# Stronger Urysohn lemma?

1. Aug 14, 2009

### 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)

2. Aug 14, 2009

### 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.

3. Aug 14, 2009

### quasar987

I see, thank you!