Is a compact union of compact spaces still compact?

Question

I'm wondering if the following statement is true :

Let $ K $ be a compact space, $ X $ a topological space and $ \forall x $ in $K$, $L(x)$ a compact subspace of $ X $. Then $ \cup_{x \in K} L(x) $ is a compact subspace of X.

I understand that the statement is false if $K$ is not supposed to be compact and true if $K$ is finite. (two cases that are already discussed in other topics).

If it is false, can it become true when adding some hypothesis ? In particular I'm interested in the case where L would be continuous. But to be accurate this means to define a topology on the set of all subspaces (or compact subspaces) of X...

Does someone have a proof of this statement (eventually in a more specific case : $X$ a metric space or a normed vector space) or a counterexample ?

Answer 1

Your conjecture is false.

Let K = X = [0,1]. For all x in K, let L(x) = {x} if x is rational and {0} otherwise.

K is compact and every L(x) is a compact subset of X.

$ \cup_{x \in K} L(x) $ is the set of rationals in [0,1] which is not compact.

To make your conjecture true, require K to be finite.
Notice that there are countable K counter examples.

For L(x) to be continuous, the Hausdorff metric could be useful.

Answer 2

Having a union indexed by $K$ is not going to work because the topology on $K$ plays no actual role in the statement of the conjecture (since $K$ is just an index set, then if the conjecture were true for any compact set, it would be true for any infinite set at all).

But instead of just taking a union, you could ask the following question: given a map $Y \to K$ such that all of the fibers are compact, is $Y$ itself compact? This is a bit more like really asking for a "compact family of compact sets." Unfortunately, it is still false, since e.g. $Y$ could be a discrete uncountable space mapping bijectively to the interval.

Answer 3

The result becomes true as soon as the map $L$ is upper semicontinuous. The precise statement would be the following:

Let $X,Y$ be complete metric spaces, and $L : X \rightrightarrows Y$ an upper semi-continuous set-valued application with nonempty compact images. Then for any compact set $K \subset X$, the union \begin{align*} U := \bigcup_{x \in K} L(x) \end{align*} is compact in $Y$.

Indeed, let $(y_n)_{n \in \mathbb{N}} \subset U$, and $x_n \in K$ be such that $y_n \in L(x_n)$. By compactness of $K$, some (non relabeled) subsequence of $(x_n)_{n \in \mathbb{N}}$ converges towards some $x \in K$. Let us show that the set \begin{align*} V := L(x) \cup \bigcup_{n \in \mathbb{N}} L(x_n) \end{align*} is compact. Let $\left(\mathcal{O}_{\alpha}\right)_{\alpha \in A}$ be any open cover of $V$. In particular, it is an open cover of $L(x)$, and we may extract a finite subcover $\left(\mathcal{O}_{\alpha}\right)_{\alpha \in A_{x}}$, with $A_x \subset A$ finite. As the union $\mathcal{Q} := \bigcup_{\alpha \in A_{x}} \mathcal{O}_{\alpha}$ is an open neighbourhood of $L(x)$, by upper semicontinuity of $L$, there exists $\overline{n} \in \mathbb{N}$ large enough so that $L(x_n) \subset \mathcal{Q}$ for all $n \in \mathbb{N}$. Taking a concatenation of finite open covers of $L(x_n)$ for $0 \leqslant n < \overline{n}$ and $\left(\mathcal{O}_{\alpha}\right)_{\alpha \in A_x}$, we get a finite open cover of $V$. Consequently, the subsequence $(y_n)_{n \in \mathbb{N}} \subset V$ admits a limit point, that belongs to $L(x) \subset U$ by upper semicontinuity.

If any knowledgy passerby is aware of a reference to this result in a textbook, I would gladly hear of it.