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Scott domain is a non-empty partially ordered set if the following holds: D is bounded complete, i.e. all subsets of D that have some upper bound have a supremum. ...

What would be an example of a set that doesn't have a supremum? I was under impression that the whole point of introducing supremum was to cover cases like 0 < a < 1 when there is no maximum value but there is list upper bond sup(a) = 1.

My reference to partially ordered sets was confusing, because my question was about both ordered and partially ordered sets without supremum.

Stepan
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    Another question you might ask is: given a poset and any subset of that poset (bounded or not), can you "complete" the poset by giving the set a supremum in some larger poset? For example, take the real numbers as both the poset and the subset, then we can construct a poset $ \mathbf{R} \cup {\infty} $ by adding an element which is by definition greater than any other element. It seems reasonable that one can do this by adding in a single, new element and extending the order to that element in an appropriate way. – Trevor Gunn May 06 '17 at 01:47
  • A set of natural numbers indeed doesn't have a supremum among natural numbers. Some extension of a set of natural numbers is required. Thanks. – Stepan May 09 '17 at 13:09

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The set $\{ x \in \mathbb{Q} \mid x^2 < 2 \}$ does not have a supremum in $\mathbb{Q}$, even though it is bounded. However, it does have a supremum if we view it as a subset of $\mathbb{R}$. This is an example of how the least upper bound property (every set bounded from above has a supremum) encodes completeness.

Joppy
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  • @dxiv: well, any set satisfying those conditions can't have what the OP is asking for, which I assume is an example of a bounded set with no supremum, i.e. a $D$ without that property. – Joppy May 06 '17 at 01:33
  • @dxiv I'm pretty sure you misunderstood the question. Also $\mathbb{Q}$ is a poset. –  May 06 '17 at 01:36
  • @NoahRiggenbach I first read OP's question as asking for an example of a partially ordered set which has subsets with an upper bound but without a least upper bound, and I took that to mean less common cases than those in a totally ordered set. On a second read, it's quite possible that you (and Joppy) are right, and the above answers the posted question, so I removed my previous comment. – dxiv May 06 '17 at 01:51
  • My question was "does every set with memebnrs not exceeding some value have a supremum?" The answer is "No". A counterexample is a set of rational numbers below 2^(1/2) or below pi or some other irrational number. It answers my question. – Stepan May 07 '17 at 21:30