# An example of a set

#### 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]?

R

#### rachmaninoff

irony of truth said:
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.

Last edited by a moderator:

#### 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

Science Advisor
Homework Helper
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

HallsofIvy said:
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" ?

R

#### rachmaninoff

irony of truth said:
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 said:
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")

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