let $\{A_i:1\leq i\leq n\}$ be collection of pairwise disjoint closed sets in a metric space . show that there exist a collection $\{U_i:1\leq i\leq n\}$ of open sets such that $\{cl(U_i)\}$ is pairwise disjoint and for each $i$ and we have $A_i\subseteq U_i$

i know every metric space have normal property ,mean for any disjoint closed sets $A$ and $B$ we have disjoint open sets $U_1$ and $U_2$ such that $A\subseteq U_1$ and $B\subseteq U_2$

i think this problem is connect to normal property .also problem demands of pairwise disjoints closure of open sets $U_i$.

any hint to approach this problem. thanks in advanced

  • $\begingroup$ You can show that this holds in any normal space. One idea is to do it for two sets first and then apply induction. $\endgroup$ – Henno Brandsma Jan 7 '18 at 13:53
  • $\begingroup$ You can use that $X$ normal means that for every closed set $A$ and open set $U$ with $A\subseteq U$ we have an open set $U$ with $A\subseteq V \subseteq \overline{V} \subseteq U$. So disjoint open sets always implies we have disjoint open sets with disjoint closures as well. Just shrink them. $\endgroup$ – Henno Brandsma Jan 7 '18 at 13:57
  • $\begingroup$ @HennoBrandsma i done it for two. but how can i apply induction to show it for $n$ closed sets $\endgroup$ – Eklavya Jan 7 '18 at 14:00
  • 1
    $\begingroup$ With $n+1$ sets do it for the union of the first $n$ and the last one and inside the open set we get for the union we apply the induction hypothesis for the $n$ sets. Combine. $\endgroup$ – Henno Brandsma Jan 7 '18 at 14:02

This fact holds in all normal spaces, in fact. So in the following $X$ is a normal space.

A handy fact to have (proof here):

$X$ is normal iff for every closed set $A \subseteq X$ and every open set $U$ of $X$ such that $A \subseteq U$ there exists an open set $V$ of $X$ such that $A \subseteq V \subseteq \overline{V} \subseteq U$.

So if we can show that for $A_1, \ldots A_n$ (pairwise disjoint) there are open sets $U_i$ with $A_i \subseteq U_i$ with the $U_i$ pairwise disjoint, then we are done, because we then apply the above fact to the $U_i$ to get smaller $V_i$ and their closures will be pairwise disjoint as $\overline{U_i} \subseteq V_i$ and the $V_i$ are pairwise disjoint.

To show the open pairwise disjoint property for $n$ sets we work by induction.

For $n=2$ this is just normality, so there's nothing to prove. Supppose we can separate $n \ge 2$ pw.d. closed sets and let $A_1, \ldots, A_n, A_{n+1}$ be pairwise disjoint closed sets.

Define $A = \cup_{i=1}^n A_i$, which is closed as a finite union of closed sets, disjoint from $A_{n+1}$, and by normality of $X$ we find disjoint open sets $O$ and $O'$ such that $A \subseteq O$ and $A_{n+1} \subseteq O'$.

Then by the induction hypothesis we find pairwise disjoint open sets $U_1, \ldots U_n$ areounf $A_1, \ldots A_n$ resp.

Then $U_1 \cap O, U_2 \cap O, \ldots, U_n \cap O, O'$ are the required pairwise disjoint open neighbourhoods of all $A_i$.


Let $d$ be a metric for the space.

For $n=2:$ For $x\in A_1$ let $r_1(x)>0$ such that $B_d(x,r_1(x))\cap A_2=\phi.$ For $y\in A_2$ let $r_2(y)>0$ such that $B_d(y,r_2(y))\cap A_1=\phi.$

Let $U_1=\cup_{x\in A_1}B_d(x,r_1(x)/3).$ Let $U_2=\cup_{y\in A_2}B_d(y,r_2(y)/3).$

Now use induction on $n$:

Suppose it holds for some $n\geq 2.$ Let $A_1,.., A_{n+1}$ be $n+1$ pair-wise disjoint closed sets.

Let $B=\cup_{j=1}^n A_j .$ Then $B,\; A_{n+1}$ are disjoint closed sets, so let $V$ and $U_{n+1}$ be open sets with disjoint closures, such that $B\subset V$ and $A_{n+1}\subset U_{n+1}$ .

By the induction hypothesis let $U'_1,.., U'_n$ be $n$ open sets with pair-wise disjoint closures, such that $A_i\subset U'_i$ for $1\leq i\leq n.$

Now for $1\leq i\leq n$ let $U_i=U'_i\cap V.$

  • $\begingroup$ This technique of reducing the inductive step to the case $n=2$ is also seen in other subjects. E.g. if $x_1,..,x_n$ are pair-wise co-prime integers and $y\equiv 1\mod {x_i}$ for each $i$ then $y\equiv 1\mod {\prod_{i=1}^nx_i}.$ $\endgroup$ – DanielWainfleet Jan 7 '18 at 18:22

Look at the image below:

enter image description here

First than all, lets define the minimum distance between $A_i$'s. Let, $$min\lbrace{d(x,y),x\in A_i , y\in A_j}\rbrace$$ be defined as distance between any two sets of $A_i$'s. Obviously $\epsilon_{ij}=\epsilon_{ji}$. Now define $$\epsilon =min_{i,j}\epsilon_{ij}$$ be the least distance of all. Since we have finite number of sets, such $\epsilon$ exists and is positive. Now we define $U_i$'s as below: $$U_i=A_i\cup B_i$$ $$B_i=\lbrace{x\in X,\exists y\in A_i,d(x,y)<\epsilon_1}\rbrace$$ where $\epsilon_1$ is an arbitrary positive number $<{\epsilon\over 2}$. We must show $U_i$ is both open and disjoint with others. For proving openness, let $u\in U_i$. Then $u\in A_i$ or $u\in B_i$. Therefore for any $w$ such that $d(u,w)<\epsilon_1$ by definition we deduce that $w\in B_i$ so belongs to $U_i$. Then $U_i$ is open. Now we want to show that $U_i$'s are disjoint. Assume is not, meanwhile there are $U_i$ and $U_j$ so that $U_i\cap U_j\ne \phi$. Then there exist $u\in U_i$ and $u\in U_j$.

Case 1: $u\in A_i$ and $u\in B_j$

Then by definition we obtain:

$$u\in B_j\to \exists v\in A_j,d(u,v)<\epsilon_1<\epsilon$$

which contradicts with $d(u,v)>\epsilon$.

Case 2: $u\in B_i$ and $u\in B_j$

Therefore we arrive at:

$$\exists p\in A_i,q\in A_j,d(u,p)<\epsilon_1,d(u,q)<\epsilon_1\to d(p,q)\le d(p,u)+d(q,u)<2\epsilon_1<\epsilon$$

which is also at the same contradiction with $d(u,v)>\epsilon$. Then $U_i$'s are disjoint and open.


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