Solutions of Fourth Homework
Solution of 7.4 Ex. 3:
Let U be an open set in Rd and
T Kn , n ∈ N, a nested family of
compact sets in Rd . Assume that ∞
n=1 Kn ⊂ U .
Put
Zn = Kn − U,
n ∈ N.
Since Kn are closed and the complement of U is closed, they are all
closed sets. Moreover, since Kn are bounded, all Zn are bounded. Since
(Kn ) is a nested family, (Zn ) is also a nested family.
Moreover, we have
∞
∞
∞
\
\
\
Kn − U = ∅
(Kn − U ) =
Zn =
n=1
n=1
n=1
by our assumption. By 7.4.6, at least one of Zn must be empty. But
this means that Kn ⊂ U .
Solution of 7.4 Ex. 5:
Let M = inf x∈K ky−xk. Consider the closed balls B̄M + 1 (y) centered
n
at y. We know that these sets are closed and bounded.
Since K is compact, it is closed and bounded by Heine-Borel theorem.
Hence the sets
Zn = B̄M + 1 (y) ∩ K,
n ∈ N,
n
are also closed and bounded. By the construction, they are nonempty.
Clearly, we have Zn ⊃ Zn+1 for any n ∈ N.TTherefore, the family Zn ,
n ∈ N, is nested. By 7.4.6,Tthe intersection ∞
n=1 Zn is nonempty.
∞
Let x0 be an element of n=1 Zn . Then x0 ∈ K and ky−x0 k ≤ M + n1
for all n. This implies that ky − x0 k = M , and ky − x0 k ≤ ky − xk for
any x ∈ K.
Solution of 7.4 Ex. 14:
Let x be a point of X. Consider the family U of all open balls Bn (x)
of radius n = 1, 2, . . . centered in x. The family U is an open cover
of X. Since X is compact, it has a finite subcover Bn1 (x), . . . , Bns (x).
Let N = max{n1 , . . . , ns }. Then Bni (x) ⊂ BN (x) and {BN (x)} covers
X, i.e., BN (x) = X. Therefore X is bounded.
Solution of 7.5 Ex. 6:
Let U be a family of connected sets with common point x. Let V
be the union of the family U. Assume that the union is also a disjoint
union of two relatively open sets A and B. For any set U in the family
U, U = (A ∩ U ) ∪ (B ∩ U ) is the partition of U into two disjoint
1
2
relatively open sets. Since U is connected one of these sets must be
empty. Assume that B ∩ U is empty. This means that U ⊂ A. In
particular, x ∈ A.
Let U 0 be another set in the family U. By the same argument, one
of the sets A ∩ U 0 and B ∩ U 0 must be empty. We already know that
x ∈ U 0 ∩ A. Hence B ∩ U 0 bust be empty and U 0 ⊂ A.
Since U 0 was arbitrary, we conclude that all connected sets in U are
subsets of A. Therefore, their union V is a subset of A and B is empty.
Hence, V is connected.
Solution of 7.5 Ex. 7:
Any closed interval [a, b] in R is connected by 7.5.4. Moreover, it
bounded and closed. Therefore it is also compact by Heine-Borel theorem.
On the other hand, if K is a connected compact set in R, it is a
interval by 7.5.4 (since it is connected). By Heine-Borel theorem it is
also bounded and closed. Hence, is a closed interval since it is closed,
and since it is bounded, it must be of the form [a, b].