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# sols21 - 21 NOVEMBER 17TH CARDINALITY III 49 21 November...

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21. NOVEMBER 17TH: CARDINALITY III 49 21. November 17th: Cardinality III 21.1 . Complete the alternative proof of Lemma 20.3 sketched in § 21.1. That is: with notation as in the end of § 21.1, prove that ψ does not equal ϕ s for any s S . Answer. Recall that ψ : S 0 , 1 is defined by ψ ( s ) = 1 if ϕ s ( s ) = 0 0 if ϕ s ( s ) = 1 Now let s S . Then either ϕ s ( s ) = 0, or ϕ s ( s ) = 1. If ϕ s ( s ) = 0, then ψ ( s ) = 1, so ψ ( s ) = ϕ s ( s ). This shows that ψ = ϕ s in this case. If ϕ s ( s ) = 1, then ψ ( s ) = 0, so ψ ( s ) = ϕ s ( s ). This shows that ψ = ϕ s in this case as well. This shows that ϕ s = ψ for all s S , which was the last step to verify in order to conclude the alternative proof of Lemma 20.3. Of course this argument mimics precisely the analogous part of the proof of Lemma 20.3. Whether you saw how to do this or not, please make sure you under- stand that the ‘diagonal argument’ is just another way to formulate the idea behind the original proof of Lemma 20.3. 21.2 . Let A , B be sets. Prove that | A | | B | if and only if there exists a surjective function B A . (Argue as in the beginning of the proof of Proposition 21.3.) Answer: By definition, | A | | B | if and only if there is an injective function f : A B . If f : A B is injective, then it has a left-inverse g : B A (Theorem 12.11); as g f = id A , g has a right inverse, hence it is surjective. Thus, if | A |

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