# If $f:X\to Y$ is continuous, then the induced map $F:K(X)\to K(Y)$ is continuous wrt the Vietoris topology

Let $X,Y$ be topological spaces, $f:X\to Y$ a continuous map, and denote by $K(X)$ [resp. $K(Y)$] the set of compact subsets of $X$ [resp. $Y$]. We can endow $K(X)$ with a topology (the Vietoris topology) generated by the sets:

$$\{K\in K(X): K\subset U\}$$

$$\{K\in K(X):K\cap U \neq\emptyset\}$$

where $U\subset X$ ranges on all open subsets of $X$. It is then natural to ask whether the induced map $F:K(X)\to K(Y)$ given by $F(K)=f[K]=\{f(x):x\in K\}$ is continuous. It suffices to see what happens to members of the sub-basis. Fixing $U$ open in $Y$,

• $F^{-1}(\{K:K\subset U\})=\{K:F(K)\subset U\}=\{K:f(K)\subset U\}=\{K:K\subset f^{-1}(U)\}$ which is obviously open in $K(X)$ since $U$ is open in $Y$ and $f$ is continuous.

• $F^{-1}(\{K:K\cap U\neq\emptyset\})=\{K:f(K)\cap U\neq\emptyset\}=\{K:K\cap f^{-1}(U)\neq\emptyset\}$. To see the last equality, note that: $$f(K)\cap U\neq\emptyset \iff \exists y(y\in f(K)\wedge y\in U)\iff \exists x (x\in K\wedge f(x)\in U)$$ and $$K\cap f^{-1}(U)\neq\emptyset \iff \exists x (x\in K\wedge f(x)\in U).$$

Is this correct? In Kechris' Classical Descriptive Set Theory, he leaves this as an exercise, but asks that $X$ and $Y$ be metrizable, which I have not used in the argument above.