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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 $\mathbb R^m$.

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    $\begingroup$ Is this equivalent to asking whether $K(\Bbb R^n)$ with the Vietoris topology is an absolute extensor? $\endgroup$ – Alessandro Codenotti Aug 3 at 21:39
  • $\begingroup$ @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? $\endgroup$ – aduh Aug 3 at 22:01

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