# Application of Urysohn's Lemma.

There is one argument of Urysohn's lemma I do not understand.

Assume we have a compact Hausdorff space, so we are in a normal space.

Assume that we have a closed set F, and an open set V, such that $F \subset V$. Then my book states that we can find a continuous function sent to $[0,1]$, such that $f(\{F\})=1$, and supp $f\subset V$.

But my problem is this:

We have two closed sets, F, and $V^c$, these are also disjoint. So Urysohn's lemma states that we can find f, such that $f(\{F\})=1$, and $f(\{V^C\})=0$. Now we have that the set $\{x: f(x) \ne 0\}\subset V$, but in order to get our result, we must have that the closure of this set is in V, and how does this also hold when we take the closure?

## 2 Answers

It doesn’t: you need an intermediate step. Use the normality of the space to conclude that there is an open set $U$ such that $F\subseteq U\subseteq\operatorname{cl}U\subseteq V$, and then take a Uryson function for $F$ and $X\setminus U$.

Instead of using the function $f$ you get from Urysohn's lemma directly (which indeed might not work), you can use $g(x)=\max(2f(x)-1,0)$.

• Thank you for your reply, but how does this work? I see that g is 1 on F, and zero on $V^C$. But I do not really see how the result follows. – user119615 Oct 14 '15 at 22:03
• The closure of the set where $g$ is nonzero is contained in the set where $f$ is $\geq 1/2$, which is contained in $V^c$. – Eric Wofsey Oct 14 '15 at 22:04
• Thank you, that was a smart solution. But you mean V not $V^C$ in your sentence? So the argument is this?: From general topology we have that $\overline{f^{-1}(A)}\subset f^{-1}(\bar{A})$. And the set where g is nonzero is $f^{-1}((1/2,\infty))$, so $\overline{f^{-1}((1/2,\infty))}\subset f^{-1}([1/2,\infty))\subset V$. – user119615 Oct 14 '15 at 22:13
• Yes, that's right. – Eric Wofsey Oct 14 '15 at 22:16