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.

  • 2
    $\begingroup$ 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. $\endgroup$ Sep 23 '14 at 14:30
  • $\begingroup$ I have added the definition of $\bar S$. Thanks for reminding. Let me also try Thomas's ideas. $\endgroup$
    – Nighty
    Sep 23 '14 at 14:32
  • 2
    $\begingroup$ 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
    Sep 23 '14 at 14:37

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 $\bar S \subset \bar J=J$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.