We have verified that \u0393 is a bijective function 66 a Let AB be sets and A \u03b1 \u03b1

# We have verified that γ is a bijective function 66 a

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We have verified that Γ is a bijective function. 6
6. (a) Let A,B be sets, and ( A α ) α I be a non-empty family of subsets of A . Let x be an object. x parenleftBigg intersectiondisplay α I A α parenrightBigg B iff ( x intersectiondisplay α I A α or x B ) iff (( x A α for any α I ) or x B ) iff (( x A α or x B ) for any α I ) iff (( x A α B ) for any α I ) iff x intersectiondisplay α I ( A α B ) It follows that parenleftBigg intersectiondisplay α I A α parenrightBigg B = intersectiondisplay α I ( A α B ). (b) Let A,B be sets and ( A α ) α I be a non-empty family of subsets of A . Let x be an object. x B \ parenleftBigg intersectiondisplay α I A α parenrightBigg iff ( x B and x / intersectiondisplay α I A α ) iff ( x B and (it is not true that x intersectiondisplay α I A α )) iff ( x B and (it is not true that ( x A α for any α I ))) iff ( x B and ( x / A α for some α I )) iff (( x B and x / A α ) for some α I ) iff ( x B \ A α for some α I ) iff x uniondisplay α I ( B \ A α ) It follows that B \ parenleftBigg intersectiondisplay α I A α parenrightBigg = uniondisplay α I ( B \ A α ). (c) Let A,B be sets, ( B β ) β J be a non-empty family of subsets of B , and f : A −→ B be a function. For any γ J , we have B γ uniondisplay β J B β . Then, for any γ J , we have f 1 ( B γ ) f 1 uniondisplay β J B β . Therefore uniondisplay δ J f 1 ( B δ ) f 1 uniondisplay β J B β . Pick any x f 1 uniondisplay β J B β . We have f ( x ) uniondisplay β J B β . Then there exists some κ J such that f ( x ) B κ . For the same κ , we have x f 1 ( B κ ) uniondisplay β J f 1 ( B β ). Therefore f 1 uniondisplay β J B β uniondisplay β J f 1 ( B β ). 7
It follows that f 1 uniondisplay β J B β = uniondisplay β J f 1 ( B β ). 7. Let ( f α ) α I be a family of functions, with f α = ( A α ,B α ,G α ) for any α I . Write A = uniondisplay α I A α , B = uniondisplay α I B α , G = uniondisplay α I G α . Consider the ordered triple f = ( A,B,G ). For any α I , we have G α A α × B α . Then G = uniondisplay α I G α uniondisplay α I ( A α × B α ) A × B . (Why?) Suppose that for any α,β I , for any x A α A β , f α ( x ) = f β ( x ). (a) We verify the following statement: ‘For any x A , there exists some y B such that ( x,y ) G .’ Pick any x A . There exists some σ I such that x A σ . For the same σ I , we have f σ ( x ) B σ B , and ( x,f σ ( x )) G σ G . We verify the following statement: ‘For any x A , y,z B , if ( x,y ) , ( x,y ) G then y = z .’ Pick any x A , y,z B . Suppose ( x,y ) , ( x,z ) G . Then there exist some σ,τ I such that ( x,y ) G σ and ( x,z ) G τ . For the same σ I , since ( x,y ) G σ , we have x A σ and y = f σ ( x ) B σ . For the same τ I , since ( x,z ) G τ , we have x A τ and z = f τ ( x ) B τ . Now, for the same σ,τ I , we have y = f σ ( x ) = f τ ( x ) = z under the assumption. It follows that f is a function. (b) Suppose f α is a surjective function for any α I .

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