Download Exercises on Separation Axioms, Products and Quotients

Exercises on Separation Axioms, Products and Quotients
(3.23) Prove that
Normal =⇒ regular =⇒ T2 =⇒ T1 =⇒ T0 .
Proof: Normal =⇒ regular. That is, every normal space is a regular space.
Suppose that X is a normal space. We show that X is regular by verifying the definition
of a regular space. Let C be a closed set in X and let x ∈ X − C. Since X is normal and by
the definition of a normal space, D = {x} is closed. By the definition of a normal space again,
there are disjoint open sets U and V such that C ⊆ U and D ⊆ V . As D = {x}, it follows
that there are disjoint open sets U and V such that C ⊆ U and x ∈ V . By the definition of a
regular space, X is regular.
regular =⇒ T2 . That is, every regular space is a T2 space.
Suppose that X is a regular space. We show that X is T2 by verifying the definition of a
T2 space. Let x, y be two distinct points in X. Since X is regular and by the definition of a
regular space, C = {y} is closed. Note that as x 6= y, x 6∈ C. By the definition of a regular
space, there are disjoint open sets U and V such that x ∈ U and C ⊆ V . As C = {y}, it follows
that there are disjoint open sets U and V such that x ∈ U and y ∈ V . By the definition of a
T2 space, X is T2 .
T2 =⇒ T1 . That is, every T2 space is a T1 space.
Suppose that X is a T2 space. We show that X is T1 by verifying the definition of a T1
space. Let x, y be two distinct points in X. Since X is T2 and by the definition of a T2 space,
there are disjoint open sets U and V such that x ∈ U and y ∈ V . As U ∩ V = ∅ (or as X and
V are disjoint), x 6∈ V and y 6∈ U . It follows by the definition of a T1 space that X is T1 .
T1 =⇒ T0 . That is, every T1 space is a T0 space.
(3.27) If X is a compact topological space and Z is a Hausdorff space, with a continuous
one-to-one function f : X 7→ Z onto Z, then f is a homeomorphism.
Proof: Since f : X 7→ Z is a one-to-one and onto function, the inverse function g = f −1 :
Z 7→ X exists. By the definition of a homeomorphism, it suffices to show that g : Z 7→ X if
continuous.
By Exercise 3.9, it suffices to show that if C is a closed set in X, then g −1 (C) is closed in
Z. Since f is a bijection, f −1 (y) ∈ C if and only if for some x ∈ C, y = f (x). Thus by the
definition of g −1 , g −1 (C) = {y ∈ Z : f −1 (y) ∈ C} = {y ∈ Z : ∃x ∈ C, y = f (x)} = f (C).
Now let C denote an arbitrary closed set in X. By Theorem 3.18, C is also compact. By
Theorem 3.19, f (C) as a topological subspace in Z is compact. By Theorem 3.25, and since Z
is T2 , g −1 (C) = f (C) is closed in Z.
If f (C) = Z, then g −1 (C) = f (C) = Z is closed in Z. Hence we assume that f (C) 6= Z, and
so Z − f (C) 6= ∅. To show that g −1 (C) = f (C) is closed, it suffices to show that Z − f (C) is
open in Z, or equivalently, that for every z ∈ Z − f (C), Z has an open set N such that z ∈ N
and N ⊆ Z − f (C).
Pick an arbitrarily z ∈ Z − f (C). Since Z is T2 , for this fixed z and for each y ∈ f (C),
there are disjoint open sets Uy , Vy such that z ∈ Uy and y ∈ Vy . Since y ∈ Vy , it follows
that O = {Vy : y ∈ f (C) is an open cover of f (C). Since f (C) is compact, O has a finite
subcover {Vy1 , Vy2 , · · · , Vym } of f (C). Since intersection of finitely many open sets is also open,
m
m
Uz = ∩m
i=1 Uyi is open. Since for each i, Uyi ∩Vyi = ∅, and since Uz = ∩i=1 Uyi , Uz ∩(∪i=1 Vyi = ∅.
−1
It follows by the definition of open sets that Z − f (C) is open. This proves that g (C) = f (C)
is closed. By Exercise 3.9, g = f −1 is continuous and so by the definition of a homeomorphism,
f : X 7→ Z is a homeomorphism.
(3.29) Show that the product of two Ti spaces is a Ti space, for i = 0, 1, 2.
