Answer to Question #148114 in Discrete Mathematics for Promise Omiponle

Question #148114
(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?
1
Expert's answer
2020-12-16T11:56:51-0500

(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.


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