Download THE PRODUCT TOPOLOGY Contents 1. The Product Topology 1 2

THE PRODUCT TOPOLOGY
GILI GOLAN
Abstract. In this paper we introduce the product topology of an arbitrary
number of topological spaces. We define the separation axioms and characterize the Tychonoff Spaces as those which can be embedded in a cube. We also
prove a sufficient condition for a space to be metrizable.
Contents
1. The Product Topology
2. Tychonoff’s Theorem
3. Separation Properties
4. Metrizability
References
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2
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1. The Product Topology
Definition 1.1. Let {(XiQ
, τi ) | i ∈ I} be a family of topological
Q spaces. The
product space, denoted by i∈I (Xi , τi ), consists of the product set i∈I Xi (whose
elements are the choice functions on {Xi | i ∈ I}) with the topology τ whose basis
is the family
Y
B = { Ui | Ui ∈ τi and Ui = Xi f or all but a f inite number of i0 s }
i∈I
Q
Lemma 1.2. Let {(Xi , τi ) | i ∈ I} be a family of topological
Q spaces and i∈I (Xi , τi )
be the Q
corresponding product space. The projections pj : i∈I (Xi , τi ) → Xj defined
by pj ( i∈I xi ) = xj are continuous open surjective maps.
Proof. Let j ∈ I be. It is clear that pj is onto. We shall prove first that pj is
Q
continuous. Let U ⊆ Xj be an open set then p−1
j (U ) =
i∈I Ui where Uj = U and
for all i 6= j Ui = Xi . Therefore, since p−1
(U
)
belongs
to
the
basis of the topology
j
Q
of i∈I (Xi , τi ), it is open and pj is continuous.
Therefore
it
remains
to show that
Q
pj is open. Let V be a set in the basis
of
(X
,
τ
).
It
suffices
to
show that
i i
i∈I
Q
pj (V ) is open. But V ∈ B ⇒ V = i∈I Ui where for all i ∈ I Ui is open in Xi .
Therefore pj (V ) = Uj is open in Xj and pj is an open mapping.
Proposition 1.3. Let {(Xi , τi ) | i ∈QI} be a family of topological spaces and f a
mapping of a topological space Y into i∈I (Xi , τi ), then f is continuous if and
Q only
if for each j ∈ I, pj ◦ f is continuous, where pj denotes the projection of i∈I Xi
onto Xj .
Date: June 5th , 2011.
Key words and phrases. Product Topology, Tychonoff’s Theorem.
1
2
Proof. (⇒) f is continuous, therefore, since by Lemma 1.2 pj is continuous for all
j ∈ I, pj ◦ f is continuous
for all j ∈ I.
Q
Q
(⇐) Let V = i∈I Ui ∈ B be a set in the basis of i∈I (Xi , τi ). It suffices to
Q
show that f −1 (V ) is open. V = i∈I Ui ∈ B ⇒ Ui ∈ τi for all i ∈ I and there
existsQa finite subset JT⊆ I such that Ui T
= Xi for all i ∈ I \ J. Therefore
f −1 (V ) =
T
−1
−1
−1
−1
−1
f ( i∈I Uj ) = f ( j∈J pj (Uj )) = j∈J f (pj (Uj )) = j∈J (pj ◦f )−1 (Uj ).
But, for all j ∈ J pj ◦ f is continuous ⇒ (pj ◦ f )−1 (Uj ) is open. Therefore f −1 (V )
is open as a finite intersection of open sets.
Definition 1.4. We say that a family of functions {fi : X → Yi |i ∈ I} from a
topological space X to topological spaces Yi separates points if for every x 6= y ∈ X
there exists i ∈ I such that fi (x) 6= fi (y).
Definition 1.5. We say that a family of functions {fi : X → Yi | i ∈ I} from
a topological space X to topological spaces Yi separates points and closed sets if
for every closed subsetset A ⊆ X and any x ∈ Ac , there exists i ∈ I such that
fi (x) ∈
/ fi (A).
