For which compact sets can the size of the finite subcover be bounded? I've been struggling to find a solution to this problem:
For which compact sets can you set an upper bound on the number of sets in a subcover of an open cover.  
My understanding is that I need to show that there are compact sets that cannot have infinite subcovers. This is different from showing that every open cover of a compact set has a finite subcover. Please let me know if I am interpreting this problem correctly and I'd appreciate any help coming up with an answer.  Thanks!  
 A: By definition (of compactness), given any compact space $X$, any open cover of $X$ has a finite subcover. In general, however, one cannot say before one has an open cover how many open sets one might need to make a subcover.
For example, for any $\epsilon > 0$, the set of $\epsilon$-balls in $[0, 1]$ is an open cover, and so admits some finite subcover. One needs $\approx \frac{1}{2 \epsilon}$ of these balls to cover the entire interval, and this number can be made arbitrarily large my making $\epsilon$ sufficiently small. So, even though we can always construct finite open subcovers of $[0, 1]$, we cannot say a priori how many subsets we'll need.
The question is asking for a condition on the compact set $X$ for which this issue doesn't occur, that is, for which you can say before you get an open cover of $X$ (at most) how many open sets you'll need to make a subcover.
A: Let $X$ be a compact Hausdorff space. Assume that there are $k$ distinct points, $x_i,i=1,2,\ldots,k$, in $X$. By Hausdorfness we can find open sets $U_i\subset X, i=1,2,\ldots,k$, such that $x_i\in U_j$ if and only if $i=j$.
Consider the closed set $K=X\setminus \bigcup_{i=1}^k U_i$. It is closed and intersects trivially with the closed set $N=\{x_1,x_2,\ldots,x_k\}$. A compact Hausdorff space is normal, so we can find non-intersecting open set $U,V$ such that $K\subset U$ and $N\subset V$.
We see that the sets $U,U_1,U_2,\ldots,U_k$ cover $X$. Furthermore, the point $x_i$ belongs to $U_i$ but does not belong to any other set in this cover. Therefore any subcover of this cover will contain at least $k$ sets.
A corollary to this is the following
Observation. If $X$ is an infinite compact subset of a Hausdorff space, then we cannot bound from above the number of sets needed in a subcover.
Caveat. I don't know what happens, if we drop the Hausdorff axiom.
A: Any $T_1$ space $X$ (points are closed) such that every open cover have a subcover of cardinality bounded by a fixed $N\in\mathbb{N}$ has at most $N$ points. Indeed, if we suppose that $X$ contains $N+1$ different points, say $x_0,\ldots,x_n$, then the open cover
$$
\bigl\{X\setminus\{x_0,\ldots,\widehat{x}_i,\ldots,x_n\}:0\leq i\leq n\bigr\},
$$
where $\widehat{x}_i$ means that the element $x_i$ must be omitted from the list, has $N+1$ members and no proper subcover.
For a general topological space $X$, I think that an equivalent condition to the above strong compactness property is that it has at most $N$ minimal closed sets (non-empty closed sets with no non-empty proper closed subsets).
Hope this will be useful.
