Download (Urysohn`s Lemma for locally compact Hausdorff spaces).

TOPOLOGY WEEK 9
Lemma 0.1 (Urysohn’s Lemma for locally compact Hausdorff spaces). Let X be a locally compact
Hausdorff space. If K, F ⊆ X disjoint such that K is compact and F is closed, then there exists a
continuous function f ∶ X → [0, 1] such that f (x) = 1 for all x ∈ K and f (x) = 0 for all x ∈ F .
(1) Let (X, τ ) be locally compact Hausdorff. The goal of this exercise is to show that
Cc (X) = {f ∶ X → R ∶ f is continuous and X ∖ f −1 ({0}) is compact} is a dense subset of
C0 (X). Cc (X) is usually called the continuous functions of compact support.
(a) Show that Cc (X) ⊆ C0 (X), which by last week, provides that ∥ ⋅ ∥∞ is a norm on Cc (X).
(Hint:X ∖ f −1 ({0}) = {x ∈ X ∶ ∣f (x)∣ > 0}). Also, show that Cc ([1, ∞)) ≠ C0 ([1, ∞)), and
if X is compact Hausdorff, then Cc (X) = C0 (X).
(b) Show that if f, g ∈ C0 (X), then f g ∈ C0 (X), where f g denotes the point-wise multiplication. Furthermore, show that if f ∈ Cc (X) and g ∈ C0 (X), then f g ∈ Cc (X).
(c) Prove that Cc (X) is dense in C0 (X) with the following outline.
(i) For f ∈ C0 (X), we find for all ε > 0, fε ∈ Cc (X) such that ∥fε − f ∥∞ < ε. The goal is
to construct fε using Urysohn’s Lemma. Let ε > 0, and use f and ε/2 with definition
of C0 (X) to find a compact set, K, which will be the compact set for Urysohn.
(ii) Use local compactness to find an open cover of K. Take the closure of each set
in the finite subcover and call their union L. If you used local compactness in the
suggested way, then L should be compact, why? Show that K ⊆ L̊.
(iii) Show that F = L ∖ L̊ is closed and disjoint from K. Use Urysohn’s lemma on K
and F . Extend this function to all of X by setting it equal to 0 on X ∖ L. Call this
function g and show that g ∈ Cc (X).
(iv) Define fε ∶= f g. Why is fε ∈ Cc (X)? Show that ∥fε − f ∥∞ ≤ ε/2 < ε.
(2) Consider {0, 1}[0,1] = {f ∶ [0, 1] → {0, 1}} with the product topology.
(a) Show that {0, 1}[0,1] is compact.
(b) For x ∈ [0, 1], letfn (x) be the nth term of the binary expansion of x. Show that this
sequence of {0, 1}[0,1] has no convergent subsequence. Thus, sequential convergence does
not determine compactness.
(c) It can be shown that a topological space is compact if and only if every net has a convergent
subnet. Since sequences are nets and {0, 1}[0,1] is compact, the sequence in the previous
problem must have a convergent subnet. Find a convergent subnet of {fn }∞
n=1 .
Date: May 20, 2015.
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