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Question: If $P\subseteq \mathbb{R}^n$, how to show that $\overline{\overline{P}}=\overline{P}$, i.e. the closure of $\overline{P}$ equals the closure of $P$.

I know that in a vector space with a Euclidean norm $\|\cdot \|$ a point $x$ is said to be the closure point of $P$ $(P\subseteq \mathbb{R}^n)$ if for all $\epsilon \gt 0$, there exists a point $p \in P$ such that $\|x-p\| \lt \epsilon$. The set of all closure points of P is called closure of P, denoted by $\overline{P}$.

But how to prove the above statement?

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If you're defining $\overline{P}$ as the intersection of all closed sets containing $P$, the result is immediate since $\overline{P}$ is closed.

Or maybe you're defining $\overline{P} = P \cup P'$ where $P'$ is the set of limit points of $P$. In this case, it suffices to show that any point $x \in \mathbb{R}^n \setminus \overline{P}$ is not a limit point of $\overline{P}$. By definition, we can find an open disk $N$ such that $x \in N \subseteq \mathbb{R}^n \setminus P$. Now if a point $p$ of $P'$ lies in $N$, we can find a smaller disk $N'$ around $p$ such that $N' \subseteq N$, but then $N$ would intersect $P$. Hence, $N \cap \overline{P} = \emptyset$, and therefore, $\overline {\overline{P}} = \overline{P}$

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The result comes from characterizing the closure as the smallest close set $C$ such that $P\subset C$ in particular if $K\subset \mathbb{R}^n$ is a close set, then $\overline{K}=K$

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  • $\begingroup$ Sorry, I didn't get you. Could you please elaborate a little. $\endgroup$ – Mathy Jun 23 '13 at 18:27
  • $\begingroup$ The other two answers sumarize what I tried to explain briefly. $\endgroup$ – DavidBecharaSenior Jun 23 '13 at 19:01
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If $x \in \overline{\overline{P}}$, then for each $\epsilon > 0$ there's an $y \in \overline{P}$ such that $\|y - x\| < \frac{\epsilon}{2}$. Since $y \in \overline{P}$, there is a $z \in P$ such that $\|z - y\| < \frac{\epsilon}{2}$. So by the triangle inequality $$ \|z - x\| \leq \|z-y\| + \|y - x\|$$ $$ < \epsilon$$ Since this holds for arbitrary $x \in \overline{\overline{P}}$ and arbitrary $\epsilon > 0$, we conclude $\overline{\overline{P}} \subset \overline{P}$. Since every set is a subset its closure, the opposite inclusion automatically holds and we have $\overline{\overline{P}} = \overline{P}$.

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