Diam A = diam Cl(A) in metric space

In summary: So, in summary, we can prove that for any subset A of a metric space (X, d), the diameter of A equals the diameter of its closure, regardless of whether A is open or closed. This is shown by considering points x, y from Cl(A) and using the triangle inequality to obtain an upper bound on the diameter of A.
  • #1
radou
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As the title suggests, I have to prove, if (X, d) is a metric space, that for any subset A of X, diam A = diam Cl(A), i.e. the diameter of A equals the diameter of its closure.

So, if A is closed, it is trivial, since Cl(A) = A. Assume A is open. Now I'm a bit lost.

If A is open in (X, d), then it is a union of open balls in X. By the way, I know that for every open ball K(x, r) in X, diam K = diam Cl(K). Also, I have shown that Cl(A) can be written as a union of closed sets in X (since the closure of a union equals the union of closures).

Also, I know that, since A is a subset of Cl(A), diam A <= diam Cl(A) must hold.

And also, I know that for subsets A, B of X, diam (A U B) <= diam A + diam B + d(A, B), which can be generalized for any finite number of subsets.

The definitions of diameters didn't get me anywhere, too.

For some reason, I can't make any use of these facts to prove it. Any suggestion?
 
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  • #2
You can't just assume A is open. More likely it is neither open nor closed.

The diameter of Cl(A) can be represented by a sequence (xn,yn) of points such that d(xn,yn) converges monotonically to the diameter. See if you can hijack this sequence to find a lower bound on the diameter of A
 
  • #3
Could you tell me some more about this sequence? I'm not exactly sure how to construct it.

By the way, here's another idea. Let A be a nonempty bounded subset of X, so diam A = r. Then there exists some point x from A such that A is contained in the open ball K(x, r/2) of diameter r. But then the closed ball K[x, r/2] has a diameter r, too. Clearly Cl(A) is contained in K[x, r/2] (since if y were from X\K[x, r/2], which is open, there would exist a neighborhood of y which does not intersect A), so diam Cl(A) <= diam K[x, r/2] = r = diam A, which, when combined with diam A <= diam Cl(A), gives diam A = diam Cl(A).
 
  • #4
radou said:
Could you tell me some more about this sequence? I'm not exactly sure how to construct it.

If B is a subset of a metric space:

[tex]diam(B)=sup\{d(x,y)|x,y\in B \}[/tex].

For a number to be a supremum of a set, it means that it is the least upper bound. So there has to be a sequence of distances in B that converge to the supremum.. those distances are based on pairs (x,y)

Let A be a nonempty bounded subset of X, so diam A = r. Then there exists some point x from A such that A is contained in the open ball K(x, r/2)

First, it should be the closed ball you're looking at. You seem to be under the impression that A is open, and it isn't,. As an example, look at the set A=[0,1].

Also, the middle of this set does not have to exist. If your metric space is just the points {0,1/4,3/4,1} and A={0,1} you can't even find a center for A inside of the metric space, let alone inside of A
 
  • #5
Another option: Let x,y be arbitrary points in the closure and consider two small open balls around them. They contain points x',y' in A. Use those points and the triangle inequality to get an upper bound on d(x,y).
 
  • #6
Fredrik said:
Another option: Let x,y be arbitrary points in the closure and consider two small open balls around them. They contain points x',y' in A. Use those points and the triangle inequality to get an upper bound on d(x,y).

Thanks, I think I got it now.

Let x, y be from Cl(A). Then, for any ε > 0, there exist open balls K(x, ε) and K(y, ε) which intersect A. Let x' and y' be points from these intersections, respectively. d(x, y) <= d(x, x') + d(x', y') + d(y', y) < 2ε + d(x', y') <= 2ε + diam A. So, for every x, y from Cl(A), d(x, y) <= diam A. Since diam Cl(A) is the least upper bound for the set {d(x, y) : x, y from Cl(A)}, and diam A <= diam Cl(A), we conclude that diam A = diam Cl(A).
 
  • #7
Perfect. That's exactly what I had in mind.
 

What is "Diam A = diam Cl(A) in metric space"?

"Diam A = diam Cl(A) in metric space" is an equation in mathematics that states the diameter of a set A is equal to the diameter of the closure of set A in a metric space. This is also known as the closure diameter property.

What is a metric space?

A metric space is a mathematical concept that describes a set of objects where the distance between any two objects can be measured using a metric function. This function is a rule that assigns a positive number to any pair of objects in the set, representing their distance.

What is the closure of a set?

The closure of a set is the set itself along with all of its limit points. A limit point is a point that can be approached arbitrarily closely by points in the set. In other words, the closure of a set contains all of its boundary points and any points that can be reached by continuously moving from the boundary points.

What does "diameter" mean in this context?

In mathematics, the diameter of a set is the maximum distance between any two points in the set. It can also be thought of as the length of the longest line segment that can be drawn within the set.

Why is "Diam A = diam Cl(A) in metric space" important?

This equation is important because it shows that the diameter of a set and the diameter of its closure are equal in a metric space. This helps us understand the behavior of sets and their boundaries in a mathematical context.

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