Question

3. Prove the following consequences of the well ordering principle:

(a) For all nonempty sets S that are bounded below (there exists an a ∈ Z such that for all s ∈ S, a ≤ s), there is a smallest element (there exists an a ∈ S so that for all s ∈ S, a ≤ s).
(b) For all nonempty sets S that are bounded above (there exists an a ∈ Z such that for all s ∈ S, a ≥ s), there is a largest element (there exists an a ∈ S so that for all s ∈ S, a ≥ s).

Answer 1

0 Solution :- (a) let s be a non empty subset of integet Z. Suppose s is bounded below and r is a lower bound of s. hence , res & SES. to show s has a smallest element Suppose if possible that s does not have smallest smallest. * &mes ; J PES such that Pem that is, if mes then ml ES now since r is finite , hence I integer i such that dcr.. how let mes then applying above argument (m-l) times we get les - which is contradiction. hence s must have smallest element. o let 5 be a non-empty set which is bounded above ands is a upper bound for s. > ass Anes. to show s has largest element Suppose if possible s does not have largest element = #mes, mtt ES. how .s is finite . I integez u such that scu how as mes ; apply above ab argument mtu times, we get UES which is contradiction, hence s must have lazgest element.