Download Some consequences of Urysohn`s lemma. Local connectedness.

Consequences of Urysohn’s Lemma
Saul Glasman
October 28, 2016
We’ve shown that metrizable spaces satisfy a number of nice topological
conditions, but so far we’ve never been able to prove a converse theorem: that a
space with good topological properties is metrizable. Urysohn’s lemma is going
to allow us to change that now.
Theorem 1 (Urysohn metrization theorem). Suppose X is a normal space with
a countable basis. Then X is metrizable.
Remark 2. This isn’t an if and only if; not every metric space has a countable
basis. But it’s something.
Proof. We know that any subspace of a metric space inherits a metric, so if we
can embed X into a metric space, we’re done. The metric space we’ll use is the
space
Ω = [0, 1]N
of sequences of elements of [0, 1]. The metric on Ω is as follows: if (an )n∈N and
(bn )n∈N are two elements of Ω, then
(|an − bn |)n∈N
is a sequence of elements of [0, 1]; in particular, it’s a bounded sequence of real
numbers, and so it has a least upper bound. We’ll write
d((an )n∈N , (bn )n∈N ) = lub((|an − bn |)n∈N ).
The proof that this is a metric is very similar to the proof that the l∞ metric
on Rn is a metric, so we’ll leave it as an exercise.
First we’ll prove a lemma:
Lemma 3. Suppose X is normal and has a countable basis. Then there is a
sequence of continuous functions
(gn )n∈N , gn : X → [0, 1]
such that for any x0 ∈ X and any neighborhood U of x0 , there is some n such
that
• gn (x0 ) > 0, and
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• gn (x) = 0 for x ∈
/ U.
Proof. Let (Bn )n∈N be a countable basis for X. For each pair m, n with Bn ⊆
Bm , use the Urysohn lemma to give a continuous function
gn,m : X → [0, 1]
such that
gn,m (Bn ) = {1}, gn,m (X \ Bm ) = {0}.
We claim that the gn,m is the required collection of functions. Indeed, suppose
x0 ∈ X and U is an open neighborhood of x0 . Then we can choose m such that
x0 ∈ Bm ⊆ U , and by using normality on the closed sets {x0 } and X \ Bm , we
can find an open set V with
x ∈ V, V ⊆ Bm .
Take n such that x0 ∈ Bn ⊆ V ; then
gn,m (x0 ) = 1,
and since Bn ⊆ U ,
gn,m (X \ U ) = {0}.
Since the collection of the gm,n is indexed on a subset of N × N and is therefore
countable, we can reindex them on N, proving the lemma.
Now let’s cook up our embedding. First we’ll set
fn = gn /n
so that
fn (X) ⊆ [0, 1/n].
ω
Define F : X → [0, 1] by
F (x) = (f1 (x), f2 (x), f3 (x), · · · ).
We’d like to prove that f is an embedding. The first thing to prove is that
f is injective; indeed, if x0 6= x1 ∈ X, then by regularity, let U be an open
neighborhood of x0 with x1 ∈
/ U . Then there is some n such that fn (x0 ) > 0
and fn (x1 ) = 0. In particular, F (x0 ) 6= F (x1 ).
Next we’ll prove that F is open onto its image. Let U ∈ X be open and
let x0 ∈ U ; we need to find an open neighborhood W of z0 = F (x0 ) such that
W ⊆ F (U ). Choose k such that fk (x0 ) > 0 and fk (X \ U ) = {0}. Then
f (x0 ) ∈ {(an )n∈N ∈ F (X) | ak > 0} ⊆ f (U ).
Finally, we have to prove that F is continuous. For any x0 ∈ X and > 0,
we have to find a neighborhood U of x0 such that
x ∈ U ⇒ d(f (x), f (x0 )) < .
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First choose N large enough that
1/N < .
For each n ∈ {1, 2, · · · , N }, choose a neighborhood Un of x0 such that
x ∈ Un ⇒ |fn (x) − fn (x0 )| < .
This is possible, of course, since fn is continuous.
Now let
N
\
U=
Un .
n=1
We claim that if x ∈ U , then d(x, x0 ) < . By the definition of d, this is
equivalent to
|fn (x) − fn (x0 )| < for all n ∈ N.
For n ≤ N , this is by definition of Un . But for n > N ,
fn (X) ⊆ [0, 1/n], 1/n < so |fn (x) − fn (x0 )| < for all x ∈ X.
Corollary 4. A compact Hausdorff space is metrizable if and only if it has a
countable basis.
Proof. A compact Hausdorff space is normal, and so by the Urysohn metrization
theorem it’s metrizable if it has a countable basis. The converse is going to be
on the next problem set.
Here’s another powerful consequence of Urysohn’s lemma:
Theorem 5 (Tietze extension theorem). Suppose X is a normal space and A
is a closed subspace of X. Then any continuous function
f :A→R
can be extended to a continuous function on all of X: there is a continuous
function
g:X→R
such that
g|A = f.
This is a super powerful theorem, and I’m not going to prove it here, since
the proof would take us too far afield, but it’s just really nice to know it exists.
OK, next week we’re going to start algebraic topology, but there’s a loose
end I have to tie off first. In algebraic topology we’re going to want to study
spaces with nice connectedness properties, and in particular we’ll usually want
to exclude things like the topologist’s sine curve and Q.
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Definition 6. Let X be a space. We say X is locally connected at a point
x ∈ X if for every open neighborhood U of x, there is a connected open V such
that x ∈ V ⊆ U . We say X is locally connected if it’s locally connected at every
point.
Local path connectedness and at a point and of a space is defined similarly.
Example 7. Any open subset of R is locally connected and locally path connected, since R has a basis given by intervals, which are connected and path
connected. Of course, some open subsets of R - for example, (0, 1) ∪ (2, 3) - are
not connected.
The topologist’s sine curve is connected but not locally connected (draw
picture).
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