Download COUNTABLE PRODUCTS 1. The Cantor Set Let us constract a very

COUNTABLE PRODUCTS
ELENA GUREVICH
Abstract. In this paper, we extend our study to countably infinite products of topological
spaces.
1. The Cantor Set
Let us constract a very curios (but usefull) set known as the Cantor Set. Consider the
closed unit interval [0, 1] and delete from it the open interval ( 31 , 23 ) and denote the remaining
closed set by
1
2
G1 = [0, ] ∪ [ , 1]
3
3
Next, delete from G1 the open intervals ( 91 , 29 ) and ( 79 , 98 ), and denote the remaining closed
set by
1
2 1
2 7
8
G2 = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , 1]
9
9 3
3 9
9
If we continue in this way, at each stage deleting the open middle third of each closed
interval remaining from the previous stage we obtain a descending sequence of closed sets
G1 ⊃ G2 ⊃ G3 ⊃ · · · ⊃ Gn ⊃ . . .
T
The Cantor Set G is defined by G = ∞
n=1 Gn , and being the intersection of closed sets, is
a closed subset of [0, 1]. As [0, 1] is compact, the Cantor Space (G, τ ) ,(that is, G with the
subspace topology), is compact.
It is useful to represent the Cantor Set in terms of real numbers written to basis 3, that is,
5
ternaries. In the ternary system, 76 81
would be written as 2211.0012, since this represents
3
2
1
0
−1
2 · 3 + 2 · 3 + 1 · 3 + 1 · 3 + 0 · 3 + 0 · 3−2 + 1 · 3−3 + 2 · 3−4
So a number x ∈ [0, 1] is represented by the base 3 number a1 a2 a3 . . . an . . . where
x=
∞
X
an
n=1
3n
an ∈ {0, 1, 2}
∀n ∈ N
Turning again to the Cantor Set G, it should be clear that an element of [0, 1] is in G if
5
/ G 81
∈
/ G but
and only if it can be written in ternary form with an 6= 1 ∀n ∈ N, so 12 ∈
1
1
1
∈
G
and
1
∈
G.
[we
denote
=
0.1111
.
.
.
,
=
0.02222
.
.
.
1
=
0.2222
.
.
.
]
3
2
3
Thus we have a function f from the Cantor Set into the set of all sequences of the form
< a1 , a2 , a3 , . . . , an , · · · >, where each ai ∈ {0, 2} and f is one-to-one and onto.
2. The Product Topology
Definition 2.1. Let (X1 , τ1 ), (X2 , τ2 ), Q
. . . , (Xn , τn ), . . . be a countably infinite family of
topological spaces. Then the product ∞
i=1 Xi of the sets Xi , i ∈ N consists of all the
infinite sequences < x1 , x2 , x3 , . . . , xn , · · · > where xi ∈ Xi for all i. The product space
Date: May 6, 2011.
1
2
Q∞
i=1 (Xi , τi )
consists of the product
Q∞
i=1 Xi
with the topology τ having as its basis the
family
∞
Y
B = { Oi : Oi ∈ τi , and Oi = Xi for all but a finite number of i0 s}
i=1
0
Proposition 2.2. Let (Xi , τi ), (Yi , τi ) i ∈ N, be countably infinite families of topological
Q∞
Q
0
mapping hi :
spaces, having product spaces ( ∞
i=1 Xi , τ ) ( i=1 Yi , τ ) respectively. If theQ
0
(Xi , τi ) → (Yi , τi ) is continuous for each i ∈ N then so is the mapping - h : ( ∞
i=1 Xi , τ ) →
Q∞
0
( i=1 Yi , τ ) given by
h(< x1 , x2 , x3 , . . . , xn , · · · >) =< h1 (x1 ), h2 (x2 ), h3 (x3 ), . . . , hn (xn ), · · · >
Q
−1
Proof. It suffices to show that if O is a basic open set in ( ∞
i=1 Xi , τ ) then h (O) is open
Q∞
0
0
in ( i=1 Yi , τ ). Consider the basic open set U1 × U2 × · · · × Un × Yn+1 × . . . where Ui ∈ τi
for i = 1, 2, . . . , n. Then
h−1 (U1 × U2 × · · · × Un × Yn+1 × . . . ) = h−1 (U1 ) × h−1 (U2 ) × · · · × h−1 (Un ) × Xn+1 × . . .
since the continuity of each hi implies that h−1
i (ui ) ∈ τi for i = 1, 2, . . . , n. So h is continuous.
