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

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In summary: The proof states that for any ε> 0 there is a point s belonging to S such that Λ-ε <s < Λ. To end up the proof, he used the definition of an accumulation point... how do I prove this...?
  • #1
irony of truth
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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]?
 
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  • #2
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
[tex] \left[0 , 1 \right] \cup \{ 2 \} [/tex].

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.
 
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  • #3
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
 
  • #4
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.
 
  • #5
So the concept of accumulation point comes in here "Given ε> 0 suppose there were no members of A between Λ-ε and Λ. ...".
 
  • #6
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" ?
 
  • #7
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.
 
  • #8
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")
 

1. What is an infinite set with a least upper bound?

An infinite set with a least upper bound is a set that has an infinite number of elements and also has a specific element that is greater than or equal to all other elements in the set. This element is known as the least upper bound, or the supremum, of the set.

2. What is an accumulation point?

An accumulation point, also known as a limit point, is a point in a set where an infinite number of elements of the set are located. This means that any neighborhood of the accumulation point contains an infinite number of elements from the set.

3. Can you give an example of an infinite set with a least upper bound and an accumulation point?

Yes, an example of such a set is the set of all real numbers greater than or equal to 0 and less than 1. The least upper bound of this set is 1, and 0 is an accumulation point since any neighborhood of 0 contains an infinite number of elements from the set.

4. How do you prove the existence of a least upper bound and an accumulation point in an infinite set?

To prove the existence of a least upper bound in an infinite set, you need to show that the set is bounded above and that there is no smaller number that can serve as the least upper bound. To prove the existence of an accumulation point, you need to show that for any point in the set, there is an infinite number of points in the set that are arbitrarily close to that point.

5. What is the significance of an infinite set with a least upper bound and an accumulation point in mathematics?

An infinite set with a least upper bound and an accumulation point is significant in mathematics because it allows for the study of infinite sets and their properties. This concept is also important in real analysis, topology, and other branches of mathematics.

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