Solution to (b) Let S={1,2,3}, and define the poset (P(S),⪯) by A⪯B if and only if A⊆B. … - Sikademy
Author Image

Archangel Macsika

(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?

The Answer to the Question
is below this banner.

Can't find a solution anywhere?


Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

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.

Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-3478-qpid-2177