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There is indeed a very similar question asked [See Proving closure of S is the smallest closed set containing S. but the content is too advance for a starter in analysis like me and the focus of question is not the same. So I have decided to raise my question here:

Let $X$ be a metric space and $S$ be a subset in $X$. Show that

$\bar S$ is the smallest closed subsets of $X$ which contains $S$.

[ Here, define $\bar S$ as the closure of $S$ in $X$, where the closure is the set containing all the adherent points of $S$ in $X$. We also define $x \in S$ an adherent point of $S$ if $B(x,r) \cap S \neq \emptyset$ for all $r>0$.]

The question is reduced to proving the following:

  1. $\bar S$ is indeed closed in X.
  2. $\bar S$ contains $S$, i.e. $S \subset \bar S$.
  3. $\bar S$ is indeed the smallest in the sense that if $J \subset X$ satisfies 1. and 2., then $\bar S \subset J$.

The first two is rather straightforward and I cannot formulate the proof of the third claim. Here, I would also like to write down my proof for 1. and 2. Feel free to comment on it or provide your proof so that I can learn from you.

Proof of 1.: Consider $X \setminus \bar S$. For any $x \in X \setminus \bar S$, $x \notin \bar S$. It implies that there exists $r_x >0$ such that $B(x,r_x) \cap S = \emptyset$. Now, for any $y \in B(x,r_x)$, choose $r=min \{ d(x,y), r-d(x,y)\}$. Then $B(y,r) \subset B(x,r_x)$. It follows that $B(y,r) \cap S = \emptyset$, proving that $X \setminus \bar S$ is open.

Proof of 2.: It is trivial since for any $x \in S$, $x \in B(x,r) \cap S$ for all $r >0$.

Please tell me your thoughts in proving the third claim. Thanks in advance.

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    $\begingroup$ Do you know that the intersection of two closed sets is s a closed set? If so, if $\overline{S}\not\subseteq J$ then $J\cap \overline{S}$ is a closed set and $S\subseteq J\cap\overline{S} \subsetneq \overline{S}$. Prove that isn't possible. $\endgroup$ Commented Sep 23, 2014 at 14:30
  • $\begingroup$ I have added the definition of $\bar S$. Thanks for reminding. Let me also try Thomas's ideas. $\endgroup$
    – Nighty
    Commented Sep 23, 2014 at 14:32
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    $\begingroup$ Prove that a closed set contains all of its adherent points. Then apply the obvious fact that if $S\subseteq T$, all points adherent to $S$ are also adherent to $T$. $\endgroup$
    – egreg
    Commented Sep 23, 2014 at 14:37

2 Answers 2

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Ad 2.) This follows since any $x\in S$ is adherent to $S$. (We have $B(x,r)\cap S\supset \{x\} \neq \emptyset$ for any $r>0$.)

Ad 1.) Let $x\in X \backslash \bar S$. Then there exists $r_x$ such that $B(x,r_x)\cap S =\emptyset$ (since otherwise $x\in\bar S$). By 2.) we have $S\subset \bar S$ thus $B(x,r)\cap \bar S \supset B(x,r)\cap S = \emptyset$. This proves $B(x,r_x)\subset X\backslash \bar S$ and therefore $X\backslash \bar S$ is open. Hence $\bar S$ is closed.

Ad 3.) Let $J\subset X$ closed such that $S\subset J$. Assume $x\in\bar S\backslash J$. Then there exists $r_x$ such that $B(x,r_x)\cap J = \emptyset$ (since $J$ is closed). By definition of $\bar S$ we have $B(x,r_x)\cap S \neq\emptyset$. We conclude $$\emptyset \neq B(x,r_x)\cap S \subset B(x,r_x)\cap J =\emptyset$$ -- a contradiction. Thus $\bar S\subset J$.

Alternative 3.) The closure operation is monotone thus $S\subset J$ implies $\bar S \subset \bar J=J$.

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Note that $J=\overline{J}$. As $J$ is closed, then $X\setminus J$ is open. Taking a point in $X\setminus J$ there would be an open ball fully contained in $X\setminus J$, hence that point would not be in $\overline{J}$. Then use the simple fact that $\overline{S}\subseteq \overline{J}$, as advised in comments.

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