Download Math 535 - General Topology Fall 2012 Homework 11 Solutions

Math 535 - General Topology
Fall 2012
Homework 11 Solutions
Problem 1. Let X be a topological space.
a. Show that the following properties of a subset A ⊆ X are equivalent.
1. The closure of A in X has empty interior: int(A) = ∅.
2. For all non-empty open subset U ⊆ X, there is a non-empty open subset V ⊆ U satisfying
V ∩ A = ∅.
A subset A ⊆ X satisfying these equivalent properties is called nowhere dense in X.
Solution. Recall that a subset B ⊆ X is dense if and only if its complement has empty
interior:
c
B = X ⇔ B = ∅ = int(B c ).
Now consider the following equivalent conditions.
A has empty interior.
c
c
⇔ A is dense. But note A = int(Ac ).
⇔ For all non-empty open subset U ⊆ X, we have U ∩ int(Ac ) 6= ∅.
⇔ For all non-empty open subset U ⊆ X, there is a point x ∈ U ∩Ac and an open neighborhood
W of x satisfying W ⊆ Ac , in other words W ∩ A = ∅.
⇔ (Taking V = U ∩ W ) For all non-empty open subset U ⊆ X, there is a non-empty open
subset V ⊆ U satisfying V ∩ A = ∅.
b. Show that the following properties of the space X are equivalent.
1. Any countable intersection of open
T dense subsets is dense. In other words, if each Un ⊆ X
is open and dense in X, then ∞
n=1 Un is dense in X.
2. Any countable union of closed subsets with empty interior has empty interior. In other
words,
S if each Cn ⊆ X is closed in X and satisfies int(Cn ) = ∅, then their union satisfies
int ( ∞
n=1 Cn ) = ∅.
A space X satisfying these equivalent properties is called a Baire space.
Solution. Consider the following equivalent conditions.
T
If each Un ⊆ X is open and dense in X, then ∞
n=1 Un is dense in X.
T
c
⇔ If each Unc ⊆ X is closed and has empty interior in X, then ( ∞
n=1 Un ) has empty interior
in X.
S
⇔ (Taking Cn := Unc ) If each Cn ⊆ X is closed and has empty interior in X, then ∞
n=1 Cn has
empty interior in X.
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Definition. Let X be a topological space. A function f : X → R is lower semicontinuous
if for all a ∈ R, the preimage f −1 (a, +∞) is open in X.
Equivalently: For all x0 ∈ X and > 0, there is a neighborhood U of x0 satisfying f (x) >
f (x0 ) − for all x ∈ U . This means that the values close to x0 can “suddenly jump up” but
not down.
Problem 2.
a. Let X be a topological space and f : X → R a continuous real-valued function. Show that
for every non-empty open subset U ⊆ X, there is a non-empty open subset V ⊆ U on which f
is bounded.
Solution. Pick a point x ∈ U . Since f is continuous at x, there is an open neighborhood W
of x satisfying f (W ) ⊆ (f (x) − 1, f (x) + 1), in particular f is bounded on W . Now the subset
V := W ∩ U is non-empty (since x ∈ V ), open, and f is bounded on V .
b. (Willard Exercise 25C) Let X be a Baire space and f : X → R a lower semicontinuous
function. Show that for every non-empty open subset U ⊆ X, there is a non-empty open subset
V ⊆ U on which f is bounded above.
c
Solution. Note that for all a ∈ R, the preimage f −1 (−∞, a] = (f −1 (a, +∞)) is closed in X.
Express X as the countable union
X = f −1 (R)
∞
[
f −1
!
(−∞, n]
n=1
∞
[
f −1 (−∞, n]
n=1
∞
[
=:
An
n=1
of closed subsets, and likewise
U=
∞
[
(An ∩ U ).
n=1
Since U is open (and non-empty) and X is Baire, U cannot be meager, so that for some m ∈ N,
Am ∩ U is not nowhere dense. Let W ⊆ X be a non-empty open subset satisfying
W ⊆ Am ∩ U ⊆ Am ∩ U = Am ∩ U .
Since W is open and satisfies W ⊆ U , it also satisfies W ∩ U 6= ∅. This subset V := W ∩ U is
non-empty, open, and contained in Am so that f is bounded above on V (by the upper bound
m).
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Problem 3. Show that a topological space X is of second category in itself if and only if any
countable intersection of open dense subsets of X is non-empty.
Solution. Consider the following equivalent conditions.
X is of second category
in itself, i.e. for any countable collection of nowhere dense subsets
S∞
An ⊆ X, we have n=1 An 6= X.
S
⇔ For any countable collection of closed nowhere dense subsets Cn ⊆ X, we have ∞
n=1 Cn 6= X.
(This implies the previous condition since A being nowhere dense implies A being nowhere
dense.)
⇔ (Taking Un = Cnc ) For any countable collection of open dense subsets Un ⊆ X, we have
T
∞
n=1 Un 6= ∅.
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Problem 4. (Uniform boundedness principle) (Willard Exercise 25D.5) (Munkres Exercise
48.10) (Bredon I.17.2)
Let X be a Baire space and S ⊆ C(X, R) a collection of real-valued continuous functions on X
which is pointwise bounded: for each x ∈ X, there is a bound Mx ∈ R satisfying
|f (x)| ≤ Mx for all f ∈ S.
Show that there is a non-empty open subset U ⊆ X on which the collection S is uniformly
bounded: there is a bound M ∈ R satisfying
|f (x)| ≤ M for all x ∈ U and all f ∈ S.
