Download Problem 3, Page 100 Show that if A is closed in X and B is closed in

Problem 3, Page 100
Show that if A is closed in X and B is closed in Y , then A × B is closed in X × Y .
Proof: Since A, B are closed in X, Y respectively, X − A, Y − B are open in
X, Y respectively. Then (X − A) × Y , X × (Y − B) are open in X × Y . We have
X × Y − A × B = (X − A) × Y ∪ X × (Y − B) are open. Thus A × B is closed in
X ×Y.
Problem 6, Page 101
Let A, B and Aα denote subsets of a space X. Prove following:
(a) If A ⊂ B, then A ⊂ B.
(b) A ∪ B = A ∪ B
S
S
(c) Aα ⊃ Aα .
Proof: (a) Since A ⊂ B and B ⊂ B, we have A ⊂ B. Since B is closed and
A is the smallest closed set containing A, we have A ⊂ B.
(b)Since A ⊂ A and B ⊂ B, we have A ∪ B ⊂ A ∪ B. Thus A ∪ B ⊂ A ∪ B. Since
A ⊂ A ∪ B, A ⊂
B ⊂ A ∪ B. Therefore,
S A ∪ B. Similarly,
S
S A∪B ⊂A
S∪ B. S
A
⊂
A
.
Thus
ASα ⊂ S Aα .
(c)Since Aα ⊂ Aα and Aα is closed, we have
αS
α
S
S
Example: let Ai = {1/i}, then Ai = Ai and Ai = Ai . But Ai = Ai {0}.
The equality fails in this case.
Problem 10, Page 101
Show that every order topology is Hausdorff.
Proof: Suppose X is a space with order topology. Let a 6= b ∈ X. Without
loss of generality, we assume a < b.
1 If there is no c, such that a < c < b. Take A = {x ∈ X | x < b} and
B = {y ∈ X | y > a}. By the definition of order topology, A, B are open
in X. It is easy to see A ∩ B = ∅.
2 If there is c such that a < c < b. Take A = {x ∈ X | x < c} and
B = {y ∈ X | y > c}. By the definition of order topology, A, B are open in
X. It is easy to see A ∩ B = ∅.
Above all, X is a Hausdorff space.
Problem 11, Page 101
Show that the product of two Hausdorff space is a Hausdorff space.
Proof: Let X, Y be two Haudorff topological space and X × Y be the product
space. Suppose (x1 , y1 ), (x2 , y2 ) are two different points in X × Y . Then x1 6= x2
or y1 6= y2 . Without loss of generality, assume x1 6= x2 . Then since X is a Hausdorff space, we can choose two open sets V1 , V2 ⊂ X, such that x1 ∈ V1 , x2 ∈ V2
1
2
and V1 ∩ V2 = ∅. Choose any two open sets U1 , U2 ⊂ Y , such that y1 ∈ U1 and
y2 ∈ U2 . Then we have (x1 , y1 ) ∈ V1 × U1 and (x2 , y2 ) ∈ V2 × U2 . Moreover,
V1 × U1 ∩ V2 × U2 = (V1 ∩ V2 ) × (U1 ∩ U2 ) = ∅. Thus for any two distinct points in
X × Y , we can find two disjoint open sets containing two points respectively. Thus
X × Y is Hausdorff.
Problem 17, Page 101
Consider the lower limit topology on R and the topology given √
by the basis C √
of Exercise 8 of § 13. Determine the closures of the intervals A = (0, 2) and B = ( 2, 3)
in this two topologies.
Proof:
Case 1: Lower limit topology.
For any open set U containing 0, there is a ² such that [0,
√ ²) ⊂ U by the√definition
of Lower limit topology. √Then we have ∅ 6= [0, ²) ∩ (0, 2) ⊂ U ∩ (0, 2).√ Thus
S
Since (−∞, 0) = [−n, 0),
0 ∈ A. We have that
√ [0, 2) ⊂ A. √
√n ∈ N and [ 2, ∞)
are open, R − √
[0, 2) = (−∞, 0) ∪ [ 2, ∞) is open. Thus [0, 2) is closed. Above
all,we have [0, 2) = A.
