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Let P ( S ) denotes the power set of the set S, the dual of the lattice ( P(S), ⊆ ) is a) Doesn't’t exist b) ( P(S), ⊆ ) c) ( P(S), ⊇ ) d) ( S,⊇)

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good question :)

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to prove that a lattice is dual we have to prove that it is a POSET

for reference -> https://en.wikipedia.org/wiki/Duality_(order_theory)

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for power set:

1)reflexive : every set is a subset of itself. so its reflexive

2)anti symmetric: if a is subset of b and b is a subset of a means a and b are equal(reflexive). so it is antisymmetric as a is subset of b doesnt means b is a subset of a .

3)transitive: a is subset of b and b is subset of c means a is subset of c.

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which is a POSET. hence dual is option C (as the posets <P,R> and <P,R-1> are called DUALS.

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