Talk:Least-upper-bound property

Latest comment: 1 year ago by Wku2m5rr in topic Proof using Cauchy sequences

Dedekind completeness

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Dedekind completeness is the property that every Dedekind cut of the real numbers is generated by a real number. I would have simply removed the comment that Dedekind completeness is sometimes used to refer to the LUB property, but I wanted to get a second opinion on whether this misnomer is actually in sufficiently common usage. Hurkyl (talk) 11:15, 2 July 2012 (

Willard describes a space as dedekind complete space as a space where any subset with an upper bound has a least upper bound, which is the LUB property. This is from page 124/125 from exercise 17E. 130.63.185.213 (talk) 18:36, 17 April 2013 (UTC)Reply
Thanks for pointing that out. Citation and quote from Willard added here. --50.53.61.240 (talk) 17:38, 9 September 2014 (UTC)Reply

Proof of Bolzano Weierstrass Theorem

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The proof seems unconvincing. Someone better than me in mathematics please verify it, and perhaps replace it with the proof in the main page involving the monotone subsequence theorem followed by monotone convergence theorem. Venkatarun95 (talk) 04:06, 10 July 2014 (UTC)Reply

Proof using Cauchy sequences

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It says: "suppose that S has an upper bound B1". But S may contain no upper bound at all, all its upper bound being real numbers but not in S. For example, in the set ]0,1[ (or {0,1} if you prefer this notation), there is no upper bound belonging to S. Even its least one, 1, is not member of S. In that case, there is no "B1". and the proof does not work. wku2m5rr (talk) 07:17, 18 July 2023 (UTC)Reply

Well, I answer myself, I am wrong because the sentence "suppose that S has an upper bound B1" does not imply that this upper bound belongs to S and, of course, it never does. Sorry for this message. wku2m5rr (talk) 07:33, 18 July 2023 (UTC)Reply