√
Use the exactly same discussion as above, we have B = [ 2, 3).
Case 2: C−basis topology.
For any open set U containing 0, there is a ² ∈ Q such that√[0, ²) ⊂ U by√the
definition of C−basis topology. Then we have √
∅ 6= [0, ²) ∩ (0, 2) ⊂ U ∩ (0, 2).
Thus 0 ∈ A.
For
any
open
set
U
containing
2, there is an interval [a, b) ⊂ U
√
containing 2 where √
a, b ∈ Q by the√ definition of
√ C−basis topology. Then we
have
∅
6
=
[a,
b)
∩
(0,
2)
⊂
U
∩
(0,
2).
Thus
2√ ∈ A. Above all, we have
√
S
S
N
and
(
2, ∞) = [an , ∞)
[0, 2] ⊂ B. Since (−∞, 0)√= (−n, 0), n ∈
√
√ are open
where {an } ⊂ Q and an → 2, we√have [0, 2] is closed. Thus A = [0, 2].
For
√ any open set U containing 2, there is an interval [a, b) ⊂ U containing
2 where
√ a, b ∈ Q by√the definition
√ of C−basis topology.
√ Then we have ∅ 6=
[a, b) ∩ ( 2,
3)
⊂
U
∩
(
2,
3).
Thus
2
∈
A.
Therefore,
[
2, 3) ⊂ B. √
Since [3, ∞)
√
S
and
(−∞,
2)
=
(−∞,
a
)
are
open,
where
{a
}
⊂
Q
and
a
→
2, we have
n
n
√
√ n
[ 2, 3) is closed. Thus [ 2, 3) = B.
Problem 18, Page 101
Determine the closures of the following subsets of the ordered square:
A = {(1/n) × 0 | n ∈ Z+ }
B = {(1 − 1/n) × (1/2) | n ∈ Z+ }
C = {x × 0 | 0 < x < 1}
D = {x × (1/2) | 0 < x < 1}
E = {(1/2) × y | 0 < y < 1}
Proof:
A = A ∪ {0 × 1}
B = B ∪ {1 × 0}
3
C = C ∪ [0, 1) × 1
D = D ∪ (0, 1] × 0 ∪ [0, 1) × 1
E = E ∪ {1/2 × 0, 1/2 × 1}
Problem 1, Page 111
Prove that for functions f : R → R, the ² − δ definition of the continuity implies
the open set definition.
Proof: We need to show that if f is continuous in sense of ² − δ definition, then it
is continuous in sense of the open set definition.
If f is continuous in sense of ² − δ definition, then we have for any point x ∈ R and
δ > 0, there is an ² > 0 such that for any y ∈ (x−², x+²) f (y) ∈ (f (x)−δ, f (x)+δ),
i.e. (x − ², x + ²) ⊂ f −1 ((f (x) − δ, f (x) + δ)). Now we need to show that for any
open set U ⊂ R, f −1 (U ) is open in R. Since U is open, for any f (x) ∈ U we have
an δx such that (f (x) − δx , f (x) + δx ) ⊂ U . Since f is continuous in sense of ² − δ
definition, we have that there is an ²x such that (x − ²x , x + ²x ) ⊂ f −1 (U ). Therefore, by exercise 1 in homework 1, we have f −1 (U ) is open. Thus f is continuous
in sense of the open set definition.
Problem 2, Page 111
Suppose that f : X → Y is continuous. If x is a limit point of the subset A of X,
is it necessarily true that f (x) is a limit point of f (A).
Proof: No.
Let X = (0, 1] ⊂ R with usual topology. And Y = {0, 1} discrete topology. Let
f (x) = 0 for all x. Then f (0) = 0 but f (0) is not the limit point of f ((0, 1]).
Problem 8, Page 111
Let Y be an ordered set in the order topology. Let f, g : X → Y be continuous.
