# sols3 - M114S, Solutions to HW #3 x3.4. First note that if...

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Unformatted text preview: M114S, Solutions to HW #3 x3.4. First note that if W is a connection of A and B , then W ⊆ P ( A ∪ B ). Thus W ∈ P ( P ( A ∪ B )). So the set Σ( A,B ) of connections of A and B is a subset of P ( P ( A ∪ B )); it exists by the Separation Axiom (since the conditions in the definition of “connection” are all definite). x3.5. For the first property we need a set C disjoint from and equinumerous with A . Let t = r ( uniontext A ) be the object guaranteed by the Russell paradox to be outside uniontext A , and take C = df {{ t,x } | x ∈ A } . This is a set (by the Powerset and doubleton axioms), and it is disjoint from A ; because if { t,x } ∈ A for some x , then t ∈ uniontext A , contradicting the choice of t . Moreover, there is an easy connection of A with C , namely W = {{ x, { t,x }} | x ∈ A } = { z ∈ P ( A ∪ C ) | ( ∃ x ∈ A )[ z = { x, { t,x }} ] } . The transitivity property also needs an argument. We are given E , F , dis- joint respectively from A , B and B , C , and the relevant connections. Choose some t / ∈ uniontext ( A ∪ C ) as above, use the given connections to define a bijection f : A ֌ → C in ordinary terms, and let M = {{ t,a,f ( a ) } | a ∈ A } . This is disjoint from both A and C , as above, and there are simple connections of it with both these sets. x4.1. Wiener’s pair is ( x,y ) = df {{∅ , { x }} , {{ y }}} . The main point of this is that ( x,y ) is a set with two elements; this is because ∅ negationslash = { x } , so {∅ , { x }} has exactly two elements, while {{ y }} has exactly one element—so {∅ , { x }} and {{ y }} cannot be equal. We first check(P1): Supposethat ( x,y ) = ( x ′ ,y ′ ) underWiener’sdefinition of pair. We need to show that x = x ′ and y = y ′ . We know that {{∅ , { x }} , {{ y }}} = {{∅ , { x ′ }} , {{ y ′ }}} . It follows that the two-element member of the first set equals the two-element member of the second. Thus, {∅ , { x }} = {∅ , { x ′ }} . Since { x } and { x ′ } are not empty, we must have { x } = { x ′ } . Thus x = x ′ . Furthermore, the one- element subsets of ( x,y ) and ( x ′ ,y ′ ) must be equal. That is, {{ y }} = {{ y ′ }} ....
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## This homework help was uploaded on 04/16/2008 for the course MATH 114S taught by Professor Moschovakis during the Winter '08 term at UCLA.

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sols3 - M114S, Solutions to HW #3 x3.4. First note that if...

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