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Let $X$ be a metric space, $x \in X$ and $S \subset X$. Then,

I have to prove that $x \in Cl(S)$ if and only if, every open ball of $X$ centered at $x$ has non-empty intersection with $S$.

I managed to do the first half of the proof but got stuck with the converse of it.

I checked my book for the converse part, it was proved as follows :

If $dist(x,S)$ is not equal to zero, then either $S$ is empty or

$b(x;dist(x,S)) \cap S$ is empty. I don't really follow this last part.

Up till " either $S$ is empty" it seems alright, but not the part after it.

Any help would be appreciated.

Also my book defines dist(x,S)=inf{d(x,s):s∈S} and Cl(S) = {$x \in X$ : dist(x,S)=0}.

johny
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  • "if and only if, every open ball of $X$ centered at $x$ has non-empty intersection with $S$." That's usually pretty much the definition of closure. What definition are you working with? – Daniel Fischer Oct 11 '13 at 12:26
  • @DanielFischer Well its given as a theorem in my book, so i have to prove it ! – johny Oct 11 '13 at 12:32
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    The definition is usually that statement with "open ball" replaced with "open set" and "centred at" replaced with "containing". It seems to be a weaker condition to require it for just open balls. The theorem johny's proving is that in fact it's not. – Callus - Reinstate Monica Oct 11 '13 at 12:34
  • I understand that. What I don't know is how the book defines the closure of a set, so I don't know how a proof should look like. It seems the book defines it in terms of the distance from the set, so is a) $\operatorname{dist}(x,S) = \inf { d(x,s) : s \in S}$, and b) $\operatorname{Cl}(S) = {x : \operatorname{dist}(x,S) = 0}$? – Daniel Fischer Oct 11 '13 at 12:35
  • Or it might be defined globally, as the smallest closed set containing it. johny, the way one would go about proving the proposition depends on the way one chooses to define the notions involved. – Jonathan Y. Oct 11 '13 at 12:36
  • @DanielFischer yes they are defined like that. – johny Oct 11 '13 at 12:40
  • Good. Now one last thing to clarify, is $\operatorname{dist}(x,\varnothing) = \infty$, or is it defined in some other way? (Or is it left undefined?) – Daniel Fischer Oct 11 '13 at 12:43
  • @DanielFischer It is defined as dist(x,∅)=∞. – johny Oct 11 '13 at 12:44

1 Answers1

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With these definitions, we have to show

$$\operatorname{dist}(x,S) = 0 \iff \bigl(\forall r > 0\bigr) \bigl(S \cap B_r(x) \neq \varnothing\bigr).$$

In the one direction, we have $0 = \operatorname{dist}(x,S) = \inf \{ d(x,s) : s \in S\}$, so that means for all $\varepsilon > 0$ there is an $s_{\varepsilon} \in S$ with $d(x,s_\varepsilon) < \varepsilon$ by the definition of $\inf$. But $d(x,s_\varepsilon) < \varepsilon \iff s_\varepsilon \in S\cap B_\varepsilon(x)$, so every ball centered at $x$ intersects $S$.

In the other direction, we have an $s_r \in S \cap B_{r}(x)$ for all $r > 0$. But $s_r \in S \cap B_r(x)$ means $d(x,s_r) < r$ and hence $\operatorname{dist}(x,S) \leqslant d(x,s_r) < r$, so we have $0 \leqslant \operatorname{dist}(x,S) \leqslant \inf \{ r \in \mathbb{R} : r > 0\} = 0$.

Regarding the converse part of the proof from the book, that shows the contrapositive

$$x \notin \operatorname{Cl}(S) \Rightarrow \bigl(\exists r > 0\bigr)\bigl(S\cap B_r(x) = \varnothing\bigr).$$

If $R := \operatorname{dist}(x,S) > 0$, then either $S$ is empty and $R = \infty$, or $S\neq\varnothing$ and $R < \infty$. In the first case, every ball centered at $x$ has empty intersection with $S$, for example $B_1(x) \cap S = B_1(x)\cap\varnothing = \varnothing$. In the second case, the assertion is that $B_R(x) \cap S = \varnothing$. For if there were an $s \in B_R(x)\cap S$, then $\operatorname{dist}(x,S) \leqslant d(x,s) < R$, contradicting the definition $R = \operatorname{dist}(x,S)$.

Daniel Fischer
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    @Fischer Thank you! got it now. I don't know why they can't write proofs like this one in books ! – johny Oct 11 '13 at 13:16