Talk:Heine-Borel Theorem/Dedekind Complete Space

Quite convinced it's correct. The lemma is awkwardly phrased - it'd be more natural to state "... Then $b \ne \sup S$ or similar. Incidentally such a lemma has settled in my mind for proving my once-more-adapted alternative proof of the other theorem (what was the title again) and it may therefore serve its purpose as a genuine result, be it with a proper title or "Lemma 1" in my newly proposed paradigm. --Lord_Farin (talk) 21:18, 11 February 2013 (UTC)

The lemma doesn't require any form of completeness, so it can't refer to a supremum. --Dfeuer (talk) 21:36, 11 February 2013 (UTC)

"... Then there is another upper bound $a \in X$ with $a \prec b$", then - nonexistence of suprema makes sense whether or not completeness plays a role. --Lord_Farin (talk) 21:41, 11 February 2013 (UTC)

Yes, but you can't phrase it as an inequality. That's all I was saying. By the way, check out Heine-Borel iff Dedekind Complete. --Dfeuer (talk) 21:46, 11 February 2013 (UTC)