Vtyjkyuk

Yesterday, I asked the question: Prove that if $A,B$ are closed then, $ exists;U,V$ open sets such that $Ucap V= emptyset$.

Here is the correct question: prove that if $A,B$ are closed sets in a metric space such that $Acap B= emptyset$, there exists $U,V$ open sets such that $Asubset U$, $Bsubset V$, and $Ucap V= emptyset$.

I am thinking of going by contradiction, that is: $forall; U,V$ open sets such that $Asubset U$, $Bsubset V$, and $Ucap Vneq emptyset$.

Let $ U,V$ open. Then, $exists;r_1,r_2$ such that $B(x,r_1)subset U$ and $B(x,r_2)subset V.$ I got stuck here!

I'm thinking of using the properties of $T^4-$space but I can't find out a proof! Any solution or reference related to metric spaces?

Sets $A, B$ are called completely separated iff there are disjoint open sets $U,V$ with $Asubset U$ and $Bsubset V.$ The definition of a normal ($T_4$) space is a $T_1$ space in which every disjoint pair of closed sets is completely separated. It appears you are trying to prove that a metric space is a normal space.

Let $(X,d)$ be a metric space and let $A,B$ be a disjoint pair of closed subsets of $X.$ For each $ain A,$ let $r_a>0$ such that $B (a,r_a)cap B=emptyset.$ For each $bin B$ let $s_b>0$ such that $B(b,s_b)cap A=emptyset.$

Let $U=cup_ain AB(a,frac 12r_a).$ Let $V=cup_bin BB(b,frac 12s_b).$

The reason $Ucap V=emptyset$ is that if we suppose $cin Ucap V$ then there exist $ain A$ and $bin B$ such that $d(c,a)<frac12r_a$ and $d(c,b)<frac 12s_b.$ But by the def'n of $r_a$ and of $s_b,$ and by the triangle inequality, $$r_aleq d(a,b)leq d(a,c)+d(c,b)<frac 12r_a+frac 12s_b$$

$$s_bleq d(a,b)leq d(a,c)+d(c,b)<frac 12r_a+frac 12s_b.$$ Adding the far left and far right expressions in the two lines above gives $r_a+s_b<r_a+s_b,$ an absurdity. So $cin Ucap V$ cannot exist.

We can also do this last part by noting that since $d(a,b)geq r_a$ and $d(a,b)geq s_b,$ we have $d(a,b)geq max (r_a,s_b).$ But $d(a,b)leq d(a,c)+d(c,b)<(r_a+s_b)/2.$ It would be absurd for real numbers $r,s$ that $max(r,s)$ is $less$ than their average $(r+s)/2.$

Let

$$
f(X,d) longrightarrow [0,1] \
x mapsto fracd(x,A)d(x,A) + d(x,B)
$$

Note that $f$ is continuous, because $d(cdot,A)$ is, and that it is in fact well defined, because $d(x,A) + d(x,B) = 0$ if and only if $d(x,A) = 0$ and $d(x,B) = 0$ which cannot happen because it would imply $x in overlineA cap overlineB = A cap B = emptyset$. Now,

$$
f(x) = 0 iff d(x,A) = 0iff x in overlineA = A
$$

and

$$
f(x) = 1 iff d(x,A) = d(x,A) + d(x,B) iff d(x,B) = 0 iff x in B
$$

Thus, $A = f^-1(0)$ and $B = f^-1(1)$. Now you can for example take $U = f^-1[0,frac14)$ and $V = f^-1(frac34,1]$ as the desired open sets: these are open because the are preimages of open sets of $[0,1]$ and $f$ is continuous, and they are disjoint and contain each closed set by construction.

If you our space is metric space, $X$, let's define

$f_S(x)=dist(x,S)=inf_yin S|x-y|$ and note that, for each closed set $S$, it is a continuous function on the space.

Let $U=leftxin X: f_A(x) < f_B(x) right=(f_A-f_B)^-1((-infty,0))$

and $V=leftxin X : f_B(x)<f_A(x) right=(f_A-f_B)^-1(0,infty)$.

And clearly, $Asubset U$ and $Bsubset V $.

Also, since $f_A-f_B$ are continuous function and $(0,infty)$ and $(-infty,0)$ are open in $mathbbR$, $U$ and $V$ are open and disjoint.

Claim. $f_S(x)$ is continuous on $X$ for each closed set $S$.

Proof) Let $epsilon>0$ and $x,yin X$ be given. Observe that, with out loss of generality, letting $f_S(x)-f_S(y)geq 0$, there exists $ain S$,

$f_S(x)-f_S(y)=inf_zin S|x-z|-inf_zin S|y-z| leq inf_zin S|x-z|-|y-a|+epsilonleq |x-a|-|y-a|+epsilon leq |x-y|+epsilon. $

Since $epsilon>0$ is arbitrary, we get $|f_S(x)-f_S(y)|<|x-y|$, so $f_S$ is Lipschitz continuous.

Since $A$ and $B$ are disjoint, we can place an open ball $mathcal B_a$ at each $a in A$ such that $B cap overlinemathcal B_a = varnothing$ (the bar denotes the closure). Then $bigcup mathcal B_a$ is an open neighbourhood of $A$. Because metric spaces are paracompact, we can choose a locally finite subcover of $A$. So assume the cover $mathcal B_a _ain A$ is already locally finite.

Let $b in B$. Then there exists an open neighbourhood $U$ of $b$ such that $U$ only intersects finitely many of the $mathcal B_a$. Let's suppose it intersects $mathcal B_a^1, dots mathcal B_a^n$. Then $U setminus (overlinemathcal B_a^1 cup dots cup overlinemathcal B_a^n)$ is an open set containing $b$. Thus we can choose an open ball $mathcal B_b$ such that $b in mathcal B_b subseteq U setminus (overlinemathcal B_a^1 cup dots cup overlinemathcal B_a^n)$. Note that $mathcal B_b$ is disjoint from $bigcup mathcal B_a$. Choosing $mathcal B_b$ in this manner for each $bin B$ yields an open neighbourhood for $bigcup mathcal B_b$ of $B$ that is disjoint from $bigcup mathcal B_a$.

Thus, $bigcup mathcal B_a$ and $bigcup mathcal B_b$ are disjoint open neighbourhoods of $A$ and $B$ respectively.