Download p. 1 Math 490 Notes 12 More About Product Spaces and

p. 1
Math 490 Notes 12
More About Product Spaces and Topological Properties
Recall that two spaces (X, τ ) and (Y, µ) are called homeomorphic iff there is a homeomorphism (a continuous bijection whose inverse is also continuous) between them. Homeomorphism between two topological spaces is often indicated by writing something like
(X, τ ) ≈ (Y, µ). Note that two homeomorphic spaces must have the same cardinality, and
homeomorphisms map open sets to open sets and closed sets to closed sets. Of further interest is the fact that homeomorphism is an equivalence relation on the class of all topological
spaces.
Some Examples: Here, we consider R and various subspaces of R, all equipped with the
usual topology (i.e. the one inherited from (R, τu )). The following functions are all homeomorphisms:
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¡
i) f : − π2 , π2 → R defined by f (x) = tan(x);
ii) g : (0, 1) → (a, b) defined by g(x) = a + (b − a)x;
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iii) h : (0, 1) → (1, ∞) defined by h(x) = ;
x
iv) j : (0, ∞) → (a, ∞) defined by j(x) = x + a;
v) k : (a, ∞) → (−∞, −a) defined by k(x) = −x.
These examples reveal that bounded open intervals, open rays, and R itself are all homeomorphic to one another (with topologies as in the examples). The function in (ii) can be used
to establish a homeomorphism between any two closed, bounded intervals of R, as subspaces
of (R, τu ).
Historically, topology had its origin over 100 years ago when Poincaré began studying the
topological equivalence of various geometric objects in Rn (such as spheres, tori, Klein bottles, etc.). An interest in whether two such surfaces are homeomorphic can lead to a desire to
prove or disprove that two topological spaces are homeomorphic. To prove that two spaces
are homeomorphic, one usually attempts to exhibit a homeomorphism between them.
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To prove that such a homeomorphism does not exist is generally more difficult. One good
reason to study topological properties is that two topological spaces which do not share a
certain topological property can not be homeomorphic to each other. We begin a study
of topological properties by examining the topological property of first-countability. Recall
that a topological space (X, τ ) is first-countable iff there is a countable τ -nbhd base at
each point.
Prop. 12.1
a) Any subspace of a first-countable space is first-countable.
Q
b) If (X, τ ) = i∈I (Xi , τi ), then (X, τ ) is first-countable iff each (Xi , τi ) is first-countable
and all but countably many of the component spaces Xi are indiscrete.
Proof :
a) Let (A, τA ) be a subspace of a first-countable space (X, τ ). For any x ∈ A, we need to
exhibit a τA -nbhd base at x. But since any x in A is also in X, we know there is a τ -nbhd
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base {Un ¯ n ∈ N} at x. The collection {Un ∩ A ¯ n ∈ N} then forms a τA -nbhd base at x.
Q
b) (=⇒) Assume the product space (X, τ ) = i∈I (Xi , τi ) is first-countable.
i) To prove that each component space is first-countable, let (Xj , τj ) be one of them, and
let xj be an arbitrary element of Xj . We need to show that there is a countable τj -nbhd
base at xj . To do this, note that there is an x ∈ X such that pj (x) = xj , where pj is the
jth projection map from X to Xj . By assumption, there is a countable τ -nbhd base at x;
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call it {Un ¯ n ∈ N}. The claim now is that the collection {pj (Un ) ¯ n ∈ N} forms a countable
τj -nbhd base at xj . To prove this, first note that each pj (Un ) is open in (Xj , τj ) because pj
is an open map. Now, let xj ∈ V ∈ τj . Then p−1
j (V ) is a τ -nbhd of x, since pj is continuous.
Since the Un ’s form a τ -nbhd base at x, there is a k ∈ N such that x ∈ Uk ⊆ p−1
j (V ).
Functions preserve containment, so this implies that xj ∈ pj (Un ) ⊆ V . So the collection
¯
{pj (Un ) ¯ n ∈ N} is a τj -nbhd base at xj .
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ii) To prove that there can be only countably-many of the component spaces which are not indiscrete, assume the contrary. That is, assume that uncountably-many of the (Xi , τi ) are not
Q
indiscrete, while maintaining the assumption that the product space (X, τ ) = i∈I (Xi , τi )
is first-countable. In each non-indiscrete (Xi , τi ), there must be a τi -nbhd Ui and a point
xi such that xi ∈ Ui ( Xi . Therefore, from every (Xi , τi ), we can choose a point xi and a
nbhd Ui such that xi ∈ Ui ∈ τi , and Ui 6= Xi whenever τi is not indiscrete. With points xi
and nbhds Ui chosen in this way for each i, there will be an uncountable number of the Ui ’s
for which Ui 6= Xi . Now, the point x = (xi )i∈I is in X, and since (X, τ ) is assumed to be
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first-countable, there is a countable τ -nbhd base {Vn ¯ n ∈ N} at x. Each Vn can be assumed
to be a basis element in (X, τ ), which means for each Vn , we have pj (Vn ) = Xj for all but
finitely many j ∈ I. For any particular n, let Fn be the (finite) set of all indices in I for
S
which pj (Vn ) 6= Xi . That is, for each n, j ∈ Fn ⇒ pj (Vn ) 6= Xj . Let S = n∈N Fn ⊆ I.
Then S is countable, and i 6∈ S ⇔ pj (Vn ) = Xj for all n ∈ N. Since S is countable, and
there are uncountably many τi which are not indiscrete, there must exist k ∈ I − S such
that xk has a τk -nbhd Uk with xk ∈ Uk 6= Xk . p−1 (Uk ) is a τ -nbhd containing x, and since
k 6∈ S, we have pk (Vn ) = Xk for all n ∈ N. But this means that there can be no Vn such
that x ∈ Vn ⊆ p−1 (Uk ) (if there was such a Vn , then we’d have xk ∈ pk (Vn ) = Xk ⊆ Uk ).
