# Problem of the Week #280 - Oct 16, 2018

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Let $f : X \to Y$ be a closed map of topological spaces such that the fibers $f^{-1}(y)$ are compact for every $y\in Y$. Prove that $f$ is a proper map, i.e., $f^{-1}(K)$ is a compact subset of $X$ for every compact subset $K$ of $Y$.
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#### Euge

##### MHB Global Moderator
Staff member
This week's problem was solved correctly by Janssens . You can read his solution below.

The given statement is quite natural from the point of view of set-valued analysis. (There was a recent question about that topic, too.)

Let $g : Y \twoheadrightarrow X$ be the inverse image correspondence for $f$,
$$g(y) := \{x \in X\,:\, f(x) = y\}, \qquad y \in Y.$$

(Note that $g$ may be empty-valued if $f$ is not surjective, which we explicitly allow.)

Closedness of $f$ implies (in fact, is equivalent to) upper hemicontinuity of $g$. Moreover, the given compactness of the fibers of $f$ is equivalent to compact-valuedness of $g$. So, $g$ is a compact-valued, upper hemicontinuous correspondence and therefore maps compact subsets of $Y$ to compact subsets of $X$. By definition of $g$, this shows that $f$ itself is proper.

Remarks:

1. In order to avoid ambiguity, I use the definition of upper hemicontinuity from $\S$17.2 of Aliprantis and Border, Infinite Dimensional Analysis, 3rd edition, 2007. This is a beautiful book for mathematicians and economists alike. Accordingly, a correspondence between two topological spaces is upper hemicontinuous if the upper inverse image of any open set is open. It is one of a number of co-existing conventional definitions.

2. The compactness theorem used above is nothing more than the direct analogue of the corresponding (!) theorem for ordinary, continuous functions. In particular, ordinary functions are (as correspondences) singleton-valued, hence compact-valued.

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