§ Proof of Heine Borel from Munkres (compact iff closed, bounded)

We wish to show that compact iff closed and bounded in R\mathbb R.

§ (1) Closed intervals are compact in the topology of a complete total order.

Let us work with a complete total order TT (in our case T=RT = \mathbb R). We equip TT with the order topology (which matches the usual topology on R\mathbb R). Let [l,r][l, r] be a closed interval. Let {Ui}\{ U_i \} be an open cover of [l,r][l, r]. Let MM (for middle) be the set of points mm such that [l,m][l, m] has a finite cover using {Ui}\{ U_i \} That is,
M{m[l,r]:[l,m] has finite cover} M \equiv \{ m \in [l, r] : [l, m] \text{ has finite cover} \}
  • CLAIM 1: lub(M)Mlub(M) \in M.
  • Pick an open lub(M)V{Ui}lub(M) \in V \in \{ U_i \}.
  • As VV is open, there is some (cV)<(lub(M)V)(c \in V) < (lub(M) \in V).
  • if cMc \in M: the finite cover of [l,c][l, c] along with VV give a finite cover for lub(M)lub(M). Thus lub(M)Mlub(M) \in M.
  • if c∉Mc \not \in M, then c<lub(M)c < lub(M) and cc is an upper bound for MM, which is absurd.
  • Thus, lub(M)Mlub(M) \in M.
  • CLAIM 2: lub(M)=rlub(M) = r. This implies that rMr \in M, and [l,r][l, r] has a finite subcover using UiU_i.
  • For contradiction of Claim 2, assume that lub(M)rlub(M) \neq r.
  • Pick some open set OO in the cover that contains lub(M)lub(M): lub(M)O{Ui}ilub(M) \in O \in \{ U_i \}_i.
  • As OO is open, OO contains some point cc (for contradiction) that is after lub(M)lub(M): lub(M)<club(M) < c.
  • Rewriting: cOc>lub(M)c \in O \land c > lub(M). So, the interval [l,c][l, c] has the same cover as [l,lub(M)][l, lub(M)].
  • Hence, [l,c][l, c] has a finite cover, thus cMc \in M.
  • CONTRADICTION: cMc>lub(M)c \in M \land c > lub(M), which is absurd. We would have c=lub(M)c = lub(M).
  • Thus, this means that lub(M)=rlub(M) = r, and thus the entire interval [l,r][l, r] has finite subcover.

§ (2) Closed subset of a compact set is compact

Let BB be a closed subset of a compact space KK. Take an open cover {Ui}\{ U_i \} of BB. See that {Bc,Ui}\{ B^c, U_i \} is a cover of the full space KK, and hence has a finite subcover. This subcover will be of the form {Bc,Uj}\{B^c, U_j\}. The {Uj}\{U_j\} are a finite subcover of BB. Thus, BB 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 SS be a compact subset of a haussdorf space XX.
  • For any point q∉Sq \not \in S, we need to show the existence of an open set qQq \in Q such that SQ=S \cap Q = \emptyset.
  • For each point sSs \in S, use Haussdorf to find separating sets sO(s;q)s \in O(s; q), qO(q;s)q \in O(q; s) such that O(s;q)Q(q;s)=O(s; q) \cap Q(q; s) = \emptyset.
  • See that the sets {O(s;q):sS}\{ O(s; q) : s \in S \} are a cover of SS.
  • Extract a finite subcover of this, say {O(si;q):sS}\{ O(s_i; q) : s \in S \}.
  • Use this finite subcover to separate qq from SS.
  • Now, pick the open set Q{O(q;si)}Q \equiv \cap \{ O(q; s_i) \}, which is open since it's a finite intersection.
  • See that this QQ separates qq from SS.
  • We have that QO(si,q)=Q \cap O(s_i, q) = \emptyset for each O(si,q)O(s_i, q).
  • Thus,Q(O(si;q))=Q \cap (\cup O(s_i; q)) = \emptyset, and thus QS=Q \cap S = \emptyset as SO(si;q)S \subseteq \cup O(s_i; q).
  • if q∉Sq \not \in S, we have an open QQ that separates qq from SS,thus qq is NOT A LIMIT --- not every open nbhd of QQ has non-empty intersection with SS.
  • Contrapositive: All limit points of SS are in SS. Thus, SS is closed. (4)

§ (4) A set with all limit points is closed (complement of open)

Let SS be a set that has all its limit points. Consider the complement set TT. We will show that TT is open.
  • all points in TT: have open that separates them from SS. Union of all of these opens is TT.TT open: infinite union of opens. SS: the complement of an open set, closed.
More elaborately:
  • for all all tTt \in T, tnSt \not in S (by defn).
  • tt is not a limit point of SS (SS has all limit points).
  • Thus, there is an open UtU_t such that tUtt \in U_t and UtS=U_t \cap S = \emptyset.
  • Define: TtTUtT' \equiv \cup_{t \in T} U_t. Claim: T=TT' = T.
  • As UtU_t contains no elements of SS, TT' contains no element of SS.
  • As UtU_t contains tt, TT' contains all tt.
  • Thus TT' contains all tTt \in T, and no element of SS. So TT' is a complement of SS. T=TT' = T.
  • TT is a infinite union of opens. Thus TT is open.
  • SS is complement of open set TT. SS is closed.