Cantor-Schröder-ernstein Theorem, Part 1 Jean. Larson and Christopher C. Porter MHF 3202 December 4, 2015
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.
Theorem (Cantor-Schröder-ernstein Theorem) Suppose and are sets. If and, then.
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
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 =?
counter-example Consider = + and = Q. + Q and Q +, but + Q. Note however that + Q. Is this an instance of a more general fact? Yes!
The Cantor-Schröder-ernstein Theorem Theorem Let and be sets. If and, then.
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.
Our pproach Next, let f : be a one-to-one function witnessing and g : be a one-to-one function witnessing. f g
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
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.
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
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
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
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 -1 1-1 and onto
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 -1 1-1 and onto
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)
Choosing the sets X, Y, W, and We want Y Ran(g). g X W = f(x) g() = Ran(g) Y g() = g()
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()
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()
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()
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()
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()
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()
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()
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()
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()
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()
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()
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()
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()
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()
Taking the union of the family X = { n n + } X Y = g()