Proving totally bounded sets are bounded.

[gottfried]

Homework Statement

Find the error in this proof and give an example in (ℝ,d_e) to illustrate where this proof breaks down.

Proof that every totally bounded set in a metric space is bounded.

The set S is totally bounded and can therefore be covered by finitely many balls of radius 1, say N balls of radius 1. Then S is a subset of any ball B(x,2N) provided X lies in S. Thus diam S≤4N so that S is bounded.

I can't see the fault in the proof and therefore don't know where to start when looking for an example in (ℝ,d_e) that illustrates how the proof breaks down.
Any suggestions?

 

[voko]

A totally bounded set is bounded. The former is a stronger property than the latter. The converse may not be true in any metric space. I am not sure what $(R, d_e)$ means.

 

[gottfried]

What I mean by (R,de) is the set of real numbes with the euclidean metric. I understand that totally bounded is a stronger property and that a proof of this exists but I don't know what is wrong with the proof given in my first post. I'm trying to find out why this proof is sufficient.

 

[micromass]

I suppose it breaks down for $S$ empty...

 

[gottfried]

That's interesting and hadn't occurred to me. Is the empty set totally bounded because it is a finite subcover of itself?

 

[voko]

Would any ball cover the empty set?

 

[gottfried]

Yes surely. But does there exists, for every r, a finite A contained in the empty set such that U{B(a,r);a in A} contains the empty set?

 

[voko]

How could anything be contained in the empty set? The only thing it contains as a subset is itself.

 

[gottfried]

Sure. So in my above message A would be the empty set then U{B(a,r): a in A} would also be the empty set and therefore A is in the empty set which is covered by U{B(a,r): a in A} . Making the empty set totally bounded?

 

[voko]

I am not sure what you are trying do. What is U{B(a, r) : a in A}? This is a union over what?

The empty set is totally bounded because of the reason given in #6, to which you said "yes surely".

 

[gottfried]

Ok cool.

The definition of being totally bounded that I have been given is that X is totally bounded if for every r>0 there is a finite subset of X say A such that the union of balls U{B(a,r):a in A} contains X.

So I'm just trying to get it into this form for my understanding

 

[voko]

Obviously, if A is empty, then the union is empty. And the empty set contains itself.