9th June 2005

Munkres §33

Ex. 33.1 (Morten Poulsen). Let r ∈ [0, 1]. Recall from the proof of the Urysohn lemma that if p < q then \overline{U}_{p} ⊂ U_{q}. Furthermore, recall that U_{q} = ∅ if q < 0 and U_{p} = X if p > 1.

Claim 1. f^{−1}({r}) = ⋂_{p>r} U_{p} − ⋃_{q<r} U_{q}, p, q ∈ Q.

Proof. By the construction of f: X → [0, 1],

⋂_{p>0} U_{p} − ⋃_{q<0} U_{q} = ⋂_{p>0} U_{p} = f^{−1}({0})

and

⋂_{p>1} U_{p} − ⋃_{q<1} U_{q} = X − ⋃_{q<1} U_{q} = f^{−1}({1}).

Now assume r ∈ (0, 1).

”⊂”: Let x ∈ f^{−1}({r}), i.e. f(x) = r = inf{ p | x ∈ U_{p} }. Note that x ∉ ⋃_{q<r} U_{q}, since f(x) = r. Suppose there exists t > r, t ∈ Q, such that x ∉ U_{t}. Since f(x) = r, there exists s ∈ Q such that r ≤ s < t and x ∈ U_{s}. Now x ∈ U_{s} ⊂ \overline{U}_{s} ⊂ U_{t}, contradiction. It follows that x ∈ ⋂_{p>r} U_{p} − ⋃_{q<r} U_{q}.

”⊃”: Let x ∈ ⋂_{p>r} U_{p} − ⋃_{q<r} U_{q}. Note that f(x) ≤ r, since x ∈ ⋂_{p>r} U_{p}. Suppose f(x) < r, i.e. there exists t < r such that x ∈ U_{t} ⊂ ⋃_{q<r} U_{q}, contradiction. It follows that x ∈ f^{−1}({r}). □

Ex. 33.4 (Morten Poulsen).

Theorem 2. Let X be normal. There exists a continuous function f : X → [0, 1] such that f(x) = 0 for x ∈ A, and f(x) > 0 for x ∉ A, if and only if A is a closed G_{δ} set in X.

Proof. Suppose A = f^{−1}({0}). Since

A = f^{−1}({0}) = f^{−1}( ⋂_{n∈Z_{+}}[0, 1/n)) = ⋂_{n∈Z_{+}} f^{−1}([0, 1/n))

it follows that A is a closed G_{δ} set.

Conversely suppose A is a closed G_{δ} set, i.e. A = ⋂_{n∈Z_{+}} U_{n}, U_{n} open. Then X − U_{n} and A are closed and disjoint for all n. By Urysohn’s lemma there exists a continuous function f_{n} : X → [0, 1], such that f_{n}(A) = {0} and f_{n}(X − U_{n}) = {1}.

Now define f : X → [0, 1] by

f(x) = ∑_{i=1}^{∞} \frac{1}{2^{i}} f_{i}(x).

Clearly f is well-defined. Furthermore f is continuous, by theorem 21.6, since the sequence of continuous functions (∑_{i=1}^{n} \frac{1}{2^{i}} f_{i}(x))_{n∈Z_{+}} converges uniformly to f, since

|∑_{i=1}^{∞} \frac{1}{2^{i}} f_{i}(x) − ∑_{i=1}^{n} \frac{1}{2^{i}} f_{i}(x)| = ∑_{i=n+1}^{∞} \frac{1}{2^{i}} f_{i}(x) ≤ ∑_{i=n+1}^{∞} \frac{1}{2^{i}} → 0

for n → ∞.

Clearly f(x) = 0 for x ∈ A. Furthermore note that if x ∉ A then x ∈ X − U_{n} for some n, hence f(x) ≥ \frac{1}{2^{n}} f_{n}(x) = \frac{1}{2^{n}} > 0. □

Ex. 33.5 (Morten Poulsen).

Theorem 3 (Strong form of the Urysohn lemma). Let X be a normal space. There is a continuous function f : X → [0, 1] such that f(x) = 0 for x ∈ A, and f(x) = 1 for x ∈ B, and 0 < f(x) < 1 otherwise, if and only if A and B are disjoint closed G_{δ} sets in X.

Proof. Suppose f : X → [0, 1] is a continuous function. Then clearly A = f^{−1}({0}) and B = f^{−1}({1}) are disjoint. Since

A = f^{−1}({0}) = f^{−1}( ⋂_{n∈Z_{+}}[0, 1/n)) = ⋂_{n∈Z_{+}} f^{−1}([0, 1/n))

and

B = f^{−1}({1}) = f^{−1}( ⋂_{n∈Z_{+}}(1 − 1/n, 1]) = ⋂_{n∈Z_{+}} f^{−1}((1 − 1/n, 1])

it follows that A and B are disjoint closed G_{δ} sets in X.

Conversely suppose A and B are disjoint closed G_{δ} sets in X. By ex. 33.4 there exists continuous functions f_{A} : X → [0, 1] and f_{B} : X → [0, 1], such that f_{A}^{−1}({0}) = A and f_{B}^{−1}({0}) = B. Now the function f : X → [0, 1] defined by

f(x) = \frac{f_{A}(x)}{f_{A}(x) + f_{B}(x)}

is well-defined and clearly continuous. Furthermore f^{−1}({0}) = A and f^{−1}({1}) = B, since

f(x) = 0 ⇔ f_{A}(x) = 0 ⇔ x ∈ A

and

f(x) = 1 ⇔ f_{A}(x) = f_{A}(x) + f_{B}(x) ⇔ f_{B}(x) = 0 ⇔ x ∈ B.
□

Ex. 33.7. For any topological space X we have the following implications:

X is locally compact Hausdorff
⇒^{Cor 29.4} X is an open subspace of a compact Hausdorff space
⇒^{Thm 32.2} X is a subspace of a normal space
⇒^{Thm 33.1} X is a subspace of a completely regular space
⇒^{Thm 33.2} X is completely regular

Ex. 33.8. Using complete regularity of X and compactness of A, we see that there is a continuous real-valaued function g: X → [0, 1] such that g(a) < \frac{1}{2} for all a ∈ A and g(B) = \{1\}. (There are finitely many continuous functions g_{1}, …, g_{k}: X → [0, 1] such that A ⊂ ⋃\{g_{i} < \frac{1}{2}\} and g_{i}(B) = 1 for all i. Put g = \frac{1}{k} ∑ g_{i}.) The continuous [Ex 18.8] function f = 2 max\{0, g − \frac{1}{2}\} maps X into the unit interval, g(A) = {0}, and g(B) = {1}.

References