Lemma 1.6 (The Embedding Lemma). Let {(Yi , τi ) | i ∈ I} be a family of topological spaces and for each
Q i ∈ I let fi be a continuous mapping of (X, τ ) into (Yi , τi ).
Let
e
:
(X,
τ
)
→
i∈I (Yi , τi ) be the evaluation map; that is ∀x ∈ X, e(x) =
Q
f
(x).
Then
if
the family {fi | i ∈ I} separates points
i∈I i
Q of X and separates
points and closed sets, then e is an embedding of (X, τ ) in i∈I (Yi , τi ).
Proof. It is sufficient to prove that the mapping e : (X, τ ) → (e(X), τ 0 ) is a homeomorphism where τ 0 is the subspace topology on e(X). It is clear that e : X → e(X)
is onto while the fact that {fi | i ∈ I} separates points of X makes it one-to-one.
Since for every i ∈ I, pi ◦e = fi is a continuous function, Proposition 1.3 implies that
e is continuous as well. Therefore, it remains to show that e : (X, τ ) → (e(X), τ 0 ) is
open. Let U ∈ τ be, it suffices to prove that e(U ) ∈ τ 0 . Let y ∈ e(U ) be. We need to
show that there exists a W ∈ τ 0 such that y ∈ W ⊆ e(U ). But y ∈ e(U ) ⇒ ∃x ∈ U
such that y = e(x). x ∈ U , U is an open set ⇒ U c is closed and x ∈
/ U c . Therefore,
since {fi | i ∈ I} separates points and closed sets, there exists a mapping fj such
c
c
that fj (x) ∈
/ fj (U c ). Put W = p−1
j ((fj (U )) ) ∩ e(X). Since pj is continuous and
c
(fj (U c )) ⊆ Yj is open W is an open subset of e(X). Clearly e(x) ∈ W . It remains
to show that W ⊆ e(U ). Let y ∈ W . There exists t ∈ X so that y = e(t). Now,
c
fj (t) ∈ (fj (U c )) ⇒ fj (t) ∈
/ fj (U c ). In particular, fj (t) ∈
/ fj (U c ) ⇒ t ∈
/ Uc ⇒
t ∈ U . So y = e(t) ∈ e(U ) and W ⊆ e(U ).
2. Tychonoff’s Theorem
Definition 2.1. Let X be a topological space. A ⊆ X is said to be compact if
every open covering of it has a finite subcovering. More explicitly
A is compact if
S
for every family of open sets {U
|
i
∈
I}
such
that
A
⊆
U
there
is a finite
i
i
i∈I
S
subset J ⊆ I such that A ⊆ i∈J Ui . If X ⊆ X is compact we say that X is a
compact topological space.
Lemma 2.2. Let (X, τ ) be a compact space and A ⊆ X be a closed subset, then A
is compact.
3
Proof. Let U = {Ui | i ∈ I} be an open covering of A then U1 = {Ui | i ∈ I} ∪ {Ac }
is an open covering of X. Since X is compact there exists a finite subcovering of
X, V1 ⊆ U1 . But then V := V1 \ {Ac } ⊆ U is a finite subcovering of A and A is
compact.
Theorem 2.3 (Tychonoff’s Theorem).
Let {(Xi , τi ) | i ∈ I} be a family of topoQ
logical spaces. The product space i∈I (Xi , τi ) is compact if and only if for each
i ∈ I (Xi , τi ) is compact.
Definition 2.4. Let A be a set and for each a ∈ A let (Xa , τa ) be a topological
space
homeomorphic to [0, 1] with its standard topology, then the product space
Q
A
(X
and refered to as a cube.
a , τa ) is denoted I
a∈A
Corollary 2.5. For any set A, The cube I A is compact.
Proposition 2.6. Let (X, d) be a metric space. Then (X, d) is homeomorphic to
a subspace of the cube I X .
