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Does a set being bounded with respect to a order imply that Sup or inf exist?

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I think your question is a bit vague. Can you formalize your question. What is a "bounded partially ordered set?" Implicitly, it would seem that a bounded partially ordered set has a maximum and minimum, otherwise there is no bound on either side, but I've never heard a poset called "bounded." –  Thomas Andrews Jun 4 '13 at 14:36

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Let $\Bbb N\cup\{\bullet_1,\bullet_2\}$ be the set, where $\bullet_i\notin\Bbb N$. We extend the natural $\leq$ on $\Bbb N$ by setting, $n\leq\bullet_i$ for all $n\in\Bbb N,i=1,2$. Note that $\{\bullet_1,\bullet_2\}$ is an antichain.

Now $\Bbb N$ is bounded but has no supremum. The same idea can be used on $\Bbb Z$ so you have no infinimum either.

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Any subset of that order that is bounded is a singleton, isn't it? Maybe I'm misunderstanding the question. –  Thomas Andrews Jun 4 '13 at 14:30
    
@Thomas: I misunderstood the question, thanks. –  Asaf Karagila Jun 4 '13 at 14:40

The question is vague, but here's an answer to one possible interpretation of it. In the set of rational numbers, with the usual order relation, the subset $\{x:x^2<2\}$ is bounded (above and below) but has no supremum or infimum.

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