Heine-Borel Theorem
Recall (from Study Guide 2) If F, K ⊆ X and if F is closed, and K is compact, then F ∩ K is compact.
Theorem (Nested closed intervals in R are nonempty.) Let · · · ⊆ I3 ⊆ I2 ⊆ I1 be a sequence of nested closed intervals in R. Let In = [ab, bn]. If m > n, then it follows by construction an ≤ am < bm ≤ bn. Then
∞
\
n=1
In 6= ∅.
Proof Let x = sup {ai|i ∈ N}. Note that any bn is an upper-bound of {ai}. Therefore x ∈ R exists. So x ≤ bn for all n. Also note that an≤ x for all n. Therefore x ∈ In for all n.
Remark Same idea works for k-cells in Rk. Note that a k-cell is of the form:
[a1, b1] × [a2, b2] × · · · × [ak, bk].
Fun Fact Alternate proof that R is uncountable. Suppose R = {x1, x2, x3, . . . }. Let In = R \ {x1, x2, . . . , xn}. Then · · · ⊆ I3 ⊆ I2 ⊆ I1. So {In}. If In is closed. Therefore:
∞
\
n=1
In 6= ∅.
Then there exists z ∈T In, hence a contradiction.
Theorem Closed intervals in R are compact. (So are k-cells in Rk.)
Proof Let [a, b] ⊆ R. Suppose [a, b] is not compact. Then there exists an open cover {Gα} with no finite subcover. Let c1 = a + b
2 . Then {Gα} covers [a, c1] and [c, b]. At least one has no finite subcover. Without loss of generality, it is I1 = [a, c1]. Let c2 = a + c1
2 . Then [a, c2] of [c2, c1] has no finite subcover. Call it I2.
Repeat to obtain · · · ⊆ I3 ⊆ I2 ⊆ I1. Notice these are nested closed intervals with no finite subcover of {Gα}. Notice also that |In| = 2|In+1|. Then there exists a point x ∈ ∩In. But x ∈ [a, b]. Thus there exists an α such that x ∈ Gα.
Notice there exists an r > 0 such that B(x, r) ⊆ Gα. For n large enough, In ⊆ B(x, r).
But In has no finite subcover, so it can’t be contained in any finite subcollection of {Gα}. Hence a contradiction.
Definition The set K ⊆ X is boundedif there exists r > 0 and q ∈ X, such that for all p ∈ K, d(p, q) < r.
Heine-Borel Theorem In R (or Rk), K is compact if and only if K is closed and bounded.
Proof Let p ∈ K. Then K ⊆ \
n∈N
B(p, n), because this covers all of X. Assume K is compact.
Then there exists a finite subcover. Therefore, K ⊆ B(p, n1) ∪ · · · ∪ B(p, nl). Thus K ⊆ B(p, r) where r = max {n1, . . . , nl}. Thus K is bounded. Furthermore, we’ve already shown that if K is compact, then K is closed.
Conversely, suppose K is closed and bounded. Then there exists r > 0 such that K ⊆ [−r, r].
Furthermore, [−r, r] is compact. Since K is closed subset of a compact set, K is compact.
Corollary Let K ⊆ R. If K is compact, then sup K exists and sup K ∈ K.
Proof K is bound, hence sup K ∈ R. K is closed, hence K0 ⊆ K. Furthermore, observe that sup K ∈ K0 because it is a limit point.
Example Let E ⊆ Q, E = {p ∈ Q | 2 < p2 < 3}. In R, is E closed? Is it bounded?
Example Let A be any set. Suppose A is infinite. Define
d(p, q) =
(0 if p = q, 1 otherwise.
Is A ⊆ A closed? Is A bounded. Yes on both accounts. Notice that B(p,12) = {p}. Thus:
A = [
p∈A
B(p,12).
Notice that this cannot be reduced to a finite subcover. Hence, A is not compact.
Theorems (from Study Guide 2) Let X be a metric space and K be a compact subset of X.
Then
(a) K is closed, i.e. Kc= X \ K is open.
(b) K is bounded, i.e. K ⊂ Nr(p) ⊂ X for some r > 0 and p ∈ X.
Remarks Let X = (Q, d) be a metric space with d(p, q) = |p − q|, and let K = {p ∈ Q | 2 < p2 < 3}.
(a) Show that X is not complete.
(b) Show that K is closed and bounded in Q, but K is not compact.
(c) Show that K is open.
Theorems Let X be a metric space.
(a) Let {Kα} be a collection of compact subsets of X such that the intersection of every finite subcollection of {Kα} is nonempty. Then\
α
Kα 6= ∅.
(b) Nested Sequence Theorem Let {Kn} be a (nested) sequence of nonempty compact subsets of X such that ∀n ∈ N , Kn ⊇ Kn+1. Then
∞
\
n=1
Kn6= ∅.
(c) Let X = Rk and {In = [an,1, bn,1] × · · · × [an,k, bn,k]} be a sequence of k−cells such that
∀ n ∈ N In ⊇ In+1, then
∞
\
n=1
In 6= ∅.
