Bounded Subsets of a Metric Space

[gajohnson]

Homework Statement

Let X be a metric space and let E be a subset of X. Show that E is bounded if and only if there exists M>0 s.t. for all p,q in E, we have d(p,q)<M.

Homework Equations

Use the definition of bounded which states that a subset E of a metric space X is bounded if there exists a q in X and an M>0 s.t. d(p,q)<M for all p in E.

The Attempt at a Solution

Proof (so far):
Let E be bounded. Then there exists M/2>0 and x in X s.t. d(p,x)<M/2 for all p in E.
Now, take arbitrary p,q in E and observe that:
d(p,q) ≤ d(p,x)+d(x,q) < M/2+M/2= M
Thus, d(p,q)<M for all p,q in E.

Now, I'm getting hung up on the second part of the proof, but I feel as if it shouldn't be hard. I think I've just been staring at this for too long at this point. Any advice as to how I ought to start going in the opposite direction would be greatly appreciated. I considered trying to show that E is compact (and would therefore be closed and, more importantly, bounded), but I'm not sure that's the best route or if it's awfully easy to do.

Thanks!

 

[micromass]

A bounded set is not compact in general. So you won't be able to show that E is compact.

Try to pick an arbitrary element p in E. What is the distance from p to the other points in E?

 

[gajohnson]

A bounded set is not compact in general. So you won't be able to show that E is compact.

Try to pick an arbitrary element p in E. What is the distance from p to the other points in E?

You're right, not sure what I was thinking there.

Yes, as I looked at it again I just reached that conclusion before I saw your response. I suppose I overlooked that because it's so obvious!

Is it as simple as picking any q in E? Then q is in X and d(p,q)<M for all p in E.

Thanks!

 

[micromass]

Doesn't work if E is empty though, so you want to treat that in a special case.

 

[gajohnson]

Doesn't work if E is empty though, so you want to treat that in a special case.

Ah, of course.

Thanks for your help!