# Proof in a Metric Space

• Jan 17th 2006, 08:36 AM
TexasGirl
Proof in a Metric Space
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'.
• Jan 18th 2006, 06:10 AM
CaptainBlack
Quote:

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
• Jan 18th 2006, 07:04 AM
TexasGirl
thanks
Many thanks! :)
• Jan 18th 2006, 08:36 AM
CaptainBlack
Quote:

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
• Jan 19th 2006, 11:07 PM
Rebesques
Not really, the proof is fine. :)

(we don't need completeness. but spoke too fast, rgep is right. :confused: )
• Jan 19th 2006, 11:21 PM
rgep
Quote:

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 $\{ x : d(x,V) > d(x,W)\}$.
• Jan 20th 2006, 12:17 AM
CaptainBlack
Quote:

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.

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 $\{ x : d(x,V) > d(x,W)\}$.