Prove x belongs to the set or is an accumulation point.

[jrsweet]

Homework Statement

Let S be a nonempty set of real numbers bounded from above and let x=supS. Prove x either belongs to the set or is an accumulation point of S.

Homework Equations

x is an accumulation point of S iff each neighborhood of x contains a member of S different from x. That is, every neighborhood of x contains infinitely many points of S.

The Attempt at a Solution

So, do I need to prove that if x is not a member of S, then it is an accumulation point? I am a little confused about how to go about this.

So, there would obviously be two possibilities. Either x is a member of S, or it is not. If not, we need to prove x is an accumulation point. Wouldn't we need to know that S is infinite though? Is so, wouldn't it be much like the proof of the Bolzano-Weierstrass theorem?

Any help would be greatly appreciated! Thanks!

 

[jrsweet]

So, I might have got it...

I tried it by contradiction.
Assume x is not a member of S and assume it is not an accumulation point of S. If there is a neighborhood (x - epsilon, x + epsilon) containing x that does not have a point of S, then (x - epsilon) is an upper bound of S that's less than x. This contradicts x being the LEAST upper bound of S. Therefore, we have found a contradiction and x is indeed an accumulation point of S.

Just a thought. Let me know if it's right!

 

[Dick]

So, I might have got it...

I tried it by contradiction.
Assume x is not a member of S and assume it is not an accumulation point of S. If there is a neighborhood (x - epsilon, x + epsilon) containing x that does not have a point of S, then (x - epsilon) is an upper bound of S that's less than x. This contradicts x being the LEAST upper bound of S. Therefore, we have found a contradiction and x is indeed an accumulation point of S.

Just a thought. Let me know if it's right!

That's it exactly.

 

[jrsweet]

Thanks Dick! :)