Proof in a Metric Space

**TexasGirl** · January 17th 2006, 07:36 AM

Let (X, d) be a metric space, and let V,W contained in X be disjoint (ie the intersection of V and W equals zero), nonempty and closed. Prove that there exist disjoint open V' contained in X and W' contained in X such that V is contained in V' and W is contained in W'.

**CaptainBlack** · January 18th 2006, 05:10 AM

Originally Posted by TexasGirl

Let (X, d) be a metric space, and let V,W contained in X be disjoint (ie the intersection of V and W equals zero), nonempty and closed. Prove that there exist disjoint open V' contained in X and W' contained in X such that V is contained in V' and W is contained in W'.

Sketch of a proof (you may have to fill in detail):

1. There exist a $\delta >0$ such that for all $v\epsilon V$ and $w\epsilon W$ $d(v,w)>\delta$ .
For if this were not the case we could construct a sequence $v_i \epsilon V, i=1,2,3..$ with
limit $w \epsilon W$ , but this contradicts $V$ closed; as closed means the limit must be in $V$ .
(A closed set contains all its limit points)

2. Let $W'$ be the union of all open balls centred on points in $W$ of diameter $<\delta/2$ ,
and similarly for $V'$ . Then $V'$ and $W'$ have the required properties.

RonL

**TexasGirl** · January 18th 2006, 06:04 AM

Many thanks!

**CaptainBlack** · January 18th 2006, 07:36 AM

Originally Posted by CaptainBlack

Sketch of a proof (you may have to fill in detail):

1. There exist a $\delta >0$ such that for all $v\epsilon V$ and $w\epsilon W$ $d(v,w)>\delta$ .
For if this were not the case we could construct a sequence $v_i \epsilon V, i=1,2,3..$ with
limit $w \epsilon W$ , but this contradicts $V$ closed; as closed means the limit must be in $V$ .
(A closed set contains all its limit points)

I think there is a hidden assumption here that the space is complete.

I will have to look at this again to see if I can get around the problem.

A possible way around this problem is to work with the completion of X?

Further thought shows that this cannot be adapted to non-complete
metric spaces, sorry

RonL

**Rebesques** · January 19th 2006, 10:07 PM

Not really, the proof is fine.

(we don't need completeness. but spoke too fast, rgep is right.

)

**rgep** · January 19th 2006, 10:21 PM

Originally Posted by CaptainBlack

1. There exist a $\delta >0$ such that for all $v\epsilon V$ and $w\epsilon W$ $d(v,w)>\delta$ .

No: consider the disjoint closed sets in $\mathbf{R}^2$ given by $V=\{(x,y) : x = 0\}$ and $W=\{(x,y) : xy=1\}$ . You're implicitly assuming the space is compact.

Define $d(x,V) = \inf\{d(x,v) : v \in V\}$ , observe that this is non-zero for $x \not\in V$ and then consider the set $\{ x : d(x,V) > d(x,W)\}$ .

**CaptainBlack** · January 19th 2006, 11:17 PM

Originally Posted by rgep

No: consider the disjoint closed sets in $\mathbf{R}^2$ given by $V=\{(x,y) : x = 0\}$ and $W=\{(x,y) : xy=1\}$ . You're implicitly assuming the space is compact.

Compactness may suffice, but since I know what I was assuming
I can assure you that my assumption was in fact completness.

I had in fact constructed your counter example while investigating
the problems with the proof.

Define $d(x,V) = \inf\{d(x,v) : v \in V\}$ , observe that this is non-zero for $x \not\in V$ and then consider the set $\{ x : d(x,V) > d(x,W)\}$ .

Yes, I had got to this point, but had not yet had time to show that this
last set is open. I'm probably missing something obvious here

RonL

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