Upper bound on number of sets in a sub cover? An exercise in Strichartz asks "For which compact sets can you set an upper bound on the number of sets in a sub cover on an open cover?"
This doesn't make sense to me. How could there ever be an upper bound? If we are allowed to take any open cover that we want, then why not choose the open cover $\{ (-n,n): n\in \mathbb{N}\}$. This covers the whole line so it covers any compact subset. But $\{(-2n,2n)\}$ is a proper sub cover that has infinitely many elements. So what does Strichartz mean?
It seems to me that this question would make more sense if it were asking for an upper bound on the sizes of minimal sub covers. 

Well, a one point set is contained in at least one open set in any of its open covers.  So an upper bound of $1$ can be imposed for cardinalities of subcovers of covers of one point sets: "Every open cover of a one point set contains a subcover containing at most one open set."

  Consider $X = [1,n] \subset \Bbb{Z} \subset \Bbb{R}$ using the usual topology on $\Bbb{R}$.  Label each set with the elements of $X$ that it covers.  Any open cover contains at least one open set whose label contains "$1$", at least one open set whose label contains "$2$", ..., and at least one open set whose label contains "$n$".  The collection of these "at least one open set"s is a subcover, covering X, containing at most $n$ open sets.
(It may contain fewer if an open set covering more than one point is chosen for more than one element of its label.  A subcover is not a multiset.)
 The same argument works for any finite set if we use cardinality to find a bijection between elements of $X$ and the labels "$1$", "$2$", ..., "$n$".  Hence every open cover of a set of cardinality $n$ has a subcover containing at most $n$ open sets.