Proof. Without loss of generality for all a and b in X d(a, b) ≤ 1. For each a ∈ X
let fa : X → [0, 1] be the continuous function fa (x) = d(a, x). Let x 6= y ∈ X be.
Then fx (y) = d(x, y) 6= 0 while fx (x) = d(x, x) = 0, therefore fx (y) 6= fx (x) and
The family {fa | a ∈ X} separates points. Moreover, given a closed set A ⊆ X
and an element x ∈ Ac , since Ac ⊆ X is open, there exists an 0 < < 1 such that
B(x, ) ⊆ Ac . Therefore for each a ∈ A d(x, a) ≥ . Therefore fx (A) ⊆ [, 1] and
fx (x) = 0 ⇒ fx (x) ∈
/ fx (A) ⊆ [, 1]. Therefore {fa | a ∈ X} separates points and
closed sets as well. By The Embedding Lemma (Lemma 1.6) the evaluation map
from X to I X is an embedding which means X is homeomorphic to a subspace of
the cube I X .
3. Separation Properties
Definition 3.1. Let (X, τ ) be a topological space. X is said to be a T0 space if
for every two distinct elements x, y ∈ X there exists an open set U ∈ τ such that
only one of the elements x, y is an element of U .
Definition 3.2. Let (X, τ ) be a topological space. X is said to be a T1 space if
for each elemnt x ∈ X and any element y 6= x, there exists a set U ∈ τ such that
x ∈ U and y ∈
/ U.
Lemma 3.3. A topological space (X, τ ) is a T1 space iff each singleton {x} ⊆ X
is a closed subset of X.
Proof. (⇒) Let X be a T1 space and let x be an element of X. It suffices to show
that {x}c ∈ τ . let y ∈ {x}c then y 6= x. There exists an open set U ∈ τ such that
y ∈ U and x ∈
/ U which means y ∈ U ⊆ {x}c . Therefore {x}c ∈ τ .
(⇐) Let x ∈ X be and let y 6= x be. Then {y}c ∈ τ and x ∈ {y}c , y ∈
/ {y}c .
Definition 3.4. A topological space (X, τ ) is said to be T2 or a Hausdorff space
iff for each x 6= y ∈ X there exist U, V ∈ τ such that x ∈ U , y ∈ V and U ∩ V = ∅.
Remark 3.5. T2 ⇒ T1 ⇒ T0 .
Lemma 3.6. Let (X, d) be a metric space then (X, τ ) is a Hausdorff space, where
τ is the induced topology.
4
Proof. Let x 6= y ∈ X be. Denote = d(x, y) > 0. Then x ∈ B(x, 2 ), y ∈ B(y, 2 ),
B(x, 2 ), B(y, 2 ) ∈ τ and B(x, 2 ) ∩ B(y, 2 ) = ∅
Proposition Q
3.7. Let {(Xi , τi ) | i ∈ I} be a family of Hausdorff spaces then the
product space i∈I (Xi , τi ) is a Hausdorff space.
Q
Q
Q
Proof. Let x = i∈I xi , y = i∈I yi be two distinct elements in i∈I Xi . There
exists a j ∈ I such that xj 6= yj . Since Xj is a Hausdorff space, there exist open
subsets U, V ⊆ Xj such that xj ∈ U , yj ∈ V and U ∩V = ∅. Note that x ∈ pj −1 (U ),
y ∈ pj −1 (V ), pj −1 (U ) ∩ pj −1 (V ) = pj −1 (U ∩ V ) = pj −1 (∅) = ∅. Since by Lemma
−1
−1
1.2 pj is a continuous
Q function pj (U ) and pj (V ) are open sets which separate
x and y. Therefore i∈I (Xi , τi ) is a Hausdorff space.
Definition 3.8. Let (X, τ ) be a topological space. X is said to be regular if for
each closed subset A ⊂ X and any element x ∈
/ A there exists a pair of disjoint
open sets U, V ∈ τ such that x ∈ U and A ⊆ V . If X is also a T2 space X is said
to be a T3 space.
