diam A = diam Cl(A) in metric space

radou

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?

Office_Shredder

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 (x_n,y_n) of points such that d(x_n,y_n) converges monotonically to the diameter. See if you can hijack this sequence to find a lower bound on the diameter of A

radou

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).

Office_Shredder

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)

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

Fredrik

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).

radou

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).

Fredrik

Perfect. That's exactly what I had in mind.