# Thread: Proof in a Metric Space

1. ## 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'.

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

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

RonL

3. ## thanks

Many thanks! 4. Originally Posted by CaptainBlack
Sketch of a proof (you may have to fill in detail):

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

5. Not really, the proof is fine. (we don't need completeness. but spoke too fast, rgep is right. )

6. Originally Posted by CaptainBlack
1. There exist a $\displaystyle \delta >0$ such that for all $\displaystyle v\epsilon V$ and $\displaystyle w\epsilon W$ $\displaystyle d(v,w)>\delta$.
No: consider the disjoint closed sets in $\displaystyle \mathbf{R}^2$ given by $\displaystyle V=\{(x,y) : x = 0\}$ and $\displaystyle W=\{(x,y) : xy=1\}$. You're implicitly assuming the space is compact.

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

7. Originally Posted by rgep
No: consider the disjoint closed sets in $\displaystyle \mathbf{R}^2$ given by $\displaystyle V=\{(x,y) : x = 0\}$ and $\displaystyle W=\{(x,y) : xy=1\}$. 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 $\displaystyle d(x,V) = \inf\{d(x,v) : v \in V\}$, observe that this is non-zero for $\displaystyle x \not\in V$ and then consider the set $\displaystyle \{ x : d(x,V) > d(x,W)\}$.
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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