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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...?

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irony of truth said:

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

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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

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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 theleastupper 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" ?

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irony of truth said:

Yes, you're right.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" ?

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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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