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I have the problem:

Let $S \subset \mathbb{R}$ be any set, and define for any $x\in\mathbb{R}$ the distance between $x$ and the set $S$ by:

$$d(x,S) = \inf\{|x-s| : x \in S\}$$

I'm to prove ($\overline{S}$ being the closure of $S$):

If $x\notin \overline{S}$, then $d(x,S) > 0$.

I can understand this from a common sense way but not sure how to prove it.

Martin Sleziak
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johnson
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  • You may want to check the various characterisations of $\overline S$. Is there one about balls intersecting $S$? –  Aug 13 '16 at 17:39
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    "Proving topology questions" is a completely uninformative title. – symplectomorphic Aug 13 '16 at 17:45
  • Closely related: http://math.stackexchange.com/questions/385127/in-a-metric-x-d-prove-that-for-each-subset-a-x-in-bara-if-and-only-i – Martin Sleziak Aug 13 '16 at 17:46
  • Sorry didn't know titles could have symbols as well. Noted for future reference. Thanks for the edits. – johnson Aug 13 '16 at 17:47
  • @MathematicianInWorks Even if you want to avoid symbols, saying something like "A point outside the closure has positive distance" would describe the question much better. In general, it is advised to write descriptive titles, see also meta. – Martin Sleziak Aug 13 '16 at 17:49

1 Answers1

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If $x\notin \overline{S}$ then there is a number $r>0$ such that the open interval $(x-r,x+r)$ doesn't intersect $S$. So any point of $S$ is at distance $\geq r>0$ of $x$, hence $d(x,S)\geq r>0$.

Other idea: (with sequences)

By contrapositive, if $d(x,S)=0$ then by definition of "inf" as greatest lower bound, we see that $\forall n\in \Bbb N,\;n\geq 1,\;\dfrac 1n$ is not a lower bound of $\{ |x-s| : s\in S\}$, so there is $s_n\in S$ s.t. $|x-s_n|<\dfrac 1n$. Thus there is a sequence $(s_n)$ of elements of $S$ converging to $x$ and $x\in\overline{S}$.

paf
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