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This contradicts the fact that {Vn ¯ n ∈ N} is a τ -nbhd base at x. Therefore, under the
Q
assumption that (X, τ ) = i∈I (Xi , τi ) is first-countable, we can not have an uncountable
number of τi which are not indiscrete.
(⇐=) Here, we assume that all component spaces (Xi , τi ) are first-countable and that all
Q
but countably many of them are indiscrete. Then the product space (X, τ ) = i∈I (Xi , τi )
is the product of countably-many non-indiscrete spaces and a number of indiscrete spaces.
So, if we denote the set of indices i for which (Xi , τi ) is not indiscrete by C, then we have
Y
Y
Y
(X, τ ) ≈
(Xi , τi ) ×
(Xi , τi ) ≈
(Xi , τi ).
i∈C
i∈I−C
i∈C
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The first ’≈’ is true because any re-ordering of a product space is homeomorphic to the
Y
original product space. The second ’≈’ is a result of the fact that
(Xi , τi ) is itself an
i∈I−C
indiscrete space, and the product of a given space with an indiscrete space is homeomorphic
to the original, given space. So, it suffices here to show that the product of countably-many
first-countable spaces is first-countable.
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Toward this end, let {(Xn , τn ) ¯ n ∈ N} be a countable collection of first-countable spaces,
Y
¯
and (X, τ ) =
(Xn , τn ). Let x = (xn ) ∈ X. For any n ∈ N, let {Un,k ¯ k ∈ N}
n∈N
be a countable τn -nbhd base at xn . For any finite subset of N × N of the form F =
m
\
{(n1 , k1 ), (n2 , k2 ), . . . , (nm , km )}, define VF =
p−1
nj (Unj ,kj ). Then the set
j=1
¯
{VF ¯ F is a finite subset of N × N}
is a countable τ -nbhd base at x. This is true because for any U such that x ∈ U ∈ τ , there
is a τ -basis element B such that x ∈ B ⊆ U , and pj (B) = Xj for all but finitely many j.
Denote the set of all indices j for which pj (B) 6= Xj by J. For each j ∈ J, select a Uj,kj such
that pj (x) = xj ∈ Uj,kj ⊆ pj (B). Then p−1
j (Uj,kj ) is a τ -nbhd of x for all j ∈ J, and so is
\
p−1
¥
j (Uj,kj ), which is one of the VF ’s and is contained by B and U .
j∈J
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Note that Prop. 12.1 tells us that any product space involving uncountably many nonindiscrete spaces is not first-countable.
Recall that a topological space is 2nd-countable iff there is a countable basis for the topology. The next proposition shows that first- and 2nd-countability exhibit essentially the same
behavior with respect to subspaces and products; the proof is similar to the preceding proof
and will be omitted.
Prop. 12.2
a) Any subspace of a 2nd-countable space is 2nd-countable.
Q
b) If (X, τ ) = i∈I (Xi , τi ), then (X, τ ) is 2nd-countable iff each (Xi , τi ) is 2nd-countable
and all but countably many of the component spaces Xi are indiscrete.
Prop. 12.3 The properties zero-dimensional, first-countable, and 2nd-countable are
all topological properties.
Proof : Let f : (X, τ ) → (Y, µ) be a homeomorphism. Recall that both f and f −1 map open
sets to open sets and closed sets to closed sets.
1) If (X, τ ) is zero-dimensional, then τ has a basis B of clopen sets. Then (as you may
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verify) B′ = {f (B) ¯ B ∈ B} is a basis of clopen sets for µ.
2) Assume (X, τ ) is first-countable and let y ∈ Y . Since f is bijective, there is a unique
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x ∈ X such that f (x) = y. By assumption, x has a countable nbhd base {Un ¯ n ∈ N}. Then
¯
(you may verify) {f (Un ) ¯ n ∈ N} is a countable nbhd base at y.
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3) Finally, if (X, τ ) is second-countable and {Bn ¯ n ∈ N} is a countable base for τ , then
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¥
{f (Bn ) ¯ n ∈ N} is a countable basis for µ.
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The Properties T0 , T1 , T2
The properties T0 , T1 , T2 are the first three of several so-called “separation axioms” (not
sure why the word “axiom” is used). The higher separation axioms T3 , T3.5 , and T4 will be
defined and discussed later. Roughly speaking, all these “axioms” (properties) measure the
degree to which points can be separated in a topological space.
Def: A topological space (X, τ ) is:
T0 iff, whenever x 6= y in X, there is a nbhd of at least one of the two points which
excludes the other point;
T1 iff whenever x 6= y in X, there is a nbhd of each point which excludes the other
point;
T2 (a.k.a Hausdorff ) iff, whenever x 6= y in X, the two points have disjoint nbhds.
Clearly (hopefully), T2 ⇒ T1 ⇒ T0 .
Prop. 12.4 A topological space (X, τ ) is T1 iff {x} is closed for all x ∈ X.
Prop. 12.5 On any set X, the cofinite topology is the coarsest T1 topology.
Proof : Exercise.
Prop. 12.6:
Let (X, τ ) be a T1 space, with A ⊆ X. Then x is a limit point of A iff U ∩ A
is infinite for every nbhd U of x.
Proof : Munkres, Thm 17.9.
HW 2 (due Monday 10/3):
§19 (p. 118): 1, 3, 5
§20 (p. 126): 2, 4, 5, 6, 8
Prove Prop 12.5 from Notes 12