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Home / Lattices and Boolean Algebras / Show that in a poset a < a for no a and a

Show that in a poset a < a for no a and a

Solution: Suppose there exist some element a in a poset P, such that a<a.

Then by definition a≤a and a≠a

By anti-symmetry a≤a, a≤a ⟹a = a

Thus we get a contradiction.

Again, a < b, b < c ⟹ a ≤ b, a ≠ b

                                       b ≤ c, b ≠ c

So a ≤ b, b ≤ c ⟹ a ≤ c [by transitivity]

If a = c, then c ≤ b, b ≤ c ⟹ b =c, a contradiction.

Hence a < c.               (Proved)

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