3. The Cantor space and the Hilbert cube
Proposition 3.1. Let (G, τ ) be the Cantor
Q space and for each i ∈ N let (Ai , τi ) be the
A , τ ) be it’s product space. Then the map
set 0, 2 with the discrete topology, and let ( ∞
P∞ ani=1 i
Q∞
0
f : (G, τ ) → ( i=1 Ai , τ ) given by f ( i=1 3n ) =< a1 , a2 , a3 , . . . , an , · · · > is a homeomorphism.
Proof. It is easy to see that f is one-to-one and onto. To prove that f is continuous it
suffices to show for any basic given set U = U1 × U2 × · · · × Un × An+1
P × a.n. . and any
point a =< a1 , a2 , a3 , . . . , an , · · · >∈ U there exists an open set W 3 ∞
i=1 3n such that
P
P∞ an
an
1
1
f (W ) ⊆ U . Consider the open interval ( ∞
−
,
+
)
and
letPW be the
i=1 3n
i=1 3n
3N +2
3N +2
xn
intersection of this open intervals with G. Then W is open in (G, τ ) and if x =
3n ∈ W
then xi = ai for i = 1, 2, . . . , N . So f (x) ∈ U1 × U2 × · · · × Un × An+1 × . . . and thus
f (W ) ⊆ U as required.
Proposition 3.2. Let (Gi , τi ), i ∈ N, be a countably infinite family of topological
spaces
Q∞
each
of
which
is
homeomorphic
to
the
Cantor
space
(G,
τ
)
then
(G,
τ
)
'
(G
,
τi ) '
i
i=1
Qn
(G
,
τ
)
for
each
n
∈
N.
i
i
i=1
Proof. First let us verify
' (G1 , τ1 )×(GQ
2 , τ2 ). This is, by proposition 3.1 equivaQ∞ that (G, τ ) Q
∞
lent to showing that i=1 (Ai , τi ) ' i=1 (Ai , τi ) × ∞
) where each (A
i=1 (Ai , τiQ
Qi ,∞τi ) is the set
∞
0,
2
with
the
discrete
topology.
Now
we
define
a
function
Θ
:
(A
,
τ
)
×
i
i
i=1
i=1 (Ai , τi ) →
Q∞
(A
,
τ
)
by
Θ(<
a
,
a
,
a
,
·
·
·
>,
<
b
,
b
,
b
,
·
·
·
>)
=<
a
,
b
,
a
,
b
,
·
·
·
>.
It’s easy to
i
i
1
2
3
1
2
3
1
1
2
2
i=1
show that Θ isQa homeomorphism, and so (G1 , τ1 ) × (G2 , τ2 ' (G, τ ). By induction we get
that (G, τ ) ' ∞
i=1 (Ai , τi ) for every positive integer n. To show the infinite product case,
define the mapping
∞
∞
∞
Y
Y
Y
Φ : [ (Ai , τi ) ×
(Ai , τi ) × . . . ] →
(Ai , τi )
i=1
i=1
i=1
by Φ(< a1 , a2 , a3 , · · · >, < b1 , b2 , b3 , · · · >, < c1 , c2 , c3 , · · · >, < d1 , d2 , d3 , · · · >, < e1 , e2 , e3 , · · · >
) =< a1 , a2 , b1 , a3 , b2 , c1 , a4 , b3 , c2 , d1 , a5 , b4 , c3 , d2 , e1 , · · · > Again it is easily verified that Φ
is homeomorphism and the proof is complete.
3
Proposition 3.3. The topological space [0, 1] is a continuous image of the Cantor space
(G, τ ).
