2
$\begingroup$

A lattice is a set $L$ with two binary operations, $\lor$ "join" and $\land$ "meet". In a complemented lattice, for every element $a$ there exists an element $a^{\perp}$ such that $a \lor a^\perp=1$ and $a \land a^\perp=0$. Note this "complement" element need not be unique. But if it is unique, as is true in distributive lattices, the lattice is called a uniquely complemented lattice.

It seems to me that in a uniquely complemented lattice, each element having a unique "complement" element could define a unary operation $\perp: L \to L$ defined by $\perp(a)=a^\perp$. But I have never seen it developed this way (the way the additive inverse is defined as unary operation in a group or the multiplicative inverse is defined as a unary operation in a field). Is there a problem with defining a uniquely complemented lattice this way: that is, a lattice with a unary operation $\perp$ in addition to its two binary operations $\lor$ and $\land$?

Along the same lines, is it OK to define $0$ and $1$ as nullary operations on the lattice, the same way $0$ and $1$ are considered nullary operations on a ring? Although it seems correct to me, I could be missing something. Have nullary and unary operations been described this way in lattice theory before? Thanks for reading!

$\endgroup$

1 Answer 1

3
$\begingroup$

Nullary (constants) and unary operations are perfectly fine in universal algebra. And indeed, top and bottom elements in a bounded lattice are defined using constants. It's a little bit surprising that you didn't meet that definition yet; actually the phrase “in the language of bounded lattices” implicitly says that you have both 0 and 1 in your signature.

I don't exactly remember the treatment there, but it is for sure that in B&S A Course in Universal Algebra this is explained this way.

$\endgroup$
1
  • $\begingroup$ Thanks for your answer, Pedro! And thanks for the book you referenced - it looks like a great resource! $\endgroup$ Commented Sep 11, 2015 at 19:45

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.