Math 516
Professor Lieberman
February 28, 2005
HOMEWORK #6 SOLUTIONS
Chapter 9
11. (a) First, if x0 ∈ X, let x00 be the successor to x0 . then x0 ∈ {x ∈ X : y < x00 }, so every
x ∈ X is in some B ⊂ B. Next, suppose x0 ∈ B1 ∩ B2 for B1 and B2 elements of B.
Then B3 = B1 ∩ B2 is an element of B. To prove this fact, we first introduce some simple
notation:
(
(
a if b ≤ a,
a if a ≤ b,
max{a, b} =
min{a, b} =
b if a < b,
b if b < a.
Then
{x : x < a} ∩ {x : x < b} = {x : x < min{a, b}}, {x : x < a} ∩ {x : b < x < c}
= {x : b < x < min{a, c}}, {x : x < a} ∩ {x : x > b}
= {x : b < x < a},
{x : a < x < b} ∩ {x : c < x < d} = {x : max{a, c} < x < min{b, d}}, {x : a < x < b} ∩ {x : x > c}
= {x : a < x < min{b, c}}, {x : x > a} ∩ {x : x > b}
= {x : x > max{a, b}}.
It then follows from Proposition 12 that B is a base for a topology.
(b) Let hxn i be a sequence in X and define a subsequence inductively as follows: xn1 is
the smallest element of {xn }. Then xn2 is the smallest element of {xj : j > n1 }, and,
in general, xnk is the smallest element of the set {xj : j > nk }. It follows that hxnj i is
an increasing sequence, so it has a least upper bound x0 by Problem 1.32. The proof that
xnj → x0 is similar to the proof of the corresponding result for real numbers: Given a < x0 ,
since a is not an upper bound for the subsequence, there is xnj with a < xnj ≤ x0 , and
hence a < xnk ≤ x0 for all k ≥ j. (If there is no a < x0 , then x0 is the smallest element of
X, and the subsequence must consist of x0 repeated infinitely often.) But every open set
containing x0 contains, as a subset, a set of the form {x : a < x ≤ x0 } (which is the same
as {x : a < x < x00 } with x00 the successor to x0 ). Therefore, given such an open set U ,
there is an index j such that xnk ∈ U for all k ≥ j so xnj → x0 .
To see that X isn’t compact, let Ua = {x : x < a} for each a ∈ X. Then {Ua }a∈X is an
open cover of X. But any finite subset {Ua1 , . . . , Uan } isn’t a cover because
n
[
Uai = Ua
i=1
for a = max{a1 , . . . , an } and this isn’t all of X.
(c) Set
β = limf = inf sup f (x).
a∈X x>a
Since X is sequentially compact, it’s countably compact and hence β is finite. Now suppose
that there is a positive integer n such that An = {x : f (x) ≤ β − 1/n} is uncountable.
1
2
Then for any a ∈ X, there is x > a such that f (x) ≤ β − 1/n and y > a such that
f (y) ≥ β − 1/(2n). We now define a sequence inductively: x1 is the smallest element of
{x : x ≤ β − 1/n}. Then, once we have x1 , . . . , xj , we define xj+1 as the smallest element
of {x : x > xj , x ≤ β − 1/n} if j is even and as the smallest element of {x : x > xj , x ≥
β − 1/(2n)} if j is odd. The sequence hxj i is increasing so, as before, it has a limit x0 .
Since f is continuous, we must have f (x0 ) = lim f (xj ) but hf (xj )i is a divergent sequence
since f (x2k ) ≥ β − 1/(2n) and f (x2k+1 ) ≤ β − 1/n. This contradiction shows that An is
countable for all n and hence {x : f (x) < β}, being a countable union of countable sets, is
countable. A similar argument shows that {x : f (x) > limf } is countable. Hence there are
elements x1 and x2 such that
f (x) ≥ limf for x > x1 ,
f (x) ≤ limf for x > x2 .
But limf ≥ limf , and if x > max{x1 , x2 }, we have
limf ≤ f (x) ≤ limf.
Hence limf = limf . Calling this number β, we infer that f (x) = β if x > x > max{x1 , x2 }.
If x0 is the successor to max{x1 , x2 }, it follows that f (x) = f (x0 ) for all x ≥ x0 .
16. Using
the notation of the problem, we conclude that g is continuous as follows. Let O =
Q
Of be an open set in I A , so each Of is open and Of = [0, 1] except for finitely many f ’s,
say f1 , . . . , fn . Then
g −1 (O) = {x ∈ X : fi (x) ∈ Ofi } = ∩ni=1 fi−1 (Ofi ).
Since each fi is continuous, fi−1 (Ofi ) is an open subset of X and hence g −1 (O) is open, so
g is continuous.
Next, we show that g is one-to-one. A compact Hausdorff space is normal by exercise 4,
so points are closed sets. Then Urysohn’s Lemma implies that, for any x 6= y, there is a
continuous function f with f (x) = 0 and f (y) = 1. Hence x 6= y implies g(x) 6= g(y), so g
is one-to-one. Since each If is Hausdorff, so is Q. Therefore, Proposition 30 implies that g
is a homeomorphism onto its image.
19. (a) From Problem 18 (which was done in class), there is an open set U ⊃ K with compact
closure, and then an open set V ⊃ U with compact closure. Urysohn’s Lemma gives a
continuous function g : V → R such that g = 1 on K and g = 0 on V ∼ U . Then define
f : X → R by
(
0
if x ∈ Ṽ ,
f (x) =
g(x) if x ∈
/ Ṽ .
it follows from Proposition 8.3 that f is continuous. The support of f is a subset of V , so
it’s compact.
(b) The only part left is to show that, if K is a Gδ , then we can take f < 1 in K̃. To this
end, let K = ∩Un with each Un open, and set Vn = Un ∩ U with U an open set with K ⊂ U
and U compact. Then each Vn is open with compact closure and K = ∩Vn . Let fn be the
function from (a) so that fn = 1 on K and fn = 0 outside Vn . We now define
f=
∞
X
n=1
2−n fn .
3
Since the series converges uniformly, f is continuous and it’s easy to check that f = 1 on
K and f = 0 on ∪Vn , so f has compact support. In addition, if x ∈ K̃, then x ∈ Vn for
some n > 1. Hence fn (x) = 0, so f (x) ≤ 1 − 2−n and therefore f (x) < 1.