0
$\begingroup$

I'm reading an article and it says that, if $l$ has a $\bar{l}$ complement in a distributive lattice $L$, then $\bar{l}$ is a pseudocomplement of $l$. Can someone point me a hint, I'm new with lattice theory.

$\bar{l}$ is complement of $l$ if $l\vee\bar{l}=\top$ and $l\wedge\bar{l}=\bot$.

A pseudocomplement of $l\in L$ is an element $l^{\star}$ such that $m\leq l^{\star}$ if and only if $m\wedge l=\bot$.

$\endgroup$
0
$\begingroup$

If $m \le \bar{l}$ then $m \land l \le \bar{l} \land l = \bot$ is immediate (in any lattice $x \le y$ implies $x \land z \le y \land z$ for any $x,y,z$). And always $\bot \le m \land l$ so $m \land l = \bot$.

OTOH, if $m \land l = \bot$, then $\bar{l}= \bar{l} \lor \bot = \bar{l} \lor (m \land l)$ and we apply distributivity to get $\bar{l}= (\bar{l} \lor m) \land (\bar{l} \lor l) = (\bar{l} \lor m) \land \top = \bar{l} \lor m$ so in conclusion

$$\bar{l} = \bar{l} \lor m \implies m \le \bar{l}$$ as required.

So a complement in a distributive lattice is a pseudocomplement.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.