Spotted Beebalm CBS Theorem J. Larson, C. Porter UF

Transcription

1 Spotted Beebalm

2 Cantor-Schröder-Bernstein Theorem, Part 2 Jean A. Larson and Christopher C. Porter MHF 3202 December 4, 2015

3 Theorem (Cantor-Schröder-Bernstein Theorem) Suppose A and B are sets. If A B and B A, then A B.

4 Opening of the Proof: Recalll that for any function F : U V and any subset D U, the image of D under a F is the set F (D) := {F (d) d D}. Assume A B and B A (o ). Let f : A B and g : B A be one-to-one functions that witness the above relations (a ). Case 1: One of f and g is onto. Then one of f and g is a witness that A B. Case 2: Neither f nor g is onto

5 Plan of the proof for Case 2: Define a set X A by recursion Set W := f (X ) = {f (x) x X } and show f (X W ) : X W is one-to-one and onto. Set Z = B \ W, Y = g(z), and prove that Y = g(z) = A \ X. Show g (Z Y ) is one-to-one and onto, so (g (Z Y )) 1 = (g (Y Z)) 1 is also one-to-one and onto. Show that h = (f (X W )) (g (Y Z)) 1 : A B is the witness that A B.

6 Definitions of R, A and W : Let R = Ran(g) A. Define X by recursion: A1 := A \ R; for every n Z +, A n+1 := g(f (A n )) = {g(f (a)) a A n }. Then X := {A n n Z + }. Set W := f (X ) = {f (x) x X } = Ran(f X ).

7 Claim 1: The function f (X W ) : X W is one-to-one and onto. Proof. Since f is one-to-one and f (X W ) is a restriction of f, it follows from Exercise 5.2: 10a, that f (X W ) is one-to-one. Since f (X ) = W and f and f (X W ) agree with f on X by Exercise 5.1: 7a, it follows that Ran(f (X W )) = f (X ) = W. Thus by Theorem 5.2.3, f (X W ) is onto.

8 Claim 2: Y = g(z) = {g(z) z Z} A \ X where Z := B \ W. Proof. Assume toward a contradiction that Y = g(z) A \ X (o ). Let g(z 0 ) g(z) A be a witness, i.e. assume z 0 Z and g(z 0 ) / A \ X (a ). Since g(z 0 ) / A \ X, it follows that g(z 0 ) X (def set difference). Since g(z 0 ) is in the range of g, it is not in A 1 = Z \ Ran(g). Since X = {A n n Z + } and g(z 0 ) / A 1, we can find a witness n 0 Z + with g(z 0 ) A n0 +1 (a ).

9 Claim 2 (proof continued) Since g(z 0 ) A n0 +1 and A n0 +1 = g(f (A n0 )), we know g(z 0 ) g(f (A n0 )) = {g(f (x) x A n0 }. Let x 0 be a witness, i.e. assume g(z 0 ) = g(f (x 0 )). Since g is one-to-one, it follows that z 0 = f (x 0 ) W, by the definition of W. Thus z 0 Z = B \ W and z 0 W, which is a contradiction, since these two sets are disjoint (c ) Thus our assumption was false and Claim 2 follows.

10 Claim 3: A \ X Y = g(z) = {g(z) z Z}. Assume toward a contradiction that A \ X g(z) (o ). Let a 0 A \ X be a witness, i.e. assume a 0 / g(z) (a ). Since y 0 A \ X, it follows that y 0 / X and in particular, y 0 / A 1 = A \ Ran(g), so y 0 Ran(g). Let b 0 B be a witness, i.e. g(b 0 ) = y 0 (a ). Since y 0 / g(z), it follows that b 0 / Z = B \ W, so b 0 W = f (X ). Let x 0 X be a witness, i.e. assume f (x 0 ) = b 0 and let m 0 be such that x 0 A m0 (a ). Thus a 0 = g(b 0 ) = g(f (x 0 )) g(f (A m0 )) = A m0 +1 X, so a 0 is in both X and A \ X which is a contradiction since these sets are disjoint (c ). So our assumption was false and Claim 3 follows.

11 Claim 4:The function g (Z Y ) : Z Y is one-to-one and onto and so is its inverse, g 1 (Y Z) : Y Z. Proof. By Claims 2 and 3, Y = g(z) = A \ X. Since g is one-to-one and g (Z Y ) is a restriction of g, it follows from Exercise 5.2: 10a, that g (Z Y ) is one-to-one. Since g(z) = Y and, by Exercise 5.1: 7a, g and g (Z Y ) agree with g on Z, it follows that Ran(g (Z Y )) = g(z) = Y, so g (Z Y ) is onto by Theorem Since g (Z Y ) is one-to-one and onto, its inverse is a function, (g (Z Y )) 1 : Y Z, and it is one-to-one and onto, by Theorem

12 Claim 5: (f (X W )) (g (Y Z)) 1 : A B is one-to-one and onto. Proof. By Claims 2 and 4, the functions f (X W ) and (g (Y Z)) 1 are one-to-one and onto. Note that X and Y = A \ X are disjoint, as are Ran(f (X W ) = W and Ran(g 1 (Y Z) = Z. By Exercises 5.1: 9a and 5.2:12, (f (X W )) (g (Y Z)) 1 is a one-to-one function from A = X Y to B = W Z. Since the range of (f (X W )) (g 1 (Y Z)) is W Z = B, by Theorem 5.2.3, it is onto and Claim 5 follows.

13 Claim 6: The function h : A B defined for all a A by h(a) = f (a) and if a X and h(a) = g 1 (a) if a A \ X is one-to-one and onto. Proof. By Claim 5, (f (X W )) (g 1 (Y Z)) has the same domain and codomain as h. Note that h, f, and f (X W ) agree on X. Also h, g 1 and g 1 (Y Z) agree on Y. Thus h and (f (X W )) (g 1 (Y Z)) agree on A = X Y, so h = (f (X W )) (g 1 (Y Z)), by Theorem

14 Closing of the Proof: By Claim 6, h : A B is one-to-one and onto, so it witnesses that A B (p ). This assertion completes Case 2. By exhaustive case analysis, A B. We assumed A B and B A and proved A B, so the implication and the theorem follow (c ).

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