§ Proof of Heine Borel from Munkres (compact iff closed, bounded)
We wish to show that compact iff closed and bounded in .
- Let be closed and bounded. We wish to show that is compact. Since is bounded, is contained in some large closed interval . (1) Closed intervals are compact in the order topology of a complete total order, and thus is compact ( is a complete total order). Then (2) is a closed subset of a compact set , and is thus compact. Hence closed and bounded implies compact.
- Let be compact. We wish to show that is closed and bounded. Cover with open balls of increasing radii. This is an open cover; Extract a finite subcover. From the finite subcover, pick the largest open ball . is entirely contained in , and is thus bounded. (3) is closed, as it is a compact subset of a Haussdorff space. Hence, we are done.
§ (1) Closed intervals are compact in the topology of a complete total order.
Let us work with a complete total order (in our case ). We equip
with the order topology (which matches the usual topology on ).
Let be a closed interval. Let be an open cover of .
Let (for middle) be the set of points such that has a finite cover using
- CLAIM 1: .
- Pick an open .
- As is open, there is some .
- if : the finite cover of along with give a finite cover for . Thus .
- if , then and is an upper bound for , which is absurd.
- Thus, .
- CLAIM 2: . This implies that , and has a finite subcover using .
- For contradiction of Claim 2, assume that .
- Pick some open set in the cover that contains : .
- As is open, contains some point (for contradiction) that is after : .
- Rewriting: . So, the interval has the same cover as .
- Hence, has a finite cover, thus .
- CONTRADICTION: , which is absurd. We would have .
- Thus, this means that , and thus the entire interval has finite subcover.
§ (2) Closed subset of a compact set is compact
Let be a closed subset of a compact space . Take an open cover of .
See that is a cover of the full space , and hence has a finite subcover.
This subcover will be of the form . The are a finite subcover of .
Thus, is compact as we have extracted a finite subcover of an open cover.
§ (3) Compact subset of Haussdorf space is closed
Morally, this is true because in a Haussdorff space, single point subsets are closed.
Compactness pushes this local property to a global property --- The entire compact set
itself becomes closed.
- Let be a compact subset of a haussdorf space .
- For any point , we need to show the existence of an open set such that .
- For each point , use Haussdorf to find separating sets , such that .
- See that the sets are a cover of .
- Extract a finite subcover of this, say .
- Use this finite subcover to separate from .
- Now, pick the open set , which is open since it's a finite intersection.
- See that this separates from .
- We have that for each .
- Thus, , and thus as .
- if , we have an open that separates from , thus is NOT A LIMIT --- not every open nbhd of has non-empty intersection with .
- Contrapositive: All limit points of are in . Thus, is closed. (4)
§ (4) A set with all limit points is closed (complement of open)
Let be a set that has all its limit points. Consider the complement set . We will
show that is open.
- all points in : have open that separates them from . Union of all of these opens is . open: infinite union of opens. : the complement of an open set, closed.
- for all all , (by defn).
- is not a limit point of ( has all limit points).
- Thus, there is an open such that and .
- Define: . Claim: .
- As contains no elements of , contains no element of .
- As contains , contains all .
- Thus contains all , and no element of . So is a complement of . .
- is a infinite union of opens. Thus is open.
- is complement of open set . is closed.