Br(x) and Bs(y) such that Br(x) ∩ Bs(y) = ∅

[G-X]

Let [MATH]x∈R^k[/MATH] and [MATH]y∈R^k[/MATH] where k= 1 or 2 and [MATH]x≠y[/MATH]. Then there exist open balls Br(x) and Bs(y) such that [MATH]Br(x) ∩ Bs(y) = ∅[/MATH].

I have a few questions with regards to starting this proof:

If [MATH]x∈R^k[/MATH] and [MATH]y∈R^k[/MATH]. Do we need two proofs? With each starting with:

1) Let the metric space be defined by [MATH](R^1, d)[/MATH] where [MATH]x, y ∈ R^1 [/MATH]2) Let the metric space be defined by [MATH](R^2, d)[/MATH] where [MATH]x, y ∈ R^2 [/MATH]
If r, s are radii they won't be members of [MATH]R^1[/MATH] or [MATH]R^2[/MATH], so how do you define these distances as existing? I know r, s are a distance, but [MATH]r, s ∉ d[/MATH] because d is a function.

 

[pka]

Let [MATH]x∈R^k[/MATH] and [MATH]y∈R^k[/MATH] where k= 1 or 2 and [MATH]x≠y[/MATH]. Then there exist open balls Br(x) and Bs(y) such that [MATH]Br(x) ∩ Bs(y) = ∅[/MATH].
I have a few questions with regards to starting this proof:
If [MATH]x∈R^k[/MATH] and [MATH]y∈R^k[/MATH]. Do we need two proofs? With each starting with:
1) Let the metric space be defined by [MATH](R^1, d)[/MATH] where [MATH]x, y ∈ R^1 [/MATH]2) Let the metric space be defined by [MATH](R^2, d)[/MATH] where [MATH]x, y ∈ R^2 [/MATH]If r, s are radii they won't be members of [MATH]R^1[/MATH] or [MATH]R^2[/MATH], so how do you define these distances as existing? I know r, s are a distance, but [MATH]r, s ∉ d[/MATH] because d is a function.

In \(\Re^1\) if \(p\ne q\) then \(B_{\varepsilon }(p)=\{x: |x-p|<\varepsilon\}\) where \(\varepsilon=\frac{1}{2}|p-q|\).
In \(\Re^2\) if \((s,t)\ne (u,v)\) then \(B_{\varepsilon }((s,t))=\{(x,y): \sqrt{(s-x)^2+(t-y)^2}<\varepsilon\}\) where \(\varepsilon=\frac{1}{2}\sqrt{(s-u)^2+(t-v)^2}\).

 

[Steven G]

You do NOT mean to be asking If x∈R^k and y∈R^k. Do we need two proofs?

You should be asking if k=1 or 2 do we need two proofs? !!!

 

[Dr.Peterson]

Do we need two proofs? With each starting with:

1) Let the metric space be defined by [MATH](R^1, d)[/MATH] where [MATH]x, y ∈ R^1 [/MATH]2) Let the metric space be defined by [MATH](R^2, d)[/MATH] where [MATH]x, y ∈ R^2 [/MATH]

You could combine these cases into one proof, but in devising that I would probably have first done them separately anyway. If I were you, I would likely start with two cases as pka did, figuring that there must be something special about these two cases or they wouldn't be singled out; and maybe realize afterward that I could use a single proof.

In other words, do whatever you can do, then rethink afterward. You don't need to see the best solution to a problem from the start!

If r, s are radii they won't be members of [MATH]R^1[/MATH] or [MATH]R^2[/MATH], so how do you define these distances as existing? I know r, s are a distance, but [MATH]r, s ∉ d[/MATH] because d is a function.

Why would you want [MATH]r, s \in d[/MATH]? What would that even mean? Tell us your thinking.

I'd be interested in [MATH]d(x,y)[/MATH], which pka used without saying so.

 

[pka]

I'd be interested in [MATH]d(x,y)[/MATH], which pka used without saying so.

Actually I did not use \(d(x,y)\) anywhere is that post. True that is the usual notation for a distance function is metric spaces.
I carefully use \(\{x: |x-p|<\varepsilon\}\) Which is the interior of an interval: \((p-\varepsilon,p+\varepsilon)\). those are basic open sets for \(\Re^1\) as a metric space. For \(\Re^2\) the basic open sets are interiors of circles; for \(\Re^3\) the basic open sets are interiors of spheres using the ordinary Euclidean metric.

 

[G-X]

I was able to figure out the solution with the extra help. If you need additional guidance there is a proof under:
Distinct Points in Metric Space have Disjoint Open Balls

What helped me was the understanding that an axiom of metric space, specifically, the triangle inequality results in a contradiction from this problem.