Q
Proof. It suffices to find a continuous mapping Φ of ∞
i=1 (Ai , τi ) onto [0, 1]. Such a mapping
is given by
∞
X
ai
Φ(< a1 , a2 , . . . , ai , · · · >) =
2i+1
i=1
It is easy to show that Φ is onto and continuous, and the proof is complete.
Definition 3.4. For each positive integer
Q∞n, let the topological space (In , τn ) be homeomorphic to [0, 1]. Then the product
Q space n=1 (In , τn ) is called Hilbert cube and is denoted
by I ∞ . The product space ni=1 (Ii , τi ) is called the n-cube, and is denoted by I n , for each
n∈N
Theorem 3.5. The Hilbert cube is compact.
0
Proof. By proposition 3.3 there is a continuous mapping Φn of (Gn , τn ) onto (In , τn where
0
for each n ∈ N, (Gn , τn ) and (In , τn are homeomorphic to the Cantor space and [0, 1] respecQ∞
Q∞
0
∞
tively.
Q∞Therefore, there is a continuous mapping Ψ of n=1 (Gn , τn ) onto ∞n=1 (In , τn = I
but i=1 (Gn , τn ) is homeomorphic to the Cantor space (G, τ ). Therefore I is a continuous
image of the compact space (G, τ ) and hance is compact.
Proposition
3.6. Let (Xi , τi ) i ∈ N be a countably infinite family of matrizable spaces,
Q∞
then i=1 (Xi , τi ) is metrizable.
Corollary 3.7. The Hilbert cube is metrizable.
4. The Cantor space and the Hilbert cube
Definition 4.1. A topological space (X, τ ) is said to be separable if it has a countable
dense subset.
Proposition 4.2. Let (X, τ ) be a compact metrizable space. Then (X, τ ) is separable.
Proof. Let d be a metric on X which induces the topology |tau. For each positive integer n,
let Sn be the family of all open balls having centers in X and radius n1 .Then Sn is an open
covering of X and so there is a finite sub covering Un = {Un1 , Un2 , . . . , Unk } for some k ∈ N.
Let ynj be the center of Un , j = 1, 2, . . . , k and YN = {yn1 , . . . , ynk }. Put Y = ∪∞
n=1 Yn , then
Y is a countable subset of X. Now, if V is any non open set in (X, τ ), then for any v ∈ V ,
V contains an open ball B of radius n1 . As Un is an open cover of X, v ∈ Unj for some j.
Thus d(v, ynj ) < n1 , and so ynj ∈ B ⊂ V . Hance V ∩ Y 6= ∅ and so Y is dense in X.
Corollary 4.3. The Hilbert cube is separable space.
Definition 4.4. A topological space (X, τ ) is said to be a T1 space if every singelton set
{x}, x ∈ X, is a closed set.
Theorem 4.5. (Urisohn’s Theorem) Every separable metric space (X, d) is homeomorphic
to a subspace of the Hilbert cube.
Proof. We need to find a countably infinite family of mappings fi : (X, d) → [0, 1] which
are a) continuous, and b) separate points and closed sets. Without loss of generality we
can assume that d(a, b) ≤ 1 for all a and b in X, since every metric is equivalent to such a
metric. As (X, d) is separable, there exists a countable dense subset Y = {yi , i ∈ N}. For
4
each i ∈ N, define fi : X → [0, 1] by fi (x) = d(x, yi ). It is clear that each fi continuous.
To see that the mappings {fi : i ∈ N} separate points and closed sets, let x ∈ X and A be
any closed set not containing x. Now X \ A is an open set, and so contains an open ball B
of radius and center x, for some > 0. As Y is dense in X, there exists a yn , such that
d(x, yn ) < 2 , thus d(yn , a) ≥ 2 for all a ∈ A. So [0, 2 ) is an open set in [0, 1] which contains
fn (x), but fn (a) ∈
/ [0, 2 ), for all a ∈ A. This implies fn (A) ⊆ [ 2 , 1]. As the set [ 2 , 1] is
/ fn (A) and thus the family {fi : i ∈ N}
closed, this implies fn (A) ⊆ [ 2 , 1]. Hance fn (x) ∈
separates points and closed spaces.
