I'm wondering how one goes about proving something is both closed and bounded. I need to prove that a closed ball of radius r about x0 is closed and bounded. I'm just not sure how these proofs typically go, and I don't have a great definition of closed or bounded. Thanks for any help.
There are a ton of ways to show a set is closed. Here are some:
1) Show the set contains all its limit points.
2) Show that the complement of the set is open.
3) Show it's a finite union, or arbitrary intersection, of other closed sets.
4) Show it equals its closure.
5) Show it's the pre-image of a closed set.
6) Show it's the image under a closed map of a closed set.
7) Perhaps it's known to be a compact subset of a Hausdorf space?
And I'm sure there are plenty more. Which to try depends on the circumstances.
If it's already been proven that the distance function in a metric space is a continuous function, then there's a very quick proof:
(Note that the distance function on R^n is obviously continuous, and so this works in R^n)
Let f = dist( *, x0):V -> R. Then D = f-inverse([0, r0]).
So since f is continous, and [0, r0] is known to be closed in R, have that D is closed.
I think that, maybe in this case, the next most simple way is to prove that D-complement is open.
Pick x not in D (that closed bal). Seek to show that x is in an open set that doesn't intersect D.
Since x not in D, dist(x, x0) > r. Let a = dist(x, x0) - r.
Let U = { x' | dist(x', x) < a/2}.
Then U open (why), x in U, and y in U implies y not in D (why? hint: triangle inequality).
So D complement is open. Therefore D is closed.
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Originally Posted by Shanter 
