Section 26: Compact Spaces

1. A compact space $X$ is a space such that every open covering of $X$ contains a finite covering of $X$ .
2. If a space is compact in a finer topology then it is compact in a coarser one.
3. If a space is compact in a finer topology and Hausdorff in a coarser one then the topologies are the same.
4. Take a compact Hausdorff space. Any strictly coarser topology is not Hausdorff. Any strictly finer topology is not compact.
5. For any set there is a topology such that it is a compact Hausdorff space.
6. It is not true in general that for a compact non-Hausdorff topology there is a finer topology that is compact and Hausdorff; as well as, it is not true in general that for a non-compact Hausdorff topology there is a coarser topology that is compact and Hausdorff.
7. The finite intersection property: a collection of subsets is said to have the finite intersection property if every finite subcollection have a non-empty intersection.
8. A space is compact iff every collection of closed subsets having the finite intersection property has a nonempty intersection.
9. Nested sequence: if there is a nested sequence of closed nonempty subsets $A_i\supseteq A_{i+1}$ of a compact space then it has a nonempty intersection.
10. Subspaces and compactness:
11. A subspace $Y\subseteq X$ is compact iff every covering by sets open in $X$ contains a finite sub-covering.
12. A closed subspace of a compact space is compact.
13. A compact subspace of $X$ need not be closed even if $X$ is compact: $\{1-1/n\}$ is not closed in $[0,1]\subset\mathbb{R}_{+\infty}$ (the topology is coarser than the standard topology, hence, since 1 is a limit point in the standard topology, it is a limit point in the given topology as well) but both are compact (the only open set containing 0 is the whole space). Another example is the cofinite topology on an infinite set: only finite subsets (and the whole space) are closed, but all subspaces (including the whole space) are compact.
14. A compact subspace of a Hausdorff space is closed.
15. Moreover, two compact subsets of a Hausdorff space can be separated by two open neighborhoods.
16. A compact subspace of a metric space is closed and bounded.
17. The finite union of compact subspaces is compact.
18. The intersection of a nested sequence of connected closed subsets of a compact Hausdorff space is connected.
19. We need the sequence to be nested: consider a unit square as a subspace of the standard plane and two disjoint points in it, and connect them by two different paths.
20. Do we need (a) the subsets to be closed and (b) compactness for the non-emptiness of the intersection only? No. They are essential. We use the properties (a) and (b) in the proof several times: the intersection is closed (a) and, therefore, compact (b), a separation consists of two nonempty closed (a) and compact (b) sets (and can be separated by two disjoint neighborhoods), and every closed and connected subset in the sequence minus two disjoint neighborhoods of the separation is non-empty and closed (a), they are still nested, thus, their intersection is non-empty (b). Let $f(x)=x(1-x)$ . Then $A_n=(0,0)\cup(1,0)\cup\{0<y<f(x)/n|0<x<1\}$ is a nested sequence of not closed but connected subsets of the compact and Hausdorff unit square with the disconnected intersection $(0,0)\cup(1,0)$ . Similarly, $B_n=(0,0)\cup(1,0)\cup\{y>=nf(x)|0<x<1\}$ are connected, nested and closed in $[0,1]x\{0\}\cup(0,1)\times \mathbb{R}+$ (which is Hausdorff but not compact), but their intersection is the same discrete disconnected two-point set.
21. What about the Hausdorff property? It seems that the only place we use the property in the proof is to find two disjoint open sets U and V in the space that each contains a set of a separation A and B of the intersection. What if we cannot find such neighborhoods? Consider the unit interval with double 0, it is compact and even connected (we don’t really need that), but not Hausdorff, though, it is T1. The sequence of subsets $x\le 1/n$ (including both 0’s) is a nested sequence of connected closed subsets with the intersection containing two 0’s only, a discrete disconnected subspace. The problem is exactly that the two 0’s have no disjoint neighborhoods.
22. Continuous functions and compactness:
23. The image of a compact set under a continuous function is compact.
24. If $f:$ compact$\to$ Hausdorff-space is continuous then it is a closed map.
25. If $f:$ compact$\to$ Hausdorff-space is bijective and continuous then it is a homeomorphism.
26. If $f:X\to Y$ is continuous and closed then the preimage of any compact set is compact iff the preimage of any point is compact.
27. If $f:$ compact$\to$ Hausdorff-space is continuous then the preimage of any compact set is compact.
28. $f:X\to Y$ where $Y$ is compact and Hausdorff is continuous iff its graph in $X\times Y$ is closed.
29. Perfect maps: a closed continuous surjective function such that the preimage of every point is compact.
30. If $f:X\to Y$ is a perfect map and $Y$ is compact, so is $X$ .
31. Product and compactness:
32. The product of finitely many compacts is compact.
33. The product of any family of spaces is compact $\Leftrightarrow$ all of them are compact.
34. $\Rightarrow$ is immediate, $\Leftarrow$ will be studied in §37.
35. The tube lemma:
36. Suppose $X$ and $Y$ are arbitrary spaces and $B\subseteq Y$ is compact, then if $U$ open in the product contains $x\times B$ then it contains $U_x\times U_B$ for some neighborhood $U_x$ of $x$ and $U_B$ of $B$ .
37. If $A,B$ are compact subspaces of arbitrary spaces $X,Y$ and an open set $U$ contains $A\times B$ then $U$ contains a basis element $U_A\times U_B$ that contains $A\times B$ .
38. If $Y$ is compact then $\pi_1:X\times Y\to X$ is a closed map.
39. A partial converse to the uniform limit theorem:
40. If a monotone increasing sequence of continuous functions on a compact domain converge (pointwise) to a continuous function then the convergence is uniform.