Let $(X,d)$ be a non-empty complete metric space, let M be the set of all non-empty compact subsets equipped with the Hausdorff metric, and $N$ be a positive integer. Is $$ \{A\subset X : 1\le \# A \le N \} $$ a closed subset of $M$?
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2$\begingroup$ This sounds like a homework question. $\endgroup$– Yuzhou GuCommented Jun 23, 2016 at 14:41
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1$\begingroup$ No, I just was stupid... I have seen this statement in an article and was wondering how to prove this.... $\endgroup$– Martinus MaximusCommented Jun 23, 2016 at 16:06
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1 Answer
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We need to prove that $\{A\subseteq X:\#A>N\}$ is open. Let $d_H$ denote the Hausdorff metric. We need to prove that for any nonempty compact $A$ with $\#A>N$, there exists $\epsilon>0$ such that any nonempty compact $B$ with $d_H(A, B)<\epsilon$, we have $\#B>N$.
Assume for the sake of contrary that there exists $N$ and $A$ that violates the condition above. Then for any $\epsilon>0$, there exists $B_\epsilon$ with $\#B_\epsilon\le N$ and $d_H(A, B_\epsilon)<\epsilon$. This implies that $A$ can be covered with no more than $N$ open balls of radius $\epsilon$.
Pick $\epsilon$ so small such that we can take $N+1$ points in $A$ such that the pairwise distance between them is $\ge 2\epsilon$. Then $A$ cannot be covered with $N$ open balls of radius $\epsilon$. Contradiction.
