SOLUTION OF HW5
MINGFENG ZHAO
March 15, 2013
1. [10 Points] Show that the following finite intersection property for a set A is equivalent to compactness: if B is any collection of closed sets such that the intersection of any finite number of them
contains a point of A, then the intersection of all of them contains a point of A.
Proof. (=⇒) Assume A is compact. Let B be any collection of closed sets such that the intersection
of any finite number of them contains a point of A. If
\
B = ∅, then A =
B∈B
[
A\B. Since B is
B∈B
closed, then A\B is open. Since A is compact, then there exists finitely many B1 , · · · , Bn ∈ B such
that A ⊂
n
[
A\Bi , that is,
n
\
Bi = ∅, contradiction.
i=1
i=1
(⇐=) Assume A satisfies the finite intersection property. Assume A is not compact, then there exists
some open cover O of A such that there is no finite subcover of A in O. Let B = {A\O : O ∈ O},
then the intersection of any finite number of sets in B contains a point of A. By the assumption, we
now that
\
B 6= ∅, that is, O can not cover A, contradiction.
B∈B
2. [10 Points] If B1 , · · · , Bn is a finite open cover of a compact set A, can the union B1
S
···
S
S
Bn =
Bn
equal A exactly?
Proof. If A = ∅, just take B1 = · · · = Bn = ∅, we are done.
Claim I: If A 6= ∅, then the union B1
S
···
S
Bn can not equal A exactly.
If there exists a finite open cover B1 , · · · , Bn of a compact set A such that the union B1
S
···
A exactly, which implies that A is both closed and open. So A = ∅ or A = R. Since A 6= ∅, then
A = R is not compact, contradiction.
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2
MINGFENG ZHAO
3. [10 Points] If A ⊂ B1
A
T
S
B2 where B1 and B2 are disjoint open sets and A is compact, show that
B1 is compact. Is the same true if B1 and B2 are not disjoint?
Proof. Let O be an open cover of A
T
B1 . Since A ⊂ B1
S
B2 , then A ⊂ (A
T
B1 )
S
B2 . Since B2 is
open, hence O and B2 form an open cover of A. Since A is compact, then there exists finitely many
O1 , · · · , On such that
n
[
A⊂
Oi
[
B2 .
Oi
[
i=1
In particular, we get
A
\
B1 ⊂
n
[
B2 .
i=1
But since B1
T
B2 = ∅, then we get
A
\
B1 ⊂
n
[
Oi
i=1
So A
T
B1 is compact.
Counterexample: Let A = [0, 3], B1 = (0, 2) and B2 = (−1, 4), then A is compact, A = [0, 3] ⊂
B
S
B2 = B2 = (−1, 4) B1
T
B2 = B1 = (0, 2), and A
T
B1 = (0, 2) is not compact.
4. [10 Points] If A is compact, show that sup A and inf A belong to A. Given an example of a
non-compact set A such that both sup A and inf A belong to A.
Proof. If A = ∅, we are done. If A 6= ∅. It is easy to see that A ⊂
∞
[
Bn (0). Since A is compact,
n=1
then there exists some N ≥ 1 such that A ⊂ BN (0), that is, A is bounded. So sup A, inf A ∈ R.
Since A is compact, then A is closed. Hence we know that sup A, inf A ∈ A.
Example: Let A = {−1}
S
S
(0, 1) {2}, then A is not compact, but sup A = 2 ∈ A and inf A = −1 ∈
A.
SOLUTION OF HW5
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5. [10 Points] Show that every infinite compact set has a limit point. Is the same true of closed sets?
of open sets?
Proof. Let A be an infinite compact set. If A has no limit point, then for any x ∈ R, there exists
rx > 0 such that A
T
Brx (x) has only finitely many points. Notice that A ⊂
[
Brx (x). Since A
x∈A
n
[
is compact, then there exists some x1 , · · · , xn ∈ A and r1 , · · · , rn such that A ⊂
Bri (xi ). But we
i=1
know that A
T
Bri (xi ) is finite, then A is finite, we get a contradiction.
Counterexample I: Let A = Z, then A is closed, but not compact, and A has no limit point.
Claim I: Let A be an infinite open set, then A has limit point.
Since A is an infinite open set, then A 6= ∅. Take x0 ∈ A, since A is open, then there exists some
δ > 0 such that (x0 − δ, x0 + δ) ⊂ A, which implies that every point of (x0 − δ, x0 + δ) is a limit point
of A.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu