Download THE INTERSECTION OF TOPOLOGICAL AND METRIC SPACES

THE INTERSECTION OF TOPOLOGICAL AND METRIC
SPACES
JAMES MAXWELL CLARKE
Abstract. In this paper we will prove the Urysohn Meterization Theorem,
which gives sufficient conditions for a topological space to be metrizable. In the
process of proving this theorem, we will the discuss separation and countability
axiom of topological spaces and prove the Urysohn Lemma.
Contents
1. The Separation Axioms
2. The Urysohn Lemma
3. A Useful Metric Space
4. The Metrization Theorem
Acknowledgments
References
1
3
5
6
8
8
1. The Separation Axioms
Before we begin investigating the relation between Metric spaces and Topological
spaces we will discuss the separation axioms.
Definition 1.1. A topological space X is Hausdorff if for all points x, y ∈ X, There
exists open sets Ox , Oy ⊂ X such that x ∈ Ox , y ∈ Oy and Ox ∩ Oy = ∅.
Hausdorff spaces are very familiar to us. Any finite dimensional real space is
Hausdorff for example, in fact, any metrizable space is Hausdorff, which we now
quickly show:
Take a metric space X with the metric topology defined on it. Pick any two
distinct points x, y and set = d(x,y)
2 . Take the open sets B (x), B (y). We claim
that B (x) ∩ B (y) = ∅. Assume that there exists a point p such that p ∈ B (x)
and p ∈ B (y). Then,
d(x, p) + d(p, y) < 2 = d(x, y)
violating the triangle inequality. Thus all metric spaces are Hausdorff. To answer the question of whether all Hausdorff spaces are metrizable we will use the notion of normality, defined later in this
section.
1
2
JAMES MAXWELL CLARKE
Definition 1.2. A topological space X is regular if for all closed sets C and points
p contained in X such that p ∈
/ C, there exists open sets O1 , O2 ⊂ X such that
C ⊂ O1 , p ∈ O2 and O1 ∩ O2 = ∅.
Being regular is a stronger condition on a space than being Hausdorff.
Definition 1.3. A topological space X is normal iff for all disjoint closed sets
C1 , C2 ⊂ X, there exists open U1 , U2 such that C1 ⊂ U1 and C2 ⊂ U2 and U1 ∩U2 =
∅.
We can show that all metric spaces are normal.
Proof. Take a topological space X with the metric topology defined upon it and two
disjoint closed sets C1 , C2 . Define d(C1 , C2 ) = inf{d(x1 , x2 ) | x1 ∈ C1 , x2 ∈ C2 }.
We now define the open sets:
[
U1 =
B d(C1 ,C2 ) (x)
x∈C1
U2 =
[
x∈C2
3
B d(C1 ,C2 ) (x)
3
We can see that U1 ∩ U2 = ∅ as follows. Assume there exists a point y ∈ U1 ∩ U2 .
Then, for some x1 ∈ C1 and x2 ∈ C2 , y ∈ B d(C1 ,C2 ) (x1 ) ∩ B d(C1 ,C2 ) (x2 ). But this
3
3
means that d(y, x1 ) < d(C13,C2 ) and d(y, x2 ) < d(C13,C2 ) . So, 2d(C31 ,C2 ) > d(y, x1 ) +
d(y, x2 ) ≥ d(x1 , x2 ) ≥ d(C1 , C2 ), an absurdity. So U1 ∩ U2 = ∅. So we have
separated the C1 and C2 with disjoint open sets.
Naturally, we wish to know whether all normal spaces are metrizable. To answer
this we need to look at the countability axioms. We will see that some of them are
necessary conditions for a topological space to metrizable.
Definition 1.4. A topological space X is first countable if for each point p ∈ X,
there exists a countable family of open sets {Un }n∈N containing p such that for
each open set V 3 p, there exists an n such that Un ⊂ V .
We can clearly see that all metric spaces are first countable, for take a space
X with a metric topology defined on it and an arbitrary point x ∈ X. As the
countable family, take:
{B n1 (x)}n∈N
Take an open set U containing x. We know by definition that there must exist a such that B (x) ⊂ U and it is clear that we can find an n such that B n1 (x) ⊂ B (x).
So, our countable set allows us to satisfy the first countability axiom.
We now know that metric spaces must be first countable and normal.
Definition 1.5. A topological space X is second countable if there exists a countable basis for the topology of X.
THE INTERSECTION OF TOPOLOGICAL AND METRIC SPACES
3
For example, we can show that Rn is second countable. Each point p ∈ Rn has
the form (p1 , . . . , pn ). Take the set of all p ∈ Rn such that for each i ∈ {1, . . . , n}
we have pi ∈ Q. For each p, we take the countable set of open balls:
{Bq (p)}q∈Q
This collection is clearly a countable basis for Rn .
We have found some necessary conditions for a topological space to be metrizable, in particular that it must be first countable and normal. The remainder of
this paper will be dedicated to proving the Urysohn Metrization Theorem, which
gives a sufficient condition for a topological space to be metrizable.
2. The Urysohn Lemma
We now state the Urysohn Lemma, which will be curcial for the proof of the
Urysohn Metrization Theorem.
Lemma 2.1. Assume X is a topological space that is both normal and that C1 , C2
are closed sets in X. Then, there exists a continuous function f : X → [0, 1] such
that f (C1 ) = 0, f (C2 ) = 1.
Proof. We begin by picking a countable dense subset D of [0, 1]. Take
m
D = {q | q ∈ [0, 1], q = n , m, n ∈ Z}
2
Using D, we will construct a countable family of open sets {Od }d∈D ⊂ P(X) with
the following property:
(1) If d < d0 , then Od ⊂ Od0 .
We start by contructing sets satisfying (1) for the subset {0, 12 , 1}. The letter d
will be reserved for members of D.
Invoking the normality of X, we can pick disjoint open sets U, V such that C1 ⊂ U
and C2 ⊂ V . Set U = O0 and set X \ C2 = O1 .
We want to show that O0 ⊂ O1 . This follows from the equality O0 ∩ V = ∅.
To prove this, assume that there exists point p such that p ∈ O0 ∩ V . We know
O0 ∩ V = ∅, so p must be a limit point of O0 . But this means that V ∩ O0 6= ∅ by
the definition of a limit point, contradicting the defining property of V and O0 .
This equality yields O0 ∩ C2 = ∅ and thus, O0 ⊂ O1 . Moving on, we define
O 12 by noting that there exists open sets U 0 , V 0 such that O0 ⊂ U 0 and C2 ⊂ V 0
(again by normality) and setting O 21 = V . For the same reasons that O0 ⊂ O1 ,
O 12 ⊂ O1 .
We construct the rest of the open sets using induction. We fix n0 and assume
that for all n such that n < n0 , there exist open sets of the form O 2mn that satisfies
4
JAMES MAXWELL CLARKE
property (1). Then we can define the next round of sets as follows:
0
If m is even then 2mn is not in lowest terms and can be rewritten as 2m
n−1 , for
which we have already constructed an open set in the collection. If m is odd, then
O m−1
,O m+1
are already defined and satisfy (1). Note that O m−1
and X \ O m+1
2n
2n
2n
2n
are disjoint closed sets. Using normality, and the same arguments used for defined
O 21 we now pick O 2mn satisfying our wanted properties. Thus we have defined these
sets for all elements of D. We now move on to defining the desired function.
f (x) = inf{d | x ∈ Od }
Clearly, if x ∈ C1 , then x ∈ O0 and thus f (x) = 0; similarly, if x ∈ C2 then
f (x) = 1. It is also clear that the image of f is contained in [0, 1]. All that remains
is to prove that the function is continuous.
Note that sets of the form [0, a), (b, 1] form a subbasis of the normal topology
on [0, 1]. Proving that the preimage of sets of this form is open is sufficient to prove
that the preimage of any open set is open and that function is continuous.
We wish to show the equality
[
(2)
Od = f −1 ([0, a))
d<a
while noting that the union of open sets is open.
S
Take x ∈ d<a Od . Clearly, f (x) < a, and moreover, f (x) ≥ 0 for the image of f
S
does not extend outside of [0, 1] so f (x) ∈ [0, a). This proves d<a Od ⊂ f −1 ([0, a)).
Take x ∈ f −1 [0, a). f (x) ∈ [0, a), so inf{d | x ∈ Od } < a. We invoke the density
of D to say that there must
S be a p such that p ∈ D and
S inf{d | x ∈ Od } < p < a.
So x ∈ Op and thus x ∈ d<a Od . Thus f −1 ([0, a) ⊂ d<a Od . This finishes the
proof of (2).
We continue with the sets of the form (b, 1]. We prove the equality:
[
(3)
(X \ Op ) = f −1 ((b, 1])
b<p≤1
Take x ∈ X \ Op for some p such that b < p ≤ 1. If p = 1 then x ∈
/ O1 . Then
S
f (x) = 1. If p < 1 then b < p < f (x) ≤ 1 showing b<p≤1 (X \ Op ) ⊂ f −1 ((b, 1]) .
Take x ∈ f −1 (b, 1].
For some p ∈ (b, 1], f (x) = p. Invoking the density of D, we note that there must
be some q such that b < q < p. It is clear by the definition of f that x ∈
/ Oq .
Furthermore, there must be an s such that b < s < q such that Os ⊂ Oq . So,
S
x ∈ X \ Os ⊂ b<p≤1 X \ Op .
Since sets of these types formed a subbasis for [0, 1], we have proved the continuity of f (x), completing the proof of the lemma.
THE INTERSECTION OF TOPOLOGICAL AND METRIC SPACES
5
3. A Useful Metric Space
Now we will take a look at the product topology defined on Rω , the countable
cartesian product of the reals with itself.
Definition 3.1. The projection onto the factor Xβ is defined as the map
πβ : ×α∈A Xα → Xβ
where πβ ((xα )α∈A ) = xβ
The preimage of open sets in the projection map will form a subbasis for a topology we will define.
Take the families of the form Sn = {πn−1 (U ) | U is open in R}. The union
[
Sn
n∈N
will from a subbasis for the product topology in Rω . We will now define a metric
on Rω that will induce the product topology.
Recall the requirements for a space to be a metric space; namely that there must
there must exist a real-valued function d(x, y), where x, y ∈ X, such that
1) d(x, x) = 0
2) If x 6= y, d(x, y) > 0
3) d(x, y) = d(y, x)
4) d(x, y) + d(y, z) ≥ d(x, z)
We define a metric on R which is known as the standard bounded metric.
d(x, y) = min{|x − y| , 1}
ω
The metric we define on R is
D(x, y) = sup
d(xi , yi )
i
Proposition 3.2. A set U is open in the product topology of Rω iff it is open in
the metric topology defined by the standard bounded metric.
Proof. Take an open set U in the metric topology. We will show that for all points
x ∈ U there exists an open set in the product topology V such that x ∈ V ⊂ U .
Thus U is the union of sets open in the product topology and is therefore open in
the product topology.
Take an arbitrary point x ∈ U and consider the open ball of radius about x,
B (x). Take an N such that N1 < . We take the open set in the product toplogy
V = (x1 − , x1 + ) × · · · × (xN − , xN + ) × R × R × · · ·
6
JAMES MAXWELL CLARKE
and we show V ⊂ U . Take y ∈ V . We want to show D(x, y) < . It is clear that
for all k such that k ≥ N
1
1
≤
<
k
N
Therefore,
d(xi , yi )
d(x1 , y1 )
d(xN , yN )
sup
= max
,...,
i
1
N
o
n
But, from our definition of V , max d(x11,y1 ) , . . . , d(xNN,yN ) < . Therefore, y ∈
B (x). So, every open set in the metric topology is an open set in the product
topology. We next prove the converse statement.
Consider a basis element U of the product topology. It has the form
Y
Uα
U=
α∈N
where for some finite list A = {α1 , . . . , αN }, Uαi is an arbitrary open set in R and for
α∈
/ A, Uα = R. Take x ∈ U . We wish to find a set V open in the metric topology
such that x ∈ V ⊂ U . For each Uαi , pick an i such that (xαi − i , xαi + i ) ⊂ Ui .
Set
n
o
i
= min
|i≤N
i
Now, we show that
V = B (x) ⊂ U
Take y ∈ V . For α ∈
/ A, it is clear that yα ∈ Uα = R. For α ∈ A, we note that
d(xα , yα )
≤ D(x, y) < α
But this implies
d(xα , yα )
α
<≤
α
α
so that |xα − yα | < α and therefore y ∈ U .
Thus, U is the union of open sets in the metric topology, and thus is open in
the metric topology. So, a set is open in the metric topology iff it is open in the
product topology, which implies the metric D induces the product topology on Rω .
4. The Metrization Theorem
Before we prove the Metrization theorem, we discuss a small lemma.
Lemma 4.1. Every second countable regular space is normal.
Proof. Take an arbitrary second countable regular space X, and consider the disjoint closed sets C1 , C2 . For each point x ∈ C1 , regularity of X gives us an open
set V 3 x such that V ∩ C2 = ∅. Using regularity, choose an open set U 3 x such
that U ⊂ V . Find a basis element B such that B 3 x and B ⊂ U . By regularity,
B ⊂ V . The collection of all these basis elements is a countable open covering of
C1 , call {Un }. We can construct a similar collection for C2 , call {Vn }.
THE INTERSECTION OF TOPOLOGICAL AND METRIC SPACES
7
Sn
We construct a new covering as follows. Define Un0 = Un \ i=1 V n . Similarly,
S
S
S
n
Vn0 = Vn \ i=1 U n . We show that Un0 ∩ Vn0 = ∅. If it is not the case then there
0
exists Vi and Uj0 such that there exists and x ∈ Vi0 ∩ Uj0 . Without loss of generality,
assume i ≤ j. Then x ∈ Vi from the definition of Vi0 . But by the definition of
Uj0 , x ∈
/ Vi0 , a contradiction. So, the coverings are disjoint. We have separated the
closed sets with disjoint open sets.
Theorem 4.2. Every second countable regular space is a topological space.
Proof. We will prove this theorem by imbedding an arbitrary second countable
regular space X inside a space that we have shown to be metrizable. The space
in which we imbed X is Rω with the product topology. Our ultimate goal is to
define a continuous bijection F : X → Z ⊂ Rω where Z = F (X). However, to construct this function we must start by defining a countable collection of functions
{fn } fn : X → [0, 1]. We require that for any point p ∈ X and neighborhood U 3 p:
(∗) There exists an index i such that fi (p) > 0 and fi (X \ U ) = {0}.
By the second countability assumption, X has a countable basis {Bn }. For
indices m, n such that Bn ⊂ Bm , define the continuous function gn,m such that
gn,m (Bn ) = {1} and gn,m (X \ Bm ) = {0}. We can do this using the Urysohn
Lemma. Note that the set of all such functions satisfy (∗). If you pick an arbitrary
point p and open set U 3 p, we can find a basis element Bm ⊂ U for which p ∈ Bm .
Using the regularity of X, we may find a basis element Bn such that Bn ⊂ Bm .
This can be seen by separating p from X \ Bm and separating the two with open
sets. This will yield the desired open set Bn . Obviously, {gn,m } is countable. We
reindex this collection with the integers, yielding the set {fn }.
So, we have the desired set of functions. We define a new function,F (x) =
(f1 (x), f2 (x), . . . ), and we show that it is an imbedding from X → Rω .
F is continuous because each fn is continuous. We can see this is the case from
observing that F is continuous if the inverse image of each subbasis element of the
product topology is open. An arbitrary subbasis element is of the form πn−1 (Un ).
Since fn = πn ◦ F , F −1 (πn−1 (Un )) = fn−1 (Un ) which is clearly open. So F is continuous.
Also, F is injective, for if x 6= y, then we can separate the points with open
sets, and find a function fi ∈ {fn } satisfying (∗) so fi (x) > 0 and fi (y) = 0 which
implies that F (x) 6= F (y). And since we are mapping X to F (X), it is surjective.
The last thing we must prove then is that F is a homeomorphism. We already have
one direction. We turn to the case of having an open set U , proving that F (U ) is
open.
We will prove this by showing that for all x ∈ U , there exists an open set
W ⊂ F (U ) containing f (x), so that F (U ) is the union of open sets and therefore
open.
8
JAMES MAXWELL CLARKE
Take an arbitrary point x ∈ U , and pick an index n such that fn (x) > 0 and
fn (Z \ U ) = {0}. We will define W = Z ∩ πn−1 (0, +∞). Clearly, F (x) ∈ W . Now
take an arbitrary p ∈ W . For some x ∈ X, F (x) = p. fn (x) > 0, so x ∈ U , for fn
vanishes outside U . So, it has been shown that F is an imbedding, which finishes
the proof that X is metrizable.
Acknowledgments. It is a pleasure to thank my mentors, Yan Zhang and Casey
Rodriguez, for their guidance through research process and their insight during the
writing and editing. I coundn’t have done it without them.
References
[1] Donald W. Kahn
Topology: An Introduction to the Point-Set and Algebraic Areas.
The Williams & Wilkins Company, Baltimore, 1975
[2] James R. Munkres.
Topology.
Pearson , 2000