2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $X\setminus U$.

$\endgroup$
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user119615 Oct 14 '15 at 22:03
  • $\begingroup$ The closure of the set where $g$ is nonzero is contained in the set where $f$ is $\geq 1/2$, which is contained in $V^c$. $\endgroup$ – Eric Wofsey Oct 14 '15 at 22:04
  • $\begingroup$ 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 $\overline{f^{-1}((1/2,\infty))}\subset f^{-1}([1/2,\infty))\subset V$. $\endgroup$ – user119615 Oct 14 '15 at 22:13
  • $\begingroup$ Yes, that's right. $\endgroup$ – Eric Wofsey Oct 14 '15 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.