Remark 3.9. T3 ⇒ T2 ⇒ T1 ⇒ T0
Lemma 3.10. Let (X, τ ) be a topological space. Then X is regular iff for each
U ∈ τ and x ∈ U there exists an open set V such that x ∈ V ⊆ V ⊆ U .
Proof. (⇒) Assume X is regular, let U ∈ τ and x ∈ U be. U c is closed and
x∈
/ U c , therefore there exist disjoint open sets V, W such that x ∈ V and U c ⊆ W .
V ∩ W = ∅ therefore V ⊆ W c . U c ⊆ W ⇒ W c ⊆ U . Therefore V ⊆ W c ⊆ U .
But W c is closed ⇒ V ⊆ W c therefore x ∈ V ⊆ V ⊆ U .
(⇐) Assume that for each U ∈ τ and x ∈ U there exists an open set V such that
x ∈ V ⊆ V ⊆ U . Let A ⊂ X be a closed subset of X and x ∈
/ A be, then x ∈ Ac
and Ac ∈ τ . Therefore there exists an open set V such that x ∈ V ⊆ V ⊆ Ac .
c
Denote U = V then U ∈ τ , A ⊆ U , and U ∩ V = ∅. Therefore U and V separate
A and x and X is regular.
Definition 3.11. Let (X, τ ) be a topological space. X is said to be completely
regular if for each closed subset A ⊂ X and any element x ∈
/ A there exists a
continuous function f : X → [0, 1] such that f (x) = 0 and f (a) = 1 for all a ∈ A.
If X is also a T2 space X is said to be a Tychonoff Space or a T3 21 space.
Remark 3.12. T3 21 ⇒ T3
Proposition 3.13. Let (X, d) be a metric space. Then (X, τ ) is a Tychonoff space
where τ is the induced topology.
Proof. Since by Lemma 3.6 X is a Hausdorff space, is suffices to prove that (X, τ )
is completely regular. Let A ⊂ X be a closed subset and x ∈ Ac be. It suffices
to show that there is a continuous function f : X → [0, 1] such that f (x) = 0 and
f (a) = 1 for all a ∈ A. x ∈ Ac and Ac is an open set, therefore there exists an
> 0 such that B(x, ) ⊆ Ac Define f : X → [0, 1] by f (y) = min{1, d(x,y)
} for all
y ∈ X. Then f is continuous, f (x) = 0 and f (A) ⊆ {1}.
Proposition 3.14. Let {(XQ
i , τi ) | i ∈ I} be a family of completely regular spaces
then the product space X = i∈I (Xi , τi ) is completely regular.
5
Q
Q
Proof. Let A ⊂ X be a closed subset and x = i∈I xi ∈ i∈I Xi be an element
c
c
c
of
1.1 of the topology of
Q A . x ∈ A and A is open, therefore by Definition Q
c
X
,
there
exists
a
finite
subset
J
of
I
such
that
x
∈
i
i∈I
i∈I Ui ⊆ A , Ui ∈ τi
for all i ∈ I and Ui = Xi for all i ∈ I \ J. As (Xj , τj ) is completely regular for
all j ∈ J there exist continuous mappings fj : Xj → [0, 1] such that fj (xj ) = 0
c
and fj (UQ
j ) ⊆ {1}. Since the projection pj is continuous for all j, the mappings
fj ◦ pj : i∈I Xi → [0, 1]
Q are continuous. Put f = max{fj ◦ pj (x)Q| j ∈ J} then f
is continuous. for x = i∈I xi , f (x) = 0. For every a ∈ A, a ∈
/ i∈I Ui ⇒ there
exists j1 ∈ J such that a ∈
/ Uj1 . Therefore fj1 (a) = 1 and f (a) = 1 as well.
Corollary 3.15.