Solution. For all n ∈ N, consider the subset of X
Cn = {x ∈ X | |f (x)| ≤ n for all f ∈ S}
\
=
{x ∈ X | |f (x)| ≤ n}
f ∈S
=
\
f −1 [−n, n]
f ∈S
which is closed in X since each f ∈ S is continuous.
Pointwise boundedness of the collection S yields x ∈ Cn whenever n ≥ Mx , or equivalently
X=
∞
[
Cn .
n=1
Since X is Baire, it is in particular of second category in itself, so that for some m ∈ N, Cm is
not nowhere dense. Let U ⊆ X be a non-empty open subset satisfying U ⊆ Cm = Cm . Then
the bound |f (x)| ≤ m holds for all x ∈ U and all f ∈ S.
4
Definition. Let X and Y be normed real vector spaces. A linear map T : X → Y is bounded
if there exists a constant C ∈ R satisfying
kT xk ≤ Ckxk
for all x ∈ X.
By linearity, this condition is equivalent to the following number being finite:
kT k :=
kT xk
x∈X\{0} kxk
sup
= sup kT xk
kxk=1
= sup kT xk.
kxk≤1
The number kT k ∈ R ∪ {∞} is called the operator norm of T .
Let
L(X, Y ) := {T : X → Y | T is linear and kT k < ∞}
denote the vector space of bounded linear maps from X to Y . It is a vector space under
pointwise addition and scalar multiplication. One readily checks that the assignment T 7→ kT k
is indeed a norm on L(X, Y ).
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Problem 5. Let T : X → Y be a linear map between normed real vector spaces. Show that
the following are equivalent.
1. T is continuous (everywhere).
2. T is continuous at some point x0 ∈ X.
3. T is continuous at 0 ∈ X.
4. T is bounded.
Solution. (1 ⇒ 2) X is non-empty since it contains 0 ∈ X.
(2 ⇒ 3) Let > 0. By continuity of T at x0 , there is a δ > 0 satisfying T Bδ (x0 ) ⊆ B (T x0 ).
For any x ∈ Bδ (0), we have
d(T x, T (0)) = kT x − 0k
= kT xk
= kT (x0 + x − x0 )k
= kT (x0 + x) − T x0 k
= d(T (x0 + x), T x0 )
<
so that T is continuous at 0.
(3 ⇒ 4) Taking = 1, since T is continuous at 0, there is a δ > 0 satisfying T Bδ (0) ⊆
B1 (T (0)) = B1 (0). Thus for any x with kxk < 1, we have
δ
x k
kT xk = kT
δ
1
= k T (δx)k
δ
1
= kT (δx)k
δ
1
< (1)
δ
1
=
δ
and linearity of T implies kT xk ≤
1
δ
whenever kxk ≤ 1. Therefore T is bounded:
1
kT k = sup kT xk ≤ .
δ
kxk≤1
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(4 ⇒ 1) If T has bound C, then T is Lipschitz continuous with Lipschitz constant C, hence
continuous. For all x, x0 ∈ X, we have
d(T x, T x0 ) = kT x − T x0 k
= kT (x − x0 )k
≤ Ckx − x0 k
= Cd(x, x0 ).
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Problem 6. Consider the Banach space
l∞ = {x ∈ RN | kxk∞ < ∞}
with the supremum norm kxk∞ = supi∈N |xi |. Consider the linear subspace of lists that are
eventually zero:
X := {x ∈ l∞ | ∃ N ∈ N such that xi = 0 for all i > N } ⊂ l∞ .
Consider the continuous linear maps Tn : X → R defined by
Tn (x) = nxn .
a. Show that the collection {Tn }n∈N is pointwise bounded but not uniformly bounded.
Solution. Pointwise bounded. Let x ∈ X and let N ∈ N be large enough so that xi = 0
for all i > N . Then for all n > N , we have
Tn x = nxn = 0
and therefore
sup|Tn x| = max |Tn x| < ∞.
1≤n≤N
n∈N
Not uniformly bounded. Consider the standard basis vectors ek ∈ X whose coordinates are
(
1 if i = k
eki =
0 if i 6= k
and note that these are unit vectors: kek k∞ = 1.
The equality Tn (en ) = n(enn ) = n(1) = n implies
kTn k =
kTn xk
x∈X\{0} kxk
sup
≥
kTn en k
ken k
=
|n|
1
= n.
It follows that the collection {Tn }n∈N is not uniformly bounded:
supkTn k = ∞.
n∈N
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b. Part (a) implies that X cannot be complete. Show explicitly that X is not complete by
exhibiting a Cauchy sequence in X that does not converge in X.
Solution. Let us denote the sequence index as a superscript. Consider the sequence (x(n) )n∈N
in X consisting of the following vectors:
(
1
if i ≤ n
(n)
xi = i
0 if i > n.
Note that each vector x(n) is eventually zero, hence a legitimate element of X.
The sequence is Cauchy. For any N ∈ N and m, n ≥ N (with m ≤ n), the distance
d(x(m) , x(n) ) = kx(m) − x(n) k
= max{
1
1
1
,
,..., }
m+1 m+2
n
1
m+1
1
<
N
=
converges to 0 as N → ∞.
The sequence does not converge in X. Let x ∈ X and let N ∈ N be large enough so that
xi = 0 for all i > N . Then for all n > N , the distance
d(x(n) , x) = kx(n) − xk
(n)
− xi |
(n)
− xi |
= sup|xi
i∈N
≥ sup|xi
i>N
(n)
= sup|xi |
i>N
(n)
= |xN +1 |
=
1
N +1
is bounded away from 0. Therefore the sequence (x(n) )n∈N does not converge to x ∈ X.
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