Distributive lattice/Proofs

Lemma 1

Every totally ordered set is a distributive lattice with max as join and min as meet.

Proof

We will show:

: $x\; vee\; (y\; wedge\; z)\; =\; (x\; vee\; y)wedge(x\; vee\; z)$

We may suppose $yle\; z$ (If not, $zle\; y$ and we may switch $y$ and $z$.)

Recall that $yle\; z$ is equivalent to $yvee\; z\; =\; z$. Hence $y\; le\; z$ implies $(xvee\; y)\; vee\; (xvee\; z)\; =\; x\; vee\; z$, i.e., $xvee\; y\; le\; xvee\; z$, so the right hand side of the equation is equal to $(x\; vee\; y)wedge(x\; vee\; z)\; =\; x\; vee\; y$. On the left hand side we have $y\; wedge\; z\; =\; y$, so equality is established.

Note that the relation $x\; vee\; (y\; wedge\; z)\; le\; (x\; vee\; y)wedge(x\; vee\; z)$is true in all lattices, as both $x$ and $ywedge\; z$ are bounded above by $(x\; vee\; y)wedge(x\; vee\; z)$.

Lemma 2

Let L be a lattice, and let x be an element of L. If x is meet-prime, then it is meet-irreducible.

Proof

Suppose x = a v b. Then x ≤ a v b. x being meet-prime, x ≤ a or x ≤ b. Without loss of generality suppose x ≤ a. Then a v b ≤ a. By definition of v, a v b ≥ a. Therefore a v b = a and x = a.

Lemma 3

Let L be a distributive lattice, and let x be an element of L. If x is join-irreducible, then it is join-prime.

Proof

Recall that in a lattice x ≤ y ⇔ x ^ y = x.
Suppose x ≤ a v b. This is equivalent to x ^ (a v b) = x which by distributivity is in turn equivalent to (x ^ a) v (x ^ b) = x. x being meet-irreducible, x = x ^ a or x = x ^ b. This is equivalent to x ≤ a or ≤ b.