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is my proof correct?

Definition:

Let $X\subset\mathbb R$ and let $x'\in\mathbb R$, we say that $x'$ is an adherent point of $X$ iff $\forall\epsilon>0\exists x\in X \text{ s.t. }d(x′,x)≤ε$. the closure of X is denoted as $\overline X$ and is defined to be the set of all the adherent points of $X$.

show: $\overline{\overline X} = \overline X$

suppose $\exists z \in \overline{\overline X}~~and~~z \notin \overline X$

then, $\exists x' \in \overline X ~~s.t.~~ |z-x'|\leq \epsilon$

$|z-x'|\leq \frac{\epsilon}{2}$

but we also know that $\exists x \in X s.t. |x'-x|\leq \frac{\epsilon}{2}$

hence, $z-x+x-x' \leq \epsilon/2$

$z-x\leq0$

hence z is an adherent point of X ($z\in \overline X$). but this is a contradiction with the above condition on $z$ hence, $\overline{\overline X} = \overline X$

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  • $\begingroup$ I guess you have a typo in the proof. ($z-x+x-x' \leq \epsilon/2$)? $\endgroup$ – user99914 Nov 11 '14 at 12:26
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An other way to prove: $\overline{\bar X}$ is the smallest close set that contain $\bar X$. But $\bar X$ is close then the smallest close set that contain $\bar X$ is $\bar X$, therefore $$\overline{\bar X}=\bar X$$

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  • $\begingroup$ this is the way to prove it. dealing with epsilon is overkill for this $\endgroup$ – nicolas Nov 11 '14 at 13:07
  • $\begingroup$ closed ${}{}{}{}$ $\endgroup$ – Pedro Tamaroff Nov 11 '14 at 14:46
  • $\begingroup$ I am not familiar yet with the concept of closed sets. $\endgroup$ – MAS Nov 11 '14 at 15:03
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    $\begingroup$ I'm sorry to say that this is a bad answer. It uses definitions the original questions does not, thus demanding knowledge the original author does not have. $\endgroup$ – 5xum Nov 11 '14 at 15:04
  • $\begingroup$ I just proposed an other proof... i didn't say that I answered to the question... $\endgroup$ – idm Nov 11 '14 at 19:49
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From $z\notin \overline{X}$, you cannot conclude that there exists such an $x'$ that $|z-x'|\leq \epsilon.$

Since $z\in \overline X$ is equivalent to $$\forall \epsilon \exists x \in X: d(z, x)\leq \epsilon$$

the statement $z\notin \overline X$ is equivalent to $$\exists \epsilon:\forall x\in X: d(z,x)>\epsilon,$$ which is not what you wrote.

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  • $\begingroup$ From $z \in \overline{\overline X}$, I concluded that $\exists x' \in \overline X ~~s.t.~~ |z-x'|\leq \epsilon$. $\endgroup$ – MAS Nov 11 '14 at 15:01
  • $\begingroup$ Yes, but how does that contradict the statement that I wrote? $\endgroup$ – 5xum Nov 11 '14 at 15:03
  • $\begingroup$ @MAS My best hint would be this: Cut the contradiction. Simply take any $x\in\overline{\overline X}$ and show that it IS an element of $\overline X$ by definition. That is, take an $\epsilon$ and find $z\in X$ such that $d(z,x)<\epsilon$. $\endgroup$ – 5xum Nov 11 '14 at 15:05
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The closure of a set is by definition the intersection of all closed sets containing the set. But any intersection of closed sets is closed, so the closure is closed, and closure of a closed set is itself.

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