Infinite Set S w/ Least Upper Bound & Acc. Point - Example?

[irony of truth]

I want to know some examples of an infinite set S with a least upper bound that is not an accumulation point of S. Is this an example... (-oo, 10]?

 

[rachmaninoff]

I want to know some examples of an infinite set S with a least upper bound that is not an accumulation point of S. Is this an example... (-oo, 10]?

No, it's not. Every point in that set is an accumulation point of it.

An example is
\left[0 , 1 \right] \cup \{ 2 \}.

edit: Another example is Z- ={-1,-2,-3,-4,-5...}, the set of negative integers. It's clearly infinite, bounded above, and does not have any accumulation points.

 

[irony of truth]

Hmmm, thank you for the help...

By the way, suppose I have Λ as my least upper bound of a set S but Λ is not in S. I want to know how this Λ is an accumulation point...

My friend told me that for any ε > 0, he can show that there is a point s belonging to S such that Λ - ε < s < Λ. To end up the proof, he used the definition of accumulation point... how do I prove this...? :D

 

[HallsofIvy]

Suppose &Lambda; is a least upper bound of a set, A, but not in the set itself. Since &Lamba; is an upper bound for A, there are no members of A larger than &lambda;. Given &epsilon;> 0 suppose there were no members of A between &Lambda;-&epsilon; and &Lambda;. Then there would be no members of A larger than &Lambda;-&epsilon;. That means that &Lamba;-&epsilon; is an upper bound for A, contradicting the fact that &Lambda; is the least upper bound.

 

[irony of truth]

So the concept of accumulation point comes in here "Given ε> 0 suppose there were no members of A between Λ-ε and Λ. ...".

 

[irony of truth]

Suppose Λ is a least upper bound of a set, A, but not in the set itself. Since &Lamba; is an upper bound for A, there are no members of A larger than λ. Given ε> 0 suppose there were no members of A between Λ-ε and Λ. Then there would be no members of A larger than Λ-ε. That means that &Lamba;-ε is an upper bound for A, contradicting the fact that Λ is the least upper bound.

May I clarify something... is this the proof for ( I have restated my problem) "Assume that Λ is the least upper bound of a set S but Λ is not in S. Show that Λ is an accumulation point of S" ?

 

[rachmaninoff]

So the concept of accumulation point comes in here "Given ε> 0 suppose there were no members of A between Λ-ε and Λ. ...".

May I clarify something... is this the proof for ( I have restated my problem) "Assume that Λ is the least upper bound of a set S but Λ is not in S. Show that Λ is an accumulation point of S" ?

Yes, you're right.

 

[irony of truth]

Ah.. ok.. I asked that because I thought the proof only shows that /\ is the least upper bound... and not that /\ is an accumulation point.

From the proof HallsofIvy stated, how did /\ turn out in the end to be an accumulation point (I apologize for being "slow")