Math 516
Professor Lieberman
February 7, 2005
HOMEWORK #3 SOLUTIONS
Chapter 8
13. First, if x ∈ X, then x ∈ X ∈ B. If x ∈ B1 ∩ B2 for some elements B1 and B2 of B, then
we can write
B1 = ∩ni=1 Ci , B2 = ∩m
j=1 Dj
for sets Ci and Dj in C and finite numbers m and n. Let B3 = B1 ∩ B2 and note that
\
B3 = ∩ni=1 Ci , ∩m
j=1 Dj ,
so B3 ∈ B. It follows that x ∈ B3 ⊂ B1 ∩ B2 . Therefore B is the base for some topology T.
If T 0 is any other topology containing C, then any finite intersection of elements of T 0 is
in T 0 . It follows that B ⊂ T 0 , and hence T ⊂ T 0 . Therefore T is the weakest topology that
contains C.
15. First, x ∈ [x, x + 1) for any x ∈ R. Also, if x ∈ [a1 , b1 ) ∩ [a2 , b2 ), then x ∈ [x, min{b1 , b2 }) ⊂
[a1 , b1 ) ∩ [a2 , b2 ). From Proposition 12, B is a base for some topology T. To show that it
satisfies the first axiom of countability, let x ∈ R. Then {[x, x + n1 ) : n ∈ N} is a countable
base at x. (It’s obviously countable and it’s a base because given [a, b) with x ∈ [a, b), there
is a positive integer n such that 1/n < b − x, so [x, x + n1 ) ⊂ [a, b).) On the other hand, if B0
is a base for T, each x must be the left endpoint of some interval [x, y) with [x, y) ⊂ B for
some B ∈ B0 . To see why, fix x ∈ R, and let B ∈ B0 be such that x ∈ B ⊂ [x, x + 1). Then
there is an interval B1 ∈ B (not B0 ) such that x ∈ B1 ⊂ B ⊂ [x, x + 1). Hence B1 = [x, y)
for some y > x.
To see that Q is dense, let x ∈ R and let U be an open set such that x ∈ U . Then
there is a real number y > x such that [x, y) ⊂ U and then a rational number q ∈ [x, y).
Finally, hX, Ti is not metrizable because it’s separable but not second countable, and every
separable metric space is second countable by Proposition 86 from Math 515.
23. (a) If X is normal, let F be a closed set and O an open set with F ⊂ O. Then Õ is a closed
set such that F ∩ Õ = ∅, so there are disjoint open sets O1 and O2 such that
F ⊂ O1 ,
Õ ⊂ O2 .
Let U = O1 . Then F ⊂ U and Õ2 is a closed set containing U so U ⊂ Õ2 . It follows that
Ũ ⊂ O.
Conversely, let F1 and F2 be disjoint closed sets and set O = F̃2 . Then there is an open
set U such that F1 ⊂ U and U ⊂ O. It follows that F2 ⊂ X ∼ U . Let O1 = U and
O2 = X ∼ U . Then O1 and O2 are disjoint open sets with F1 ⊂ O1 and F2 ⊂ O2 .
(b) First take U1/2 from part (a). Then, given a positive integer n, we assume that Ur has
been constructed for r = s2−m+1 with s a positive integer less that 2m−1 . If ρ = p2−m , we
consider two cases: if p is even, then Uρ has already been constructed. Otherwise, p = 2s+1
for some s. Use part (a) with F replaced by Us2−m+1 and O replaced by U(s+1)2−m+1 to
obtain Ur .
(c) To show that f is continuous, note that Ur = {x : f (x) < r}. (clearly f (x) ≤ r on Ur
1
2
and, if x ∈ Ur , then )
(c) =⇒ This is exactly Urysohn’s Lemma.
⇐= Given two disjoint closed sets A and B, let U = f −1 ((−1, 12 )) and V = f −1 ( 12 , 2)).
These are disjoint and open because f is continuous with A ⊂ U and B ⊂ V , so X is
normal.
Chapter 10
31. (a) Suppose it is continuous at x0 . Let x ∈ X and let U be an open neighborhood of f (x).
Then U 0 = f (x0 ) − f (x) + U is an open neighborhood of f (x0 ) because the topology is
translation invariant. Moreover, V 0 = f −1 (U 0 ) is an open neighborhood of x0 because f
is continuous there. It follows that V = x − x0 + V 0 is an open neighborhood of x and
f −1 (U ) = V , so f is continuous at x.
(b)
37. We just have to show that x1 , . . . , xn form a spanning set for X. So fix x ∈ X. Then there
is a β such that βx ∈ V . Set v1 = βx. Then there is an index j(1) and a vector v2 ∈ V so
that v1 = xj(1) + 31 v2 . We now argue by induction that there, for each positive integer i,
there are a vector vj in V and an integer j(i) in {1, . . . , n} with
1
vj = xj(i) + vj+1 .
3
Hence, for each j ∈ {1, . . . , n}, there is a sequence hm(j, i)i of nonnegative numbers such
that, for any positive integer N , we have


n N
1 X X
m(j, i)xj  ∈ 3−N V.
x−
β
j=1 i=1
P
Moreover m(j, i) = 0 or m(j, i) = 3−i for any i and j. It follows that ∞
i=1 m(i, j) converges
for any j. We denote this infinite sum by αj , and we set
y=
n
1X
αj xj .
β
j=1
It follows that β(x − y) ∈ 3−N V for any positive integer N . To prove that x = y, we
suppose that ξ = x − y isn’t zero. Then there is an open set W ⊂ V such that 0 ∈ W ,
ξ∈
/ W , and αW ⊂ W for all α ∈ (−1, 1). Now let ξj = jξ for each positive integer j. It’s
easy to check that W + ξj is a neighborhood of xj which contains only one ξi , and hence
hξj i is an infinite sequence with no cluster points. However, the sequence is in V , so it must
have a cluster point. This contradiction shows that ξ = 0 and hence x = y, which means
that x is a finite linear combination of the xi ’s. In other words, x1 , . . . , xn form a spanning
set for X.