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I'm looking for a set $M$ which is partially ordered by $\subseteq$. $M$ should have a lower bound but no infimum. Is that possible?

A lower bound is an element $x \in N$ with $M \subset N$ such that for all elements $y \in M$, $x \subseteq y$ holds.

An infimum is a maximal lower bound with respect to $\subseteq$.

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    What does "have a lower bound but no infimum" mean? Lower bound in a larger order? Or for every set there should be a lower bound, but not an infinimum? Or should there be a set without an infimum but that has a lower bound?2012-11-04
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    Now your revision has me confused: you call $x$ and $y$ elements of sets, then imply they too are sets (given your notation $y \subseteq y$, unless you mean $\subseteq$ in some other context?)2012-11-04
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    Yes that exactly right. $x,y$ are elements of $M$ but there are sets themself partially ordered by $\subseteq$.2012-11-04
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    @amWhy: It’s okay, except that *minimal* should be *maximal* in the last line. The partial order is $\langle N,\subseteq\rangle$, and $M\subseteq N$ is to be a set with a lower bound but no infimum. The elements of $N$ **are** sets.2012-11-04
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    @Brian and joachim: Got it...2012-11-04

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It doesn’t make sense to say that $M$ has a lower bound but no infimum if $M$ is the partially ordered set, so I’m assuming that you want a set $N$ partially ordered by $\subseteq$ in which there is a subset $M$ with a lower bound but no infimum. (This is now clear from the revised version of your question.)

That is possible. Let $N=\{(x,\to):x\in\Bbb Q\}$, the set of all open rays unbounded on the right; $N$ is even linearly ordered by $\subseteq$. Let $M=\{(x,\to)\in N:x>\sqrt2\}$.