(a)Show that the set {x | f (x) ≤ g(x)} is closed in X.
(b)Let h : X → Y be the function
h(x) = min{f (x), g(x)}.
Show that h is continuous.
Proof:
a) We prove that {x : f (x) > g(x)} = X − {x : f (x) ≤ g(x)} is open in X.
Given x0 ∈ {x : f (x) > g(x)}
1) If {y : f (x0 ) > y > g(x0 )} is not empty, then choose
y0 ∈ {y : f (x0 ) > y > g(x0 )}. Thus, the set U1 = f −1 ((y0 , ∞)) ∩
g −1 ((−∞, y0 )) contains x0 . Since f and g are both continuous. U1 is
opn in X. For ∀x ∈ U , we have f (x) > y0 > g(x).
So U1 ⊂ {x : f (x) > g(x)}
2) If {y : f (x0 > y > g(x0 ))} is empty.
4
Set U2 = f −1 ((g(x0 ), ∞)) ∩ g −1 ((−∞, f (x0 ))). So U2 is an open set in
X, containing x0 . For ∀x ∈ U2 , f (x) > g(x0 ) and g(x) < f (x0 ). Since
{y : f (x0 ) > y > g(x0 )} is empty, we have f (x) ≥ f (x0 ). It follows
f (x) > g(x). So x ∈ {x : f (x) > g(x)}. Therefore, {x | f (x) > g(x)}
is open in X. So x | f (x) ≤ g(x) is closed in X.
b) Set X1 = {x | f (x) ≥ g(x)}, X2 = {x | f (x) ≤ g(x)}. So X = X1 ∪ X2 . By
the conclusion of Part a, we have both X1 and X2 are closed. Moreover,
h|X1 = g(x) and h|X2 = f (x), both of which are continuous. Additionally,
X1 ∩ X2 = {x | f (x) = g(x)}, so h|X1 ∩X2 = g|X1 ∩X2 . By pasting Lemma,
h is continuous.
Problem 10, Page 111
Let f : A → B and g : C → D be continuous functions. Let us define a map
f × g : A × B → C × D by equation:
(f × g)(a × c) = f (a) × g(c)
Show that f × g is continuous.
Proof:Given U × V ⊂ B × D, where U is open in B and V is open in D.
Then (f × g)−1 (U × V ) = {(a, c) ∈ A × C | f (a) ∈ U and g(c) ∈ V }. That is,
(f × g)−1 (U × V ) = {(a, c) ∈ A × C | a ∈ f −1 (U ) and c ∈ g −1 (V )}.
So (f × g)−1 (U × V ) = f −1 (U ) × g −1 (V ). Since both f and g are continuous, we
have f −1 (U ) and g −1 (V ) are both open. So f −1 (U ) × g −1 (V ) are open in A × C.
So f × g is continuous.
Problem 12, Page 111
Let F : R × R → R be defined by the equation
(
xy/(x2 + y 2 ), x × y =
6 0 × 0,
F (x × y) =
0,
x × y = 0 × 0.
(a)Show that F is continuous in each variable separately.
(b)Compute the function g : R → R defined by g(x) = F (x × y).
(c) Show that F is not continuous.
Proof:
a) Fix y0 ∈ R. Prove that F (x, y0 ) as a map from R → R is continuous.
• If y0 = 0 then F (x, y0 ) = 0 for ∀x ∈ R. So F (x, y0 ) is continuous.
0
• If y0 6= 0, then F (x, y0 ) = x2xy
for ∀x ∈ R. Since xy0 and x2 + y02
+y02
0
are both continuous in R, we have F (x, y0 ) = x2xy
is continuous.
+y02
Similarly, if fix x0 ∈ R. F (x0 , y) as a map from R to R is continuous
as well.
Therefore, F is continuous in each variable separately.
5
b) As y = x
(
g(x) = F (x, y) =
1
2
0
x 6= 0
x = 0.
c) By the part b, we have along line x = y, F (x, y) is not continuous at (0, 0).
So F (x, y) is not continuous at (0, 0).