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Definition. A subset of $\mathbb{R}$ is open if it is union of open intervals.

Definition. A subset of $\mathbb{R}$ is closed if it's complement is open.

Definition. $\overline {A}=\bigcap${ $B\subseteq \mathbb{R}$ : $B$ is closed and contains $A$}.

Proposition Let $A\subseteq \mathbb{R}$ and $a\in\mathbb{R}$. Show that $A$ is closed if and only if $A=\overline {A}$.

My proof trying. ($\Rightarrow$) Assume $A$ is closed. We need to show that $A=\overline {A}$. Since $A$ is closed $\mathbb{R} \backslash A$ is open. By the third definition, there is a $B$ such that $B$ is closed and contains $A$. So, what should I do?... Can you help?

2 Answers 2

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It's in fact irrelevant how the topology was defined, all that matters is that it obeys the axioms of a topology. These axioms imply (using de Morgan) that arbitrary intersections of closed sets are closed.

So when we define $$\overline {A}=\bigcap\{ B\subseteq \mathbb{R} : B \text{ is closed and contains } A\}\text{,}$$

the latter intersection is well-defined (we have at least $\mathbb{R}$ as one of the sets we intersect) and closed by the intersection property. Also as we take an intersection of sets that all contain $A$, $A \subseteq \overline{A}$

So $\overline{A}$ is a closed set. So if $A = \overline{A}$ then $A$ is a closed set as $\overline{A}$ is closed by its definition. And if $A$ is closed, we can take $A$ to be one of the $B$ that we take the intersection of (this is ok , because we intersect all closed sets that contain $A$ and now $A$ is one of them), so $\overline{A} \subseteq A$ (as the intersecion of a family is contained in any of its members), and so $A = \overline{A}$ in that case.

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Hints:

$\;A\;$ is closed iff it is one of the subsets in $\;\{B\subset\Bbb R\;/\;B\;\text{is closed and contains}\;A\}...$

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    Namely, $A$ is closed. Is the $\mathbb{R} \backslash A\equiv\;\{B\subset\Bbb R\;/\;B\;\text{is closed and contains}\;A\}$ true? – 2017-01-15
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    @Kahler No, of course not. Unless you're using notation I can't understand, for me $\;\Bbb R\setminus A\;$ is the complement of $\;A\;$ in the reals, thus: it is the set of all elements *not in* $\;A\;$ .This cannot obviously be the right side you wrote... – 2017-01-15
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    Yes, you are right. I wrote false. I'm editing, $A$ is closed if it's complement, $\mathbb{R} \backslash A$ is open. Also, $A$ is closed iff it is one of the subsets in $\;\{B\subset\Bbb R\;/\;B\;\text{is closed and contains}\;A\}$. So, how can I use your hint? – 2017-01-15
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    Well, observe that (1) since $\;A\subset B\;$ for any subset $\;B\;$ in that set, and (2) since $\;A\;$ is **one of those** subsets $\;B\;$ , their intersection is simply $\;A\;$ ... and now: what do you know about **any** intersection of closed sets? – 2017-01-15
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    İt is closed, ,is it? – 2017-01-15
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    @Kahler Yes, it is. – 2017-01-15