Prove that the set of finite sets with bounded cardinal is closed with the hausdorff topology.

Solution 1:

Choose $A \in H \setminus H_s$. Then $A$ has at least $s+1$ distinct points, $\{ a_1, \ldots, a_{s+1} \}$. Let $d$ be the minimum distance between any two of these points and let $\varepsilon = \frac d3 \gt 0$. Then the triangle inequality tells us the $\varepsilon$-balls $B_i=B(a_i, \varepsilon) \subseteq E$ are mutually disjoint.

Choose $K \in H_s$. Then because the $B_i$ are mutually disjoint, no point of $K$ can be in more than one $B_i$ so by the pigeonhole principle, for some $i, K \cap B_i = \varnothing$. This shows that $\forall K \in H_s ~(h(K, A) \geq \phi(A, K) \geq d(a_i, K) \gt \varepsilon)$; in other words, $\{ B \in H \mid h(A, B) \lt \varepsilon \} \cap H_s = \varnothing$, or $\{ B \in H \mid h(A, B) \lt \varepsilon \} \subseteq H \setminus H_s$.

This shows that $A$ is in the interior of $H \setminus H_s$. But $A$ was an arbitrary element of $H \setminus H_s$, so $H \setminus H_s \subseteq H$ is open and $H_s \subseteq H$ is closed.