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Use an Urysohn function to give a solution of this problem:

Prove that if $F_1$ and $F_2$ are disjoint closed sets in $\mathbb{R}^n$, then there exist disjoint open sets $G_1$ and $G_2$ such that $F_1\subset G_1$ and $F_2\subset G_2$.

The hard part of this is that I must use an Urysohn function in the proof. I know how to prove this by just using the definition of open set without resorting to any Urysohn function: we know $G_1^c$ is open, so for each $x\in G_2$ there exists $r(x)>0$ such that $B(x,r(x))\subset G_1^c$. Let $S=\cup_{x\in G_2} B(x,r(x)/3)$. Similarly define $T=\cup_{y\in G_1}B(y,r(y)/3)$. Then it's not hard to show that $G_1=T, G_2=S$ works.

Any ideas on how to prove this using Urysohn functions? Thank you.

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1 Answer 1

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HINT: Let $f$ be a Uryson function for $F_1$ and $F_2$, say with $f(x)=0$ for $x\in F_1$ and $f(x)=1$ for $x\in F_2$. Now look at the sets $$f^{-1}\left[\left[0,\frac12\right)\right]\qquad\text{and}\qquad f^{-1}\left[\left(\frac12,1\right]\right]\;.$$

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  • $\begingroup$ Brian: What is $f(x)$ when $x$ belongs to neither $F_1$ nor $F_2$? Thanks. $\endgroup$
    – user70520
    Commented Sep 12, 2013 at 5:21
  • $\begingroup$ @user70520: All we know in that case is that $f(x)\in[0,1]$, simply because $f:X\to[0,1]$. We don’t need to know anything more than that, however. $\endgroup$ Commented Sep 12, 2013 at 5:22
  • $\begingroup$ Is there a lemma I need to know? In my textbook (Lebesgue Integration on Euclidean Space by Frank Jones) it proves if $F$ is closed and $G$ is open with $F\subset G\subset\mathbb{R}^n$, then there exists continuous $f:\mathbb{R}^n\to[0,1]$ with $f(x)=1$ if $x\in F$ and $f(x)=0$ if $x\in G^c$. This $f$ is called an Urysohn function for the pair $F\subset G$. I don't see how this theorem is useful for this problem... Also I don't see how to use your hint. I was thinking of continuity of $f$ but didn't get too far... Thank you! $\endgroup$
    – user70520
    Commented Sep 12, 2013 at 5:30
  • $\begingroup$ @user70520: Yes, continuity of $f$ is crucial. Let $F=F_2$ and $G=\Bbb R^n\setminus F_1$; then $G$ is open, and $F\subseteq G\subseteq\Bbb R^n$, so you have a continuous $f:\Bbb R^n\to[0,1]$ such that $f(x)=1$ if $x\in F_2$ and $f(x)=0$ if $x\in\Bbb R^n\setminus G=F_1$, just as I suggested. Now look at the two inverse images that I suggested in my answer; clearly $F_1$ is a subset of the first, $F_2$ is a subset of the second, and they are disjoint, so all that remains is to check that they’re open. That’s a fairly straightforward consequence of the continuity of $f$. $\endgroup$ Commented Sep 12, 2013 at 5:38
  • $\begingroup$ I see. If $U$ is open then we know by continuity of $f$ that $f^{-1}(U)$ is also open. But $[0,1/2)$ isn't open... so maybe consider $f^{-1}(0,1/2)$? Thank you very much! $\endgroup$
    – user70520
    Commented Sep 12, 2013 at 5:41

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