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  • Proposition. Closure of a set is closed (in topology of $\mathbb{R}$
  • Theorem. $X\subseteq\mathbb{R}$ is closed if and only if $X$ contains all its limit points. (The theorem is proved in the class)

Proof-trying of proposition. Let $\overline {X}$ be closure of $X\subseteq\mathbb{R}$. By the definition, $\overline {X}=X\cup X'$ (as $X'$ is set of limit points of $X$). $\overline {X}$ is called closure of $X$. So how can I use the theorem, can you help?

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    $\begingroup$ Actually, you need to show that the limit points of $\overline X$ are all in $\overline X$. You've only shown that the limit points of $X$ are in $\overline X$. @Test123 $\endgroup$ – Thomas Andrews Nov 3 '17 at 16:54
  • $\begingroup$ the key is to prove $(X\cup X')'\subseteq X'$ $\endgroup$ – Jorge Fernández Hidalgo Nov 3 '17 at 16:54
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Suppose $p$ is a limit point of $X\cup X'$. This means that every open set $U$ containing $p$ contains a point $x\in X\cup X'$ other than itself.

If the point $x$ is in $X'$ then we have that $U$ contains a point in $X$ because $x$ is a limit point of $X$.

This proves that $p$ is a limit point of $X$.

So $X\cup X'$ contains all of its limit points.

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  • $\begingroup$ Right, and this is true in any topological space, not just $\mathbb{R}$ with the usual topology. $\endgroup$ – Michael Lee Nov 3 '17 at 18:55
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The answer already posted is correct, but I'd like to post a version with all basic details and all logical possibilities explicitly dealt with.

We are given:

(1) $X$ closed $\Leftrightarrow$ $X'\subset X.$

(2) $\overline{X}=X\cup X'.$

(3) $x\in X' \Rightarrow \forall U\ni x, \exists y\in U, y\neq x, y\in X.$

Prove: $\overline{X}$ is closed, that is, prove that $(\overline{X})'\subset \overline{X}.$

We have that $x\in(\overline{X})' \Rightarrow \forall U \ni x, \exists y\in U, y\neq x, y\in\overline{X}.$

There are two possibilities, either (i) $y\in X$ or (ii) $y\in X'$ since $\overline{X}=X\cup X'.$

(i) If $y\in X$, then

$$U \text{ contains a point in $X$ other than } x \text{, namely $y$}.$$

(ii) If $y\in X'$, then $y$ is a limit point of $X$ and $U$ is a neighborhood of $y$ thus $U\ni z, z \neq y, z \in X.$

There are two further possibilities, either this new point (ii.a) $z=x$ or (ii.b) $z \neq x$.

(ii.a) If $z=x$, then because $z\in X$ we have that

$$x\in X.$$

(ii.b) If $z \neq x,$ then

$$U \text{ contains a point in $X$ other than } x \text{, namely $z$}.$$

So we conclude that for any $x\in(\overline{X})'$ and any neighborhood $U$ of $x$, either $x\in X$ or $U$ contains a point in $X$ other than $x.$ Thus $x\in X$ or $x\in X'$ $\Rightarrow$ $x\in\overline{X}=X\cup X'.$

Since $x$ was originally an arbitrary element in $(\overline{X})'$, and the argument shows that this implies that $x$ is in $\overline X$, we have that $(\overline{X})'\subset \overline{X}.$ Hence $\overline{X}$ is closed by given (1).

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