CBS Theorem
J. Larson UF
Cantor-Schröder-Bernstein Theorem
Jean A. Larson
MHF 3202
April 20, 2015
CBS Theorem
J. Larson UF
Theorem (Cantor-Schröder-Bernstein Theorem)
Suppose A and B are sets. If A B and B A, then A ∼ B.
CBS Theorem
J. Larson UF
Opening of the Proof:
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∃).
Definition: 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}.
CBS Theorem
J. Larson UF
Plan of the proof:
I
Define a set X ⊆ A by recursion
I
Set W := f (X ) = {f (x) | x ∈ X } and show
f ∩ (X × W ) : X → W is one-to-one and onto.
I
Set Z = B \ W , Y = A \ X , and prove that
g (Z ) := {g (z) | z ∈ Z } = Y .
I
Show g ∩ (Z × Y ) is one-to-one and onto, so
(g ∩ (Z × Y ))−1 = g −1 ∩ (Y × Z ) is also one-to-one and onto.
I
Show that h = (f ∩ (X × W )) ∪ (g −1 ∩ (Y × Z )) : A → B is
the witness that A ∼ B.
CBS Theorem
J. Larson UF
Definitions of R, A and W :
I
Let R = ran(g ) ⊆ A.
I
Define X by recursion:
I
I
A1 := A \ R;
for every n ∈ Z+ , An+1 := g (f (An )) = {g (f (a)) | a ∈ An }.
Then X =
I
CBS Theorem
S
{An | n ∈ Z+ }.
Set W := f (X ) = {f (x) | x ∈ X } = ran(f X ).
J. Larson UF
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.
CBS Theorem
J. Larson UF
Claim 2: g (Z ) = {g (z) | z ∈ Z } ⊆ Y where Z := B \ W and
Y := A \ X .
Proof.
Assume toward a contradiction that g (Z ) 6⊆ Y (o∗).
Let g (z0 ) ∈ g (Z ) ⊆ A be a witness, i.e. assume g (z0 ) ∈
/ Y (a∃).
Since g (z0 ) ∈
/ Y = A \ X , it follows that g (z0 ) ∈ X (def set
difference).
S
Since X = {An | n ∈ Z+ }, we can find a witness n0 ∈ Z+ with
g (z0 ) ∈ An0 (a∃).
CBS Theorem
J. Larson UF
Claim 2 (proof continued)
Since g (z0 ) ∈ ran(g ), and A1 = A \ ran(g ), it follows that
g (z0 ) ∈
/ A1 (def set difference).
Thus n0 > 1 and g (z0 ) ∈ An0 = g (f (An0 −1 )). Let x0 ∈ An0 −1 be a
witness, i.e. assume g (z0 ) = g (f (x0 )).
Since g is one-to-one, it follows that z0 = f (x0 ) ∈ W , by the
definition of W . Thus z0 ∈ Z = B \ W and z0 ∈ W , which is a
contradiction, since these two sets are disjoint (c∗)
Thus our assumption was false and Claim 2 follows.
CBS Theorem
J. Larson UF
Claim 3: Y ⊆ g (Z ) = {g (z) | z ∈ Z }.
Assume toward a contradiction that Y 6⊆ g (Z ) (o∗).
Let y0 ∈ Y be a witness, i.e. assume y0 ∈
/ g (Z ) (a∃).
Since y0 ∈ Y = A \ X , it follows that y0 ∈
/ X and in particular,
y0 ∈
/ A1 = A \ ran(g ), so y0 ∈ ran(g ). Let b0 ∈ B be a witness, i.e.
g (b0 ) = y0 (a∃).
Since y0 ∈
/ g (Z ), it follows that b0 ∈
/