Q Let {(Xi , τi ) | i ∈ I} be a family of Tychonoff spaces then the
product space i∈I (Xi , τi ) is a Tychonoff space.
Corollary 3.16. For every set X the cube I X is a Tychonoff space.
Proposition 3.17. Let (X, τ ) be a Tychonoff space, Then X can be embedded in
a cube.
Proof. Let F be the family of all continuous finctions f : X → [0, 1]. Since (X, τ )
is a Tychonoff Space the family F separates points and closed sets. Since X is
Hausdorff and therefore T1 every singleton {x} is a closed subset. Therefore the
family F separates points as well. According to the Embedding Lemma (Lemma
1.6), (X, τ ) can be embedded in the cube [0, 1]F .
Lemma 3.18. Every subspace of a Tychonoff space is also a Tychonoff space.
Proof. Let (X, τ ) be a Tychonoff space and (Y, τ 0 ) be a subspace of it. Let A ⊂ Y
be a closed subset of Y and x ∈ Y \ A be. A is closed in Y ⇒ there exists a closed
set P ⊆ X such that A = P ∩ Y . x ∈ Y and x ∈
/ A ⇒ x ∈
/ P . Since (X, τ )
is a Tychonoff space there exists a continuous function f : X → [0, 1] such that
f (x) = 0 and f (a) = 1 for all a ∈ P . Therefore f |Y : Y → [0, 1] is also a continuous
function and f |Y (x) = 0, f |Y (a) = 1 for all a ∈ A.
Corollary 3.19. (X, τ ) is a Tychonoff space iff it is homeomorphic to a subspace
of a cube.
Definition 3.20. Let (X, τ ) be a topological space. X is said to be normal if for
each two disjoint closed sets P, S ⊆ X there exist two disjoint open sets U, V ∈ τ
such that P ⊆ U and S ⊆ V . If X is also a T2 space X is said to be a T4 space.
Remark 3.21. T4 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0
Lemma 3.22 (Urysohn’s Lemma). Let (X, τ ) be a topological space. Then X is
normal iff for each pair of disjoint closed sets P, S ⊆ X, there exists a continuous
function f : X → [0, 1] such that f (p) = 0 for each p ∈ P and f (s) = 1 for each
s ∈ S.
Corollary 3.23. If (X, τ ) is a T4 space then it is a T3 21 space as well.
Remark 3.24. T4 ⇒ T3 12 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0
Proposition 3.25. Let (X, τ ) be a compact Hausdorff space, then (X, τ ) is normal.
6
Proof. We shall show first that X is regular. Let A ⊂ X be a closed subset and
x ∈
/ A be. For each a ∈ A x 6= a ⇒ there exists
a pair of disjoint open sets
S
Ua , Va such that x ∈ Ua and a ∈ Va . A ⊆ a∈A Va therefore {Va | a ∈ A}
is an open covering of A. Since A is a closed subset of a compact space X, by
Lemma 2.2 A isTcompact. STherefore there exists a finiteT subcovering {VS
ai | 1 ≤
n
n
n
n
i ≤T
n} of A. Si=1 Uai , S
V
are
open
sets,
x
∈
U
,
A
⊆
a
a
i
i
i=1
i=1
i=1 Vai
Sn
Tn
n
n
n
and
U
∩
V
⊆
U
∩
V
=
∅
=
∅.
Therefore
U
a
a
a
a
i
i
i
i=1
i=1
i=1
i=1 ai and
Sn i=1 i
V
separate
x
and
A
and
X
is
regular.
a
i
i=1
Now, we can prove that X is normal. Let A, B be disjoint closed sets. For each
a ∈ ASthere exist a pair of disjoint open sets Ua , Va such that a ∈ Ua and B ⊆ Va .
A ⊆ a∈A Ua therefore {Ua | a ∈ A} is an open covering of A. Since A is a closed
subset of a compact space X, by Lemma 2.2SA is compact.
Tn Therefore there exists
n
aSfinite subcovering
{U
|
1
≤
i
≤
n}
of
A.
