,\infty)$

$\overline {A^c} = \overline {(-\infty, 0)\cup [1,\infty)} = (-\infty, 0]\cup [1,\infty)$

Not equal.

iv) $A = \mathbb R$. $\overline {A^c} = \overline {\emptyset} = \emptyset$

$(\overline A)^c = \mathbb R^c = \emptyset$.

Are equal.

Fifth example. In $\mathbb Q$ the interval $(\sqrt{2}, \sqrt{3})^\* = [\sqrt{2}, \sqrt{3}]\*\*$ is both open and closed and ... yeah, it's easy to see that this equality holds then.

$\overline{(\sqrt{2}, \sqrt{3})^c} = \overline {(-\infty, \sqrt{2})\cup (\sqrt{3}, \infty)} = (-\infty, \sqrt{2})\cup (\sqrt{3}, \infty)$

$\overline{(\sqrt{2}, \sqrt{3})}^c = (\sqrt{2}, \sqrt{3})^c = (-\infty, \sqrt{2})\cup (\sqrt{3}, \infty)$

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But $\overline A^c$ (which is open) $\subseteq \overline {A^c}$ (which is closed).

And $\overline{\overline A^c} = \overline {A^c}$.

$\*$ Yeah, I know. I'm abusing notation because $\sqrt{2}$ and $\sqrt{3}$ are not in the rationals so this notation isn't correct. Sue me.

$\*\*$ Yeah, I really abusing notation. Some days are like that...

fleablood
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