This question already has an answer here:

Given two disjoint closed sets, $F_1$ and $F_2$ in $\mathbb{R}$, prove that there exist disjoint open sets $G_1$ and $G_2$ such that $F_1 \subset G_1$ and $F_2 \subset G_2$

The answer is not trivial in that the distance between $F_1$ and $F_2$ could be going towards zero.


marked as duplicate by bof, Community Oct 25 '15 at 5:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Can you precise on which kind of spaces you're working? I imagine metric spaces as you speak of distance. No additional hypothesis? $\endgroup$ – mathcounterexamples.net Oct 25 '15 at 0:20

I'm outlining a proof from Baby Rudin. Let $X$ be any metric space an $\delta_A:X\to \Bbb R$ be defined by $$ \delta_A(x)=\inf\{d(x,a)\,|\,a\in A\} $$ where $A$ is a subset of $X$. You can verifiy that

  1. $\delta_A$ is a continuous function on $X$ (actually, it's even uniformly continuous).
  2. If $A$ is a closed set, then $\delta_A(x)=0$ iff $\ x\in A$.

Now for any disjoint closed set $A,B \subset X$, we define $\mu:X\to[0,1]$ by $$ \mu(x)=\frac{\delta_A(x)}{\delta_A(x)+\delta_B(x)} $$ Verify that

  1. $\mu$ is well-defined on $X$, i.e. $\delta_A(x)+\delta_B(x)\ne 0$ all $x\in X$(Hint: Use that fact that $A$ and $B$ are closed and disjoint.)
  2. $0\le\mu(x)\le 1$ all $x\in X$.
  3. $\mu(a)=0$ for all $a\in A$ and that $\mu(b)=1$ for all $b\in B$.

Since $\delta_A,\delta_B$ are continuous, $\mu$ is also continuous so that $\mu^{-1}[0,\frac 12)$ and $\mu^{-1}(\frac 12,1]$ are both open sets in $X$ and that they are disjoint. It's not hard to see that $A\subset\mu^{-1}[0,\frac 12)$, $B\subset\mu^{-1}(\frac 12,1]$.

Now let $A=F_1, B=F_2, G_1=\mu^{-1}[0,\frac 12)$ and $G_2=\mu^{-1}(\frac 12,1]$ and we're done.

  • $\begingroup$ Why do you have to define $\mu$? What's wrong with just defining $G_1=\{x:\delta_{F_1}(x)\lt\delta_{F_2}(x)\}$ and $G_2=\{x:\delta_{F_2}(x)\lt\delta_{F_1}(x)\}$? $\endgroup$ – bof Oct 25 '15 at 5:58
  • 1
    $\begingroup$ Actually it's not me, it was Rudin's idea. I followed the idea he outlined in an exercise that I remember doing when I was a sophomore. If I have to make a guess, I'd say that introducing $\mu$ simplifies the exposition and makes it easier to see that $G_1, G_2$ are both open. $\endgroup$ – BigbearZzz Oct 25 '15 at 6:16
  • $\begingroup$ I have a question: Why do we know that $\mu^{-1}[0,1/2)$ and $\mu^{-1}(1/2,1]$ are open sets in $X$? $\endgroup$ – user128422 Nov 17 '15 at 1:41
  • $\begingroup$ @user128422 Recall that $[0,1]$ is a subspace of $\Bbb R$ with usual topology so the sets $[0,1/2)$ and $(1/2,1]$ are open in the relative topology. Since $\mu$ is continuous, the inverse image of $[0,1/2)$ and $(1/2,1]$ are open. $\endgroup$ – BigbearZzz Nov 17 '15 at 1:59
  • $\begingroup$ You´re right! I was thinking that the space was $\mathbb R$ Thank you $\endgroup$ – user128422 Nov 17 '15 at 2:00

Such a space is called T4 space. Not all space is T4 space, so there should be some assumtion about the space.

  • $\begingroup$ I just added Rn to it $\endgroup$ – More water plz Oct 25 '15 at 0:36
  • $\begingroup$ Rn is a metric space and I found a simple proof for the case here: link $\endgroup$ – aerile Oct 25 '15 at 0:48

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