Cantor-Schröder-Bernstein Theorem, Part 1

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2 Cantor-Schröder-ernstein Theorem, Part 1 Jean. Larson and Christopher C. Porter MHF 3202 December 4, 2015

3 Proving Equinumerousity Up to this point, the main method we have had for proving that two sets and are equinumerous is to show that there is a function f : that is one-to-one and onto. In some cases, finding such a bijection can be rather difficult. Today we will prove a theorem that will provide a new and simpler method for showing that two sets are equinumerous.

4 Theorem (Cantor-Schröder-ernstein Theorem) Suppose and are sets. If and, then.

5 preliminary definition Let and be sets. We say is dominated by, in symbols, if there is a one-to-one function f :. few examples: If, then If, then P( + ) R

6 Question: Is a partial order? It is not too hard to show that is reflexive and transitive. Is antisymmetric? That is, if and, then does it follow that =?

7 counter-example Consider = + and = Q. + Q and Q +, but + Q. Note however that + Q. Is this an instance of a more general fact? Yes!

8 The Cantor-Schröder-ernstein Theorem Theorem Let and be sets. If and, then.

9 Our pproach To help us understand the general strategy of the proof, we will make use of a series of diagrams. First, we will represent the sets and as follows.

10 Our pproach Next, let f : be a one-to-one function witnessing and g : be a one-to-one function witnessing. f g

11 Our pproach Note that if either f or g is onto, it immediately follows that. So need to consider the possibility that neither f nor g are onto. f g

12 The plan Our goal is to use f and g 1 to define a one-to-one and onto function h : : To do so, we will 1. split into two pieces X and Y ; 2. split into two pieces W and ; 3. X will be matched up with W by f ; and 4. Y will be matched up with by g.

13 The plan Here is a schematic diagram in which the splits have been made the functions map in their usual directions. f X W g Y

14 The plan If we know what X is, we let W = f (X ) = {f (x) x X }. Then we let = \ W. We know what is, so we let Y = g() = {g(z) z }. f X W g Y

15 The plan It follows that f X : X W is one-to-one and onto and g : Y is one-to-one and onto. X f 1-1 and onto W = f(x) Y = g() g 1-1 and onto

16 The plan Consequently, f X : X W is one-to-one and onto and (g ) 1 : Y is one-to-one and onto. X f 1-1 and onto W = f(x) Y = g() g and onto

17 The desired function h Therefore h = f X (g ) 1 : X Y W is one-to-one and onto. We know W =, so if X Y =, then h is our witnessing function. X f 1-1 and onto h W = f(x) Y = g() g and onto

18 Choosing the sets X, Y, W, and First we recall that we assumed g is not onto, since otherwise g ia a witness that. g g() = Ran(g)

19 Choosing the sets X, Y, W, and We want Y Ran(g). g X W = f(x) g() = Ran(g) Y g() = g()

20 Choosing the sets X, Y, W, and If we let 1 = \ Ran(g), then we must have 1 X. 1 = \ Ran(g) g X W = f(x) g() = Ran(g) Y g() = g()

21 Choosing the sets X, Y, W, and Given an arbitrary a 1, since a X, it follows that f (a) W. 1 = \ Ran(g) f(a) a X W = f(x) g() = Ran(g) Y g() = g()

22 Choosing the sets X, Y, W, and For every z = \ W, z f (a) W. So, since g is one-to-one, for all z, g(f (a)) g(z). Thus g(f (a)) X. 1 = \ Ran(g) f(a) a X g(f(a)) g W = f(x) g() = Ran(g) Y g() = g()

23 Choosing the sets X, Y, W, and Since a was arbitrary, we have f (a) W and g(f (a)) X for all a 1. That is, f ( 1 ) W and g(f ( 1 )) X. 1 = \ Ran(g) X g f(1) W = f(x) g() = Ran(g) Y g() = g()

24 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )). 1 = \ Ran(g) 2 = g(f(1)) X g() = Ran(g) f f(1) f(2) W = f(x) Y g() = g()

25 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )). 1 = \ Ran(g) 2 = g(f(1)) g(f(2)) X g() = Ran(g) g f(1) f(2) W = f(x) Y g() = g()

26 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )). 1 = \ Ran(g) 2 = g(f(1)) 3 = g(f(2)) f f(1) f(2) f(3) g() = Ran(g) Y g() = g()

27 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )). 1 = \ Ran(g) 2 = g(f(1)) 3 = g(f(2)) 4 = g(f(3)) g f(1) f(2) f(3) g() = Ran(g) Y g() = g()

28 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )). 1 = \ Ran(g) 2 = g(f(1)) 3 = g(f(2)) 4 = g(f(3)) f f(1) f(2) f(3) g() = Ran(g) Y g() = g()

29 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )). 1 = \ Ran(g) 2 = g(f(1)) 3 = g(f(2)) 4 = g(f(3)) g f(1) f(2) f(3) g() = Ran(g) Y g() = g()

30 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )). 1 = \ Ran(g) 2 = g(f(1)) 3 = g(f(2)) 4 = g(f(3)) f f(1) f(2) f(3) g() = Ran(g) Y g() = g()

31 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )). 1 = \ Ran(g) 2 = g(f(1)) 3 = g(f(2)) 4 = g(f(3)) g() = Ran(g) g f(1) f(2) f(3) Y g() = g()

32 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )). 1 = \ Ran(g) 2 = g(f(1)) 3 = g(f(2)) 4 = g(f(3)) f f(1) f(2) f(3) g() = Ran(g) Y g() = g()

33 Using Recursion 1 = \ Ran(g); n+1 = g(f ( n )); and X = { n n + } 1 = \ Ran(g) 2 = g(f(1)) 3 = g(f(2)) 4 = g(f(3)) g() = Ran(g) g f(1) f(2) f(3) Y g() = g()

34 Taking the union of the family X = { n n + } X Y = g()