S 5 S C x x M M t N We now claim that the collection C x x M forms a parti tion

# S 5 s c x x m m t n we now claim that the collection

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S 5: S { C x : x M } M t N We now claim that the collection { C x : x M } forms a parti- tion of M t N (recall that this means that any two given C x s are either identical or disjoint—see Definition A.3.2.9). If C x 1 is disjoint from C x 2 we are done, so instead suppose that there is some element x 0 that is in both C x 1 and C x 2 . First, let us do the case in which x 0 M . From the definition of C x , we then must have that [ g f ] k ( x 1 ) = x 0 = [ g f ] l ( x 2 ) (A.5.1.11) for some k , l Z . Without loss of generality, suppose that k l . Then, applying f - 1 g - 1 to both sides of this equation k times, e we find that x 1 = [ g f ] l - k ( x 2 ) . (A.5.1.12) In other words, x 1 C x 2 . Not only this, but f ( x 1 ) ∈ C x 2 as well because f ( x 1 ) = f ( [ g f ] l - k ( x 2 ) ) . Similarly, g - 1 ( x 1 ) ∈ C x 2 , and so on. It follows that C x 1 C x 2 . Switching 1 2 and applying the same arguments gives us C x 2 C x 1 , and hence C x 1 = C x 2 . Thus, indeed, { C x : x M } forms a partition of M t N . In particular, it follows that C x = C x 0 for all x 0 C x . (A.5.1.13) S 6: D X 1 , X 2 , Y 1 , Y 2 Now define A B x M s.t. C x N f ( M ) C x (A.5.1.14) A.5 Cardinality, countability, and the naturals 565 as well as X 1 B M A , Y 1 B N A , (A.5.1.15a) X 2 B M A c , Y 2 B N A c . (A.5.1.15b) Note that, as { C x : x M } is a partition of M t N , we have that A c = x M s.t. C x N * f ( M ) C x . (A.5.1.16) S 7: S f | X 1 : X 1 Y 1 We claim that f | X 1 : X 1 Y 1 is a bijection. First of all, note the it x X 1 , then in fact f ( x ) ∈ Y 1 , so that this statement indeed does make sense. Of course, it is injective because f is. To show surjectivity, let y Y 1 B N A . From the definition of A (A.5.1.14) , we see that y C x N for some C x with C x N f ( M ) , so that y = f ( x 0 ) for some x 0 M . We still need to show that x 0 X 1 . However, we have that x 0 = f - 1 ( y ) , and so as y C x , we have that x 0 = f - 1 ( y ) ∈ C x as well. We already had that C x N f ( M ) , so that indeed x 0 A , and hence x 0 X 1 . Thus, f | X 1 : X 1 Y 1 is a bijection. S 8: S g | Y 2 : Y 2 X 2 We now show that g | Y 2 : Y 2 X 2 is a bijection. Once again, all we must show is surjectivity, so let x X 2 = M A c . From the definition of A (A.5.1.14) , it thus cannot be the case that C x N is contained in f ( M ) , so that there is some y C x N such that y < f ( M ) . By virtue of (A.5.1.13) , we have that C x = C y , and in particular x C y . From the definition of C y , it follows that either (i) x = y , (ii) x is in the image of f - 1 , or (iii) x is in the image of g (the other possibilities are excluded because x M ). Of course it cannot be the case that x = y because x M and y N . Likewise, it cannot be the case that x is in the image of f - 1 because x A c . Thus, we must A Basic set theory 566 have that x = g ( y 0 ) for some y 0 N . Once again, we still must show that y 0 Y 2 . However, we have that y 0 = g - 1 ( x ) , so that y 0 C x . Furthermore, as C x N is not contained in f ( M ) , from (A.5.1.16) it follows that C x A c . Thus, y 0 C x A c , and so y 0 Y 2 . Thus, g | Y 2 : Y 2 X 2 is a bijection.  #### You've reached the end of your free preview.

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