Let $X$ be separable and completely metrizable. Let the hyperspace of compact sets be denoted $H(X)$.

I want to show that $H(X)^2 \rightarrow H(X)$ defined by $(H,G)\rightarrow H \cup G$ is continuous.

I need to begin with open (or closed) balls in $H(X)$ and show that their preimage in $H(X)^2$ is open (or closed). This will show that my function is continuous.

I know the hyperspace it is the set of compact subsets of X with the Vietoris topology. I know what the Vietoris topology is and the basis for it. I know that the Hausdorff metric is compatible with the Vietoris topology. I know that since $X$ is Polish then $H(X)$ is Polish, but I don't know how that will help me.

Would I start my proof with:

Let our function be defined as above and let $X$ be Polish.

Let $H \in H(X).$ Then $H$ is nonempty closed compact subset of $X$. Let $\epsilon >0$. Let $O$ be an open set in $H(X)$. Since $H \in H(X)$ and $H(X)$ is generated by sets $\{ H \in K(X) | K \subset U \}$ and $\{ H \in K(X) | K \cap U \neq \emptyset \}$ for $U$ open in $X$.

Since $O$ is open in $X$, then $\{ H \in K(X) | K \subset O \}$ and $\{ H \in K(X) | K \cap O \neq \emptyset \}$. - BUT I can't actually say this because $O$ is an open set in $K(X)$ not in $X$. I start with an open set in $K(X)$ because I'm trying to show the function is continuous. What exactly does $O$ look like and where do I go from here? How do I show the preimage of $O$ is open in $H(X)^2$? How do I use the mapping they gave me? Should I approach this by showing the preimage of closed sets is closed, instead?


1 Answer 1


Since you already know that the space is metrizable and the Hausdorff metric gives the topology you want, why not just use the metric? Writing $d$ for the Hausdorff metric, we have $$ d(H\cup G, H'\cup G') \le \max(d(H, H'), d(G,G')) \tag{1} $$ which gives not only continuity, but Lipschitz continuity. The proof of $(1)$ goes like this: let $\rho$ be the right hand side. Then $H'$ is contained in $\rho$-neighborhood of $H$, and $G'$ is contained in the $\rho$-neighborhood of $G$. Express this as $H'\subset H_\rho$ and $G'\subset G_\rho$.

It is straightforward to prove that $(H\cup G)_\rho=H_\rho\cup G_\rho$. Thus, $H'\cup G'\subset (H\cup G)_\rho$.

Similarly in the other direction: $ (H\cup G)_\rho\subset H'\cup G' $. This proves $d(H\cup G, H'\cup G')\le\rho$ as claimed.


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