Question

(b) Let S={1,2,3}, and define the poset (P(S),⪯) by A⪯B if and only if A⊆B. Verify that this poset is a lattice. Is it a total ordering?
(c) Using your work in part (b), is every lattice necessarily a total ordering?

Answer 1

(b) A poset is a lattice if their exist sup and inf in P(S).

Sup {A,B} = $A \cup B$

Inf {A,B} = $A\cap B$

P(S) = {ɸ, {1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}}

Let A = {1}

B = {1,2}

$A \subseteq B$

Sup {A,B} = {1} $\cup$ {1,2} = {1,2} = A $\cup$ B

Inf {A,B} = $A \cap B$ = {1} $\cap$ {1,2} = {1}

Sup {A,B} and Inf {A,B} exist in P(S).

Poset $(P(S), \leqslant )$ is a lattice.

$(P(S), \leqslant )$ is totally ordering. Since, every element in P(S) is comparable.

(c) Yes, every lattice necessarily a total ordering.

Since, by definition of lattice: a poset $(P(S), \leqslant )$ is a lattice if for every $A,B \in P(S)$ , $\exists$ last upper bound and greatest lower bound. Hence, in $(P(S), \leqslant )$ , $\exists$ supremum and infimum for $A,B \in P(S)$ .

Hence, it is totally ordering. Every lattice is necessarily a totally ordering.