Diameter of an open ball = 2*radius

In summary, the homework statement is that the diameter of a subset of a metric space is equal to the sum of the distances inside the subset. The homework equation is that this sum is bounded by a certain constant. The attempt at a solution is to show that this equation holds for a given metric space, but the problem is proving it for arbitrary spaces. If the metric space is a normed vector space, the statement is false.
  • #1
Fredrik
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Not really homework, but I guess it's a "textbook-style question", so I'm putting it here.

Homework Statement



Suppose (X,d) is a metric space, and consider an open ball [itex]B(x,r)=\{y\in X|d(x,y)<r\}[/itex]. The diameter of a subset [itex]E\subset X[/itex] is defined as [itex]\text{diam\,}E=\sup\{d(x,y)|x,y\in E\}[/itex]. I want to show that [itex]\text{diam\,}B(x,r)=2r[/itex].

Homework Equations



Only the above, and the definition of a metric.

The Attempt at a Solution



If [itex]p,q\in B(x,r)[/itex], we have [itex]d(p,q)\leq d(p,x)+d(x,q)<r+r=2r[/itex], so 2r is an upper bound for the set [itex]\{d(x,y)|x,y\in E\}[/itex], and diam B(x,r) is the least upper bound of that set, so [itex]\text{diam\,}B(x,r)\leq 2r[/itex].

That means that it's sufficient to show that [itex]\text{diam\,}B(x,r)\geq 2r[/itex]. My idea is to show that for every [itex]\epsilon[/itex] such that [itex]0<\epsilon<2r[/itex], there exist points [itex]p,q\in B(x,r)[/itex], such that [itex]d(p,q)>2r-\epsilon[/itex]. This looks like it should be really easy, and it probably is, but for some reason I can't see the solution. Maybe we can assume that [itex]d(p,q)\leq 2r-\epsilon[/itex] for all p,q in B(x,r), and show that this implies that B(x,r) is a subset of a smaller open ball (which of course would be a contradiction that proves the assumption false).

If the metric space is a normed vector space, this gets really easy, but I'd like to prove it (or disprove it) for an arbitrary metric space.
 
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  • #2
If you just have a metric space the statement is false.

For example, X is the integers, with the normal norm d(a,b)=|a-b|. B(0,.5) is just {0} which has a diameter of 0
 
  • #3
D'oh. Thank you. By the way, that was the fastest useful response I ever got. :approve:
 

1. What is the formula for calculating the diameter of an open ball?

The formula for calculating the diameter of an open ball is 2 times the radius of the ball.

2. How is the diameter of an open ball related to its radius?

The diameter of an open ball is equal to twice the radius of the ball. This means that the diameter is always double the size of the radius.

3. Why is the diameter of an open ball important in scientific calculations?

The diameter of an open ball is important in scientific calculations because it is a measure of the size of the ball and can be used to determine other important properties such as volume and surface area.

4. Can the diameter of an open ball be negative?

No, the diameter of an open ball cannot be negative. It is a measurement of distance and therefore must always be a positive value.

5. How is the diameter of an open ball different from the diameter of a closed ball?

The diameter of an open ball is the distance between two points on the boundary of the ball, while the diameter of a closed ball is the distance between two points on the surface of the ball. This means that the diameter of an open ball is always smaller than the diameter of a closed ball.

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