Is it true that every distributed lattice is also bounded?

Question

As far as i understand, bounded lattice had identity elements (0 and I) and distributive lattice also has them. Is that enough to say that every distributive lattice is bounded? If not, please give a counter example. thank you.

Answer 1

A distributive lattice is simply a lattice where the meet and join operations distribute over each other: $a \vee ( b \wedge c ) = ( a \vee b ) \wedge ( a \vee c )$ and $a \wedge ( b \vee c ) = ( a \wedge b ) \vee ( a \wedge c )$.

Any linearly ordered set is a distributive lattice (using $\max$ and $\min$ as the join and meet, respectively), but not all have maximum/minimum elements.

Answer 2

Not every distributive lattice is bounded, as the lattice arising from $(\mathbb{N},\leq)$ shows. Nevertheless, we have the following useful results:

• Every non-empty finite lattice is bounded.

• Every unbounded lattice can be embedded in a bounded one: just add two elements $0$ and $1$ with the needed properties. If the original lattice is distributive, then the bounded one is distributive too.