A partially ordered set $\langle P,\leq\rangle$ is separative iff it satisfies the following condition: \[ \neg x\leq y\Rightarrow\exists z(z\leq x\wedge z\bot y) \] where: \[ x\bot y\iff\neg\exists z(z\leq x\wedge z\leq y). \] If $\langle B,\leq\rangle$ is a complete Boolean lattice (every subset of $B$ has supremum), then $B^+=B\setminus\{0\}$ is separative.

My question is: what if $B$ is not complete? Is it separative or does not have to be separative?


Recall that $\lnot x\leq y$ is the same as saying that $x\cdot y\neq x$. Let $z=x\cdot\bar y$ (where $\bar y$ is the complement of $y$).

Then $z\leq x$, and of course $z\leq\bar y$. But also $z\cdot y=x\cdot\bar y\cdot y=x\cdot 0=0$. Therefore $z\perp y$ as wanted.

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    $\begingroup$ Thanks a lot for a quick response! $\endgroup$ – Mad Hatter Oct 25 '13 at 12:27

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