Given $ (X, d)$ a metric space, $ A, B \subset X$, show that $ d(A, B)=d (\overline {A}, B) $.
I'm not being able to show that $ d(A,B) \leq d (\overline {A}, B) $. Can anybody help me? The set $\overline {A}$ is the closure of $A$
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Joaquin Liniado
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What's $\overline{A}$? – May 03 '15 at 22:59
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I think $\overline{A}$ is the closure of $A$. – Salomo May 03 '15 at 23:01
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Shouldn't it be $A\subseteq X$ then? – May 03 '15 at 23:38
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I got the \in wrong! – Joaquin Liniado May 04 '15 at 00:37
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See also: Distance between two sets in a metric space is equal to the distance between their closures – Martin Sleziak Dec 29 '23 at 11:42
2 Answers
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That is the trivial direction. The distance is an infimum, so if you take an infimum over a larger set, then the infimum can never increase. More precisely, $d(C,B)\le d(A,B)$ for all $A,B,C\subseteq X$ with $A\subseteq C$. To prove that simply note simply look at the sets over which the infima are taken, and note which is contained in which.
Ittay Weiss
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Hint:
Use the triangle inequality to show that for any $\varepsilon>0$, $$d(\overline A,B)\le d(A,B)\le d(\overline A,B)+\varepsilon.$$
Bernard
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I was wondering, to show that $ d(A, B) \leq d (\overline {A}, B)$ wouldn't it be necessary to show $d (A, B) < d (\overline{A}, B) + \varepsilon $? – Joaquin Liniado May 04 '15 at 15:16
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Still not able to solve it. When you say try using the triangle inequality, you mean, given $ A, B, C \subset X $ then $ d (A, B) \leq d (A, C) + d (C, B) $? Im not sure that statement is true. If you didn't mean that, then I don't know what triangle inequality to use. – Joaquin Liniado May 04 '15 at 20:29
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I meant the ordinary triangle inequality, with elements of $X$, then taking the infimum, &c. – Bernard May 04 '15 at 20:56
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Still not able to solve it, I have difficulties when taking the infimum on both sides of the inequality, since I should apply "taking infimum of a certain set" to both sides. And I don't know how to proceed with that. – Joaquin Liniado May 05 '15 at 20:14