Remark The intersection \
α
Kα may not be a subset of X. Consider X = (Q, d) as a metric space with d(p, q) = |p − q|, and Kn = [pn− 1
10n, pn+ 1
10n], where pn is the decimal expansion of √
2 to the nth decimal place. Note that
(a) The intersection of every finite subcollection of {Kn} is nonempty.
(b)
∞
\
n=1
Kn= {√
2} 6⊂ Q = X.
Theorem Let E be a subset of the Euclidean metric space Rk. Then the following statements are equivalent.
(a) E is closed and bounded.
(b) E is compact.
(c) Every infinite subset of E has a limit point in E.
Theorem (Weierstrass) Every bounded infinite subset of Rk has a limit point in Rk.
Definition Let X be a metric space.
(a) Two subsets A and B of X are separated if A ∩ ¯B = ∅ and B ∩ ¯A = ∅.
(b) A set E ⊂ X is connected if E is not a union of two nonempty separated sets, i.e. @ A 6=
∅ , B 6= ∅ such that E = A ∪ B , A ∩ ¯B = ∅ and B ∩ ¯A = ∅.
(c) A subset E ⊂ R is connected iff it has the following property: If x, y ∈ E and x < z < y, then z ∈ E, i.e. E ⊂ R is connected iff E is an interval.
Bolzano-Weierstrass Theorem
Theorem A metric space X is compact if and only if every infinite subset E of X has a limit point in X (note that it is not necessarily in E).
Remark Think of sequences and limit points.
Proof
(⇒) Assume X is compact. If no point of X is a limit point of E, then each q ∈ E has a
neighborhood Vq with no other points of E. Notice then that the collection {Vq} where q ∈ X is an open cover of X since no point of X is a limit point of E. But by construction, each Vq contains at most one point of E and |E| = ∞. Thus this set has no finite subcover and hence a contradiction since we assumed X was compact.
(⇐) (Specifically in Rk, the general proof is HW # 26) Assume every infinite subset E of X has a limit point in X. We will show X is closed and bounded and thus our result will follow from Heine-Borel. Suppose X is not closed. Then there exists a point z 6∈ X but z is a limit point of X. Choose E = {xn: |xn− z| < 1
n}. Note that E ⊆ X because z is a limit point. Then E is infinite, so it has a limit point, but has a limit point at z and no other. This is because if there were another limit point y ∈ X:
|xn− y| ≥ |z − y| − |xn− z| ≥ |z − y| − 1 n ≥ 1
2(z − y).
for large n. But this is a contradiction because we’ve assumed infinite subsets of X have limit points in X. Suppose X is not bounded. Choose yn such that |yn| > n. Notice that {yn} is an infinite set. But this has no limit point since the points are getting further and further from
Bolzano-Weierstrauss Theorem Every infinite bounded subset of Rk has a limit point in Rk. Proof If E ⊆ Rk is infinite and bounded, then E is a subset of a k-cell in Rk. Thus E is an infinite subset of a compact set, and E is bounded. Thus E has a limit point in Rk and specifically in the k-cell.
Example (The Cantor Set) Let K0 be [0, 1]. Let K1 be K0 minus the middle third. Let K2 be K1 minus the middle third of the two interval of K1, etc. The cantor set C is
C =
∞
\
i=1
Ki.
Properties of the Cantor Set:
• C is nonempty since it is an intersection of nested closed intervals which we showed to be nonempty.
• C is closed because arbitrary intersections of closed set is closed.
• C is perfect: closed and every point of C is a limit point of C.
• C is bijective with {0, 1}∞ and therefore C is uncountable.
• C has no interior.
• C is totally disconnected.
Connected Sets Definitions
(a) Let A, B ⊆ X, then A and B are separated if A ∩ ¯B = ∅ and ¯A ∩ B = ∅.
(b) E ⊆ X is connected if it is not the union of two nonempty separated sets. If E is not connected, it is called disconnected and A and B are a separation of E.
Example Let E = {(x, y) ∈ R2|x, y ∈ Q}. Consider the line x = √
2. We know all the points are on the left and right of this line. In other words, let A = {(x, y) ∈ E|x < √
2} and B = {(x, y) ∈ E|x >√
2}.
Theorem (Example) [a, b] ⊆ R is connected.
Proof If not, there exists a separation A, B of [a, b] with a ∈ A. Let s = sup A. S exists and s ≤ b. Then s ∈ ¯A because supremums are limit points. Thus s 6∈ B. Then s ∈ A because A ∪ B = [a, b]. Thus s 6∈ ¯B. Thus there exists > 0 such that B(s, ) ⊆ [a, b] \ B = A. Thus (s − , s + ) ⊆ A.
Remark Rudin proves that E ⊆ R is connected if and only if it has the “interval property”, i.e., x, y ∈ E, x < z < y ⇒ z ∈ E.