U
,
a
a
i
Tn
Sn
Tn i=1 i Sn i=1 Vai are open
Snsets, A ⊆
n
U
,
B
⊆
V
and
U
∩
V
⊆
U
∩
V
=
a
a
a
a
a
a
i
iT
i
i
i
i
i=1
i=1
i=1
i=1
i=1
i=1 ∅ = ∅.
Sn
n
Therefore i=1 Uai and i=1 Vai separate A and B and X is a normal space. Proposition 3.26. Let (X, d) be a metric space, then (X, τ ) is a T4 space, where
τ is the induced topology.
Proof. By Lemma 3.6 X is a Hausdorff space. Therefore it remains to prove that
X is normal. Let A, B be disjoint closed subsets of X. A ⊆ B c and B c is open
⇒ for each a ∈ A there exists an a > 0 such that B(a, a ) ⊆ B c . B ⊆ Ac and
Ac is open
an b > 0 such that B(b, b ) ⊆ Ac . Put
S ⇒ foraeach b ∈ BSthere exists
b
U := a∈A B(a, 2 ), V := b∈B B(b, 2 ). Then U, V ∈ τ , A ⊆ U and B ⊆ V .
It remains to show that U ∩ V = ∅. Assume by contradiction that U ∩ V 6= ∅
and let x ∈ U ∩ V be. There exist a ∈ A and b ∈ B such that x ∈ B(a, 2a ) and
x ∈ B(b, 2b ). WLOG a ≥ b , then d(a, b) ≤ d(a, x) + d(x, b) < 2a + 2b ≤ 2 2a = a
⇒ b ∈ B(a, a ) by contradiction to B(a, a ) being a subset of B c .
4. Metrizability
Definition 4.1. A topological space (X, τ ) is said to be metrizable if there is a
metric d on X such that the topology induced by d is equal to τ .
Definition 4.2. A topological space (X, τ ) is said to be second countable if τ has
a countable basis.
Proposition 4.3. Let (X, τ ) be a second countable T4 space, then X metrizable.
Proof. Since Hilbert’s cube I ∞ is metrizable it suffices to show that X can be
embedded in it. By the Embedding Lemma (Lemma 1.6) it is enough to find a
countable family F of continuous functions f : X → [0, 1] such that F separates
points and closed sets. (Note that since X is Hausdorff and therefore T1 , if F
separates points and closed sets it separates points as well). Let B be a countable
basis of τ and let S be the set of all the pairs (V, U ) such that U, V ∈ B and V ⊆ U .
Since B is countable, S ⊆ B2 is countable as well. Now, for each pair (V, U ) ∈ S,
by Urysohn’s Lemma (Lemma 3.22) there is a continuous function fV U : X → [0, 1]
such that f (x) = 0 for each x ∈ V and f (y) = 1 for each y ∈ U c . Let F be the
family of functions thus obtained, then F is countable. It suffices to show that F
separates points and closed sets. Therefore, let x ∈ X and A be a closed set such
that x ∈
/ A. x ∈ Ac ⇒ ∃U ∈ B such that x ∈ U ⊆ Ac . Since X is regular, by
Lemma 3.10 there exists an open set P such that x ∈ P ⊆ P ⊆ U . P ∈ τ ⇒
7
there exists a set V ∈ B such that x ∈ V ⊆ P . But V ⊆ P ⇒ V ⊆ P . Therefore
x ∈ V ⊆ V ⊆ U and (V, U ) ∈ S. Now, fV U (A) ⊆ fV U (U c ) ⊆ {1} while fV U (x) = 0.
Therefore, F separates x and A.
Lemma 4.4. Let (X, τ ) be a second countable regular space. Then X is normal.
