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 $M$?



Is the following set closed with respect to the Hausdorff metric?
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. 


