# Does a generalization of Tietze's extension theorem hold for set-valued functions?

Let $$X$$ be a normal topological space. Tietze's extension theorem says that if $$A \subset X$$ is closed, then a continuous function $$f: A \to \mathbb R^n$$ can be extended to a continuous function whose domain is all of $$X$$. If $$X$$ is assumed to be a metric space, then the theorem holds for functions taking values in any locally convex linear space (see this).

I am wondering if the theorem holds for certain set-valued functions.

In particular, let $$\phi$$ be a function from closed $$A \subset X$$ into compact subsets of $$\mathbb R^n$$. To say that $$\phi$$ is continuous on $$A$$ means that the following conditions are met for every $$x \in A$$:

(1) For every neighborhood $$U$$ of $$\phi(x)$$, there is a neighborhood $$V$$ of $$x$$ such that $$\phi(y) \subset U$$ for all $$y \in V$$;

(2) For every open subset $$U$$ of $$\mathbb R^n$$ for which $$\phi(x) \cap U \neq \emptyset$$, there is a neighborhood of $$V$$ of $$x$$ such that $$\phi(y) \cap U \neq \emptyset$$ for all $$y \in V$$.

Can $$\phi$$ be extended to a continuous set-valued function whose domain is all of $$X$$?

In principle, I don't mind assuming that $$X$$ is actually a subset of $$\mathbb R^m$$.

• Is this equivalent to asking whether $K(\Bbb R^n)$ with the Vietoris topology is an absolute extensor? – Alessandro Codenotti Aug 3 at 21:39
• @AlessandroCodenotti After looking up "absolute extensor", I believe the answer is yes by Martin's comment to this answer. Does an affirmative answer to my question then follow from the result mentioned in the first paragraph? – aduh Aug 3 at 22:01