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

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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!
 

Answers and Replies

  • #2
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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!
 
  • #3
Dick
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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.
 
  • #4
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Thanks Dick! :)
 

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