Proof. Let A, B be two disjoint closed subsets of X and B be a countable basis of
τ . Since B c is open and X is regular, for each a ∈ A there exists a set Ua ∈ B such
the elements of the set
that a ∈ Ua ⊆ Ua ⊆ B c . B is countable, therfore we can list
S∞
{Ua | a ∈ A} thus obtained by Ui , i ∈ N. That is, A ⊆ i=1 Ui and Ui ∩ B = ∅
for all i ∈ N. WLOG we can assume that the sequence {Ui }i∈N is an ascending
sequence, since otherwise we can switch our attention to the ascending sequence
U1 ⊆ U1 ∪ U2 ⊆ U1 ∪ U2 ∪ U3 ... which has the same properties.
S∞ Similarly, there is
an ascending sequence of open sets {Vi }i∈N such that B ⊆ i=1 Vi and Vi ∩ A = ∅.
Now, define two
open sets: Wn := Un \ Vn and Sn := Vn \ Un
S∞other sequences of
S∞
and put U = n=1 Wn and V = n=1 Sn . It suffices to show that the open sets
U and V separate A and B. For each a ∈ A, there exists i ∈ I such that a ∈ Ui
and Vi ∩ B = ∅. Therefore a ∈ Wi ⊆ U and A ⊆ U . Similarly B ⊆ V . Therefore,
it remains to show that U ∩ V = ∅. Assume by contradiction that there exists an
element x ∈ U ∩ V . x ∈ U ⇒ x ∈ Wn for some n ∈ N. Therefore x ∈ Un and
x∈
/ Vn . In particular x ∈
/ Vn . Since {Vi }i∈N is an ascending sequence x ∈
/ Vi for all
i ≤ n. x ∈ V ⇒ x ∈ Sm for some m ∈ N. Therefore x ∈ Vm and x ∈
/ Um . Since for
all i ≤ n x ∈
/ Vi , m > n. But then since x ∈ Un , x ∈
/ Um is a contradiction to {Ui }
being an ascending sequence.
Corollary 4.5 (Urysohn’s Metrization Theorem). Every second countable T3 space
is metrizable.
Proposition 4.6. Let (X, τ ) be a compact space. X is metrizable iff X is a second
countable Hausdorff space.
Proof. (⇐) X is compact Hausdorff and therefore by Lemma 3.25 X is a T4 space.
Therefore since X is also second countable, by Lemma 4.3 X is metrizable.
(⇒) Since X is metrizable, by Proposition 3.6 X is a T2 space. Therefore it suffices
to prove that X is second countable. For this we will construct a countable basis
of τ . Led d be a metric which induces τ , then for each n ∈ N and a ∈ X the open
ball B(a, n1 ) ∈ τ . Note that for each n ∈ N, Cn = {B(a, n1 ) | a ∈ X} is an open
covering of X. Since X is compact, for eachSn ∈ N Cn has a finite subcovering
∞
FC n = {B(ani , n1 ) | 1 ≤ i ≤ kn }. Let B = n=1 FC n , then B is countable as a
countable union of finite sets. It suffices to prove that B is a basis of τ . Note that
each set in B is open. Let U ∈ τ be and let x be an element of U . It suffices to
show there exists a set W ∈ B such that x ∈ W ⊆ U . But x ∈ U , U ∈ τ ⇒
∃n ∈ N such that B(x, n1 ) ⊆ U . Since FC 2n covers X there exists a2ni ∈ X such
1
1
1
) ∈ FC 2n . But x ∈ B(a2ni , 2n
) ⇒ d(a2ni , x) < 2n
⇒ for
that x ∈ B(a2ni , 2n
1
1
1
1
all z ∈ B(a2ni , 2n ) d(x, z) ≤ d(x, a2ni ) + d(a2ni , z) < 2n + 2n = n . Therefore
1
1
B(a2ni , 2n
) ⊆ B(x, n1 ) and for W = B(a2ni , 2n
), W ∈ B and x ∈ W ⊆ U .
References
[1] Sydney A. Morris, Topology Without Tears, 2011.
[2] D. Leibovich, Set Topology. The Open University 2007.