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Let $(X, p)$ be a compact space with metric $p$. Let $H(X) = \{A \subset X | A=[A]\}$, $p'= max_{a \in A} min_{b \in B} (p(a, b)) + max_{b \in B} min_{a \in A} ((a, b))$. Prove that $(H(X), p')$ is complete metric space.

I've already understood how to prove that $p'$ is metric. But I have question about how to prove its completeness. I think I should use the theorem about completeness of a metric space (A metric space X is complete if and only if every decreasing sequence of non-empty closed subsets of X, with diameters tending to 0, has a non-empty intersection)

Khan
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  • By $[A]$ do you mean the closure of $A$? – Demophilus Oct 25 '17 at 22:48
  • I think you'll find that for a Cauchy sequence ${A_n}$, the set $L = { x \in X\mid (\exists n)(\forall k > n)(x \in A_k)}$ will serve as a limit. – Paul Sinclair Oct 26 '17 at 01:02
  • In general, a metric space is compact iff it is complete and totally bounded. So this question is a bit superfluous. – Math1000 Oct 26 '17 at 01:30
  • See answers to this question. – Alex Ravsky Oct 26 '17 at 04:44
  • @Demophilus yes – Khan Oct 26 '17 at 19:50
  • @PaulSinclair but how to prove that $L$ is closed? – Khan Oct 26 '17 at 20:11
  • @PaulSinclair i've tried to prove that its closed: $X$ is compact, so L is covered with union of $A_k$, for $k>=0$. So we can get a finite subcover. Finite union of closed sets is closed. Am I missing something? – Khan Oct 26 '17 at 20:29
  • @PaulSinclair and showint that L is limit isn't obvious. $\forall \epsilon > 0,\exists N=N(\epsilon), \forall m, n > N, p(A_m, A_n) < \epsilon$ how do I prove from this that $p(L, A_n) = max_{l \in L} min_{a_n \in A_n} p(l, a_n) + max_{a_n \in A_n} min_{l \in L} p(l, a_n) < \epsilon$ – Khan Oct 26 '17 at 20:48
  • Concerning your argument that $L$ is closed: you are missing 2 things: Compactness only guarantees that covers by open sets have finite subcovers. And, it is $X$ that is known to be compact and so has this property. If we knew $L$ was compact, then there would be nothing more to prove as compactness implies closed in metric spaces. – Paul Sinclair Oct 26 '17 at 23:25
  • And concerning the limit, note that if $l\in L$, then there exists $m > N$ such that $l \in A_m$. – Paul Sinclair Oct 26 '17 at 23:56

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