2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.