Is There a Stronger Urysohn Lemma for Banach Spaces?

[quasar987]

Hello all,

I am reading an article and there is something I find odd. The setting is a Banach space E and we have two disjoint closed subsets A and B of E. There is no additional assumption on E, A or B. The author then says,

"Let f:E-->[0,1] be a Urysohn's function such that f(x)=0 if and only if x is in A, and f(x)=1 on B."

But never have I seen a version of Urysohn's lemma that guarantees that f(x)=0 if and only if x is in A.

Does someone have an explanation? (I would ask my advisor but she had gone on vacation for 3 weeks)

 

[rasmhop]

In the exercises (exercise 5 on pg. 213) of Munkres' topology he states and asks the reader to prove the following theorem which he refers to as the strong form of the Urysohn lemma:
Let X be a normal space. There is a continuous function f : X \to [0,1] such that f(x)=0 for x \in A, and f(x) = 1 for x\in B, and 0 < f(x) < 1 otherwise, if and only if A and B are disjoint closed G_\delta sets in X.

In a metrizable space every closed set is G_\delta and metrizable spaces are normal so we obtain the corollary:
Let X be a metrizable space. Then there exists a continuous function f : X \to [0,1] such that f(x)=0 for x \in A, and f(x) = 1 for x\in B, and 0 < f(x) < 1 otherwise, if and only if A and B are disjoint closed sets in X.

 

[quasar987]

I see, thank you!