Proof: Firstly, we show that if X and Y are both T0 spaces, then X × Y is also a T0 space.
Let (x1 , y1 ) and x2 , y2 ) be two distinct points in X × Y . Then either x1 6= x2 or y1 6= y2 .
Without loss of generality, we assume that x1 6= x2 . Since X is T0 , X has an open set U
containing one of x1 and x2 but not both. Without loss of generality, we assume that x1 ∈ U
and x2 6∈ U . By the definition of product spaces and since Y is always open in Y , U × Y is
open in X × Y such that (x1 , y1 ) ∈ U × Y but (x2 , y2 ) 6∈ U × Y . By the definition of a T0 space,
X × Y is T0 .
Then, we show that if X and Y are both T1 spaces, then X × Y is also a T1 space.
Let (x1 , y1 ) and x2 , y2 ) be two distinct points in X × Y . Then either x1 6= x2 or y1 6= y2 .
Without loss of generality, we assume that x1 6= x2 . Since X is T1 , X has open sets U and V
such that x1 ∈ U −V and x2 ∈ V −U . By the definition of product spaces and since Y is always
open in Y , both U × Y and V × Y are open in X × Y . As x1 ∈ U − V , (x1 , y1 ) ∈ U × Y − V × Y .
Similarly,(x2 , y2 ) ∈ V × Y − U × Y . By the definition of a T1 space, X × Y is T1 .
Finally, we show that if X and Y are both T2 spaces, then X × Y is also a T2 space.
Let (x1 , y1 ) and x2 , y2 ) be two distinct points in X × Y . Then either x1 6= x2 or y1 6= y2 .
Without loss of generality, we assume that x1 6= x2 . Since X is T2 , X has disjoint open sets U
and V such that x1 ∈ U , x2 ∈ V . By the definition of product spaces and since Y is always
open in Y , both U × Y and V × Y are open in X × Y . As x1 ∈ U , (x1 , y1 ) ∈ U × Y . Similarly,
(x2 , y2 ) ∈ V × Y . Since U ∩ V = ∅, we also have (U × Y ) ∩ (V × Y ) = ∅. By the definition of a
T2 space, X × Y is T2 .
(3.33) Describe the space X/ ∼ for the following spaces and equivalence relations:
(1) Find I/ ∼ for X = I = [0, 1] and the equivalence classes defined by
(
[x] =
{x}
0∼1
if 0 < x < 1
.
if x = 0, 1
(2) X is the unit square {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the equivalence classes defined by
(
[(x, y)] =
{(x, y)}
(0, y) ∼ (1, 1 − y)
if x 6= 0, 1 and 0 ≤ y ≤ 1
.
if x = 0, 1 and 0 ≤ y ≤ 1
(3) X is the disc D2 = {(x, y) : x2 + y 2 ≤ 1}. Let S 1 denote the boundary circle S1 = {(x, y) :
x2 + y 2 = 1}. Define equivalence classes of a point (x, y) by
(
[(x, y)] =
{(x, y)}
(x, y) ∼ (1, 0)
if (x, y) 6∈ S 1
.
if (x, y) ∈ S 1
(4) X = S1 = {(x, y) : x2 + y 2 = 1} and (x, y) sin(−x, −y) for each x, y ∈ S 1 .
Solutions: (1) Let f : I/ ∼7→ S 1 = {(x, y), x2 + y 2 = 1} given by f (x) = (cos(x), sin(x)).
Then f is a homeomorphism (need to verify this fact) and so I/ ∼ is homeomorphic to S 1 .
(2) This is the Mobius strip. (This can be done by first twisting the side {(1, y) : 0 ≤ y ≤ 1}
180o and then identifying the twisted side with the untwisted side {(0, y) : 0 ≤ y ≤ 1}, resulting
a Mobius strip.
(3) This is homeomorphic to the sphere S 2 = {(x, y, z) : x1 + y 2 + z 2 = 1}. Basically the space
can be obtained from D2 by identifying the boundary S 1 into a single point, resulting a sphere.
(4) This is still homeomorphic to S 1 . (It can be done by first identifying the two points (1, 0)
and (−1, 0), which S 1 becomes a figure 8. Then twist the upper half circle of the figure 8 180o
and then folded to be identified with the untwisted lower half circle. This results again a circle,
and so is homeomorphic to S 1 ).