HW set 3- solutions.
2. Let X be a compact topological space, C ⊂ X closed. Take a covering
{Uλ }λ∈Λ of C by relatively open subsets; so Uλ = Vλ ∩ C, where Vλ is open
in X. Then {Vλ , X \ C}λ∈Λ is an open covering of X, from which (by
compactness) we may extract a finite sub covering {Vi }N
i=1 , X \ C (add the
last set in if needed). Then the Vi , i = 1, . . . , N cover C, so we have a
finite subcovering {Ui }N
i=1 of the initial covering {Uλ }λ∈Λ . This shows C is
compact.
3. (a) Let (Cj )j≥1 be a decreasing sequence of closed sets in the compact
space X: Cj+1 ⊂ Cj . If each Cj 6= ∅, then their intersection ∩∞
j=1 Cj 6= ∅.
For if it were empty, its complement would satisfy:
X=(
∞
\
c
Cj ) =
j=1
∞
[
Cjc ,
j=1
so we would have a covering of X by open sets {Cjc }j≥1 . Extracting a finite
c
c
c
c
subcovering we have X ⊂ ∪N
j=1 Cj , and since Cj ⊂ Cj+1 this says X ⊂ CN ,
or CN = ∅, contradiction.
(b) Now let (Bi )i≥1 be a sequence of closed sets in the compact topological space X, with the property: Cj = B1 ∩ . . . ∩ Bj 6= ∅, for each j ≥ 1.
Note the Cj are closed and decreasing: Cj+1 ⊂ Cj . By part (a), we have
∞
\
i=1
Bi =
∞
\
Cj 6= ∅.
j=1
P
7. (i) Let A, B be n × n matrices. Then (AB)ij = k aik bkj , so:
X
X
XX
X
XX
X
b2kj ,
(AB)2ij =
a2ik
[
aik bkj ]2 ≤
[
a2ik
b2kj ] =
i,j
ij
k
ij
k
k
ik
jk
by the Cauchy-Schwarz inequality. So ||AB||2 ≤ ||A||2 ||B||2 , as claimed.
P
(ii)Note that 1 =P(AT A)ii = k a2ki for each i, if A is orthogonal. Adding
over i, we find n = ik a2ik .
(iii)
||AT A−B T B|| = ||(AT −B T )A+B T (A−B)|| ≤ ||(A−B)T A||+||B T (A−B)|| ≤ (||A||+||B||)||A−B||.
Note that this shows that if ||A − C|| ≤ R and ||B − C|| ≤ R (in particular
||A|| ≤ ||C|| + R) and ||B|| ≤ ||C|| + R) we have that the map A 7→ AT A in
1
the space of n × n matrices is Lipschitz in the ball BR (C), with Lipschitz
constant 2(||C|| + R). Being locally Lipschitz, this map is continuous.
(iv) The set of orthogonal matrices is the preimage of {In } under this
continuous map, hence is closed.
x
8. Let f : X → S n be the restriction to X of the map x 7→ |x|
of
n+1
n+1
R
\ {0}. This map is continuous at any x0 ∈ R
\ {0}, as shown by
the estimate:
|
x
x0
|x − x0 |
1
1
2
−
|=
+ |x||
−
|≤
|x − x0 |,
|x| |x0 |
|x0 |
|x| |x0 |
|x0 |
using ||x| − |x0 || ≤ |x − x0 | (from the triangle inequality).
Thus its restriction to X is continuous in the induced topology. It is
surjective, since given v ∈ S n the intersection rv ∩ X of the ray through v
with X (a unique point) maps to v under f . It is injective, for if two points
x1 6= x2 mapped to the same v ∈ S n , the ray through v would intersect X
at two points, contradicting the hypothesis.
Finally, as a continuous bijective map from a compact set, f is a homeomorphism from X to S n .
10. If f is not Lipschitz in K, for each M > 0 we can find points x, y in
K so that ||f (x) − f (y)|| ≥ M ||x − y||. In particular for each integer N ≥ 1
we find xN , yN as in (i) of the hint.
Since K is compact, these sequences have convergent subsequences xNj →
x0 , yNj → y0 (as j → ∞) . As a continuous function with compact domain,
f is bounded (say, by M > 0), so we find: N ||xN − yN || ≤ 2M for each N ,
which implies ||xN − yN || → 0. So we must have x0 = y0 .
By hypothesis, f is Lipschitz in some ball BR (x0 ), with Lispchitz constant (say) L. For all j sufficiently large, xNj , yNj will be in BR (x0 ), and
we’ll have the inequalities:
Nj ||xNj − yNj || ≤ ||f (xNj ) − f (yNj )|| ≤ L||xNj − yNj ||,
implying Nj ≤ L for all j sufficiently large, when in fact Nj → ∞ with j.
This contradiction shows f is Lipschitz on K.
Alternative solution, avoiding a contradiction argument. From the local
Lipschitz property and a compactness, we may find a finite set of balls
{BRi (xi ); i = 1, . . . , N } covering K, with xi ∈ K for all i, and L > 0 so that
2
f is Lipschitz with constant L in each ball. Letting R > 0 be the minimum
of the Ri , we find that if ||x − y|| < R/2 and x, y ∈ K, then they have to be
in the same ball BRi (xi ) (for some i = 1, . . . , N ). So we have:
x, y ∈ K, ||x − y|| < R/2 ⇒ ||f (x) − f (y)|| < L||x − y||.
Being locally Lipschitz on a compact set, f is bounded: ||f (x)|| ≤ M for all
x ∈ K. Let C = 6M/R. Then if x, y ∈ K with ||x − y|| ≥ R/3, we have:
||f (x) − f (y)|| ≤ 2M =
CR
≤ C||x − y||.
3
We conclude f is Lipschitz on K, with constant min{C, L}.
3