Corollary 4.6. Every compact metrizable space is homeomorphic to a closed subspace of
the Hilbert cube.
Q
Corollary 4.7. If for each i ∈ N, (Xi , τi ) is compact metrizable space, then ∞
i=1 Xi , τi is
compact and metrizable.
Definition 4.8. A topological space (X, τ ) is said to satisfy a the second axiom of
countability (or to be second countable) if there exists a basis B for τ such that B consists
of only a countable number of sets.
Proposition 4.9. Let (X, d), be a matric space and τ the induced topology. Then (X, τ ) is
a separable space if and only if it satisfies the second axiom of countability.
Proof. Let (X, τ ) be separable. Then it has a countable dense subset Y = {yi : i ∈ N}. Let
B consists of all the balls (in the metric d) with center yi , and radius n1 , for some positive
integer n. Clearly B is countable, and we will show that it is a basis for τ . Let V ∈ τ . Then
for any v ∈ V , V contains an open ball, B, of radius n1 about r. As Y is dense in X, there
0
1
exists a ym ∈ Y such that d(ym , v) < 2n
. Let B be the open ball with center ym and radius
0
0
1
2n . Then the triangle inequality implies B ⊆ B ⊆ V Also B ∈ B. Hance B is a basis for
τ . Conversely, let (X, τ ) second countable, having a countable basis B = {Bi : i ∈ N}. For
each Bi 6= ∅ let bi be any element of Bi , and put Z equal to the set of all such bi . Then Z
is countable set. Further, if V ∈ τ , then Bi ⊆ V for some i, and so bi ∈ V . Thus V ∩ Z 6= ∅
Hance Z is dense in X. Consequently (X, τ ) is separable.
Theorem 4.10. (Urisohn’s theorem and its converse) Let (X, τ ) be a topological space.
Then (X, τ ) is separable and metrizable if and only if it is homeomorphic to a subspace of
the Hilbert cube.
Proof. If (X, τ ) is separable and metrizable then Urisohn’s theorem says that it is homeomorphic to a subspace of the Hilbert cube. Conversely, let (X, τ ) be homeomorphic to the
subspace (Y, τ1 ) of the Hilbert cube I ∞ , I ∞ is separable and metrizable, hance it is second
countable and metrizable (since any subspace of a second countable and metrizable space
is second countable and metrizable) (Y, τ1 ) is second countable and metrizable. Therefor it
is separable. Hance (X, τ ) is also separable and metrizable.
5. Peano’s Theorem
Proposition 5.1. Every separable metrizable space (X, τ1 ) is a continuous image of a
subspace of the cantor space (G, τ ). Further if (X, τ1 ) is compact, then the subspace can be
chosen to be closed in (G, τ ).
Proposition 5.2. Let (Y, τ1 ) be a (non empty) closed subspace of the Cantor space (G, τ ).
Then there exists a continuous mapping of (G, τ ) onto (Y, τ1 ).
Theorem 5.3. Every compact metrizable space is a continuous image of the Cantor space.
5
Proposition 5.4. let f be a continuous mapping of a compact metric space (X, d) onto a
Hausdorff space (Y, τ1 ). Then (Y, τ1 ) is compact and metrizable.
Theorem 5.5. (Peano) For each positive integer n, there exists a continuous mapping φn
of [0, 1] onto the n-cube I n .
Proof. There exists a continuous mapping Φn of the Cantor space (G, τ ) onto the n-cube
I n . As (G, τ ) is obtained from [0, 1] by successively dropping out middle thirds, we extend
Φn to a continuous mapping Ψn : [0, 1] → I n by defining Ψn to be linear on each omitted
interval, that is, if (a.b) is one of the open intervals comprising [0, 1]
G, then Ψn is defined on (a.b) by
Ψn (αa + (1 − α)b) = αΦn (a) + (1 − αΦn (b)) 0 ≤ α ≤ 1
It is easily verified that Ψn is continuous.
References
1. S. Morris, Topology without tears, 2007.