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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 $X$.

Problem: let $X$ be any subset of the real line, and let $Y$ be a set such that $X \subseteq Y \subseteq \overline{X}$ show that $\overline X=\overline Y$

I would appreciate if you can give me a hint on the proof strategy.

Edit: To show $A \subseteq B => \overline A \subseteq \overline B$:we say $A \subseteq B \subseteq \overline B$. then we say let $b'\in \overline B $, $\exists b \in A \subseteq B s.t. |b'-b|\leq \epsilon$.Hence, $\overline A \subseteq \overline B$. Is this correct? is it well written?

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Hint: try to show that $X \subseteq Y$ implies $\bar{X} \subseteq \bar{Y}$, and that $Y \subseteq \bar{X}$ implies $\bar{Y} \subseteq \bar{X}$.

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  • $\begingroup$ To show $A \subseteq B => \overline A \subseteq \overline B$:we say $A \subseteq B \subseteq \overline B$. then we say let $b'\in \overline B $, $\exists b \in A \subseteq B s.t. |b'-b|\leq \epsilon$.Hence, $\overline A \subseteq \overline B$. Is this correct? is it well written? $\endgroup$ – MAS Nov 11 '14 at 11:32
  • $\begingroup$ any hint how to show $\overline \overline A = \overline A$? $\endgroup$ – MAS Nov 11 '14 at 11:33
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It's enough to prove the following two facts:

  1. Whenever $A \subseteq B$, we have $\overline{A} \subseteq \overline{B}$.

  2. We always have $\overline{\overline{A}} = \overline{A}$.

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  • $\begingroup$ I tried to prove the first but wasnt successful. $\endgroup$ – MAS Nov 11 '14 at 10:12

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