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Home / Lattices and Boolean Algebras / Show that product of two posets is a poset

Show that product of two posets is a poset

Proof: Let A and B be two posets . We have to show that A × B = {(a, b) | a ∈ A, b ∈ B}

Forms a poset under the relation defined by (a1, b1) ≤ (a2, b2)

⇔ a1 ≤ a2 in A and b1 ≤ b2 in B

It is clear that the tree relations ≤ occurring above are different, being the respective relations in A × B, A and B.

Reflexivity: (a, b) ≤ (a, b) ∀ (a, b) ∈ A × B

as a ≤ a in A and b ≤ b in B ∀ a ∈ A, b∈ B

Anti- symmetry: Let (a1, b1) ≤ (a2, b2) and (a2, b2) ≤ (a1, b1)

Then a1 ≤ a2, b1 ≤ b1 and a2 ≤ a1, b2 ≤ b1

⟹ a1 = a2, b1 = b2

⟹ (a1, b1) = (a2, b2)

Transitivity: Let (a1, b1) ≤ (a2, b2) and (a2, b2) ≤ (a3, b3)

Then a1 ≤ a2, b1 ≤ b2 and a2 ≤ a3, b2 ≤ b3

⟹ a1 ≤ a3 and b1 ≤ b3

⟹ (a1, b1) ≤ (a3, b3

Thus we can conclude that product of two posets is a poset. (Showed)

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