# Is there bounded partially ordered sets which sup doesn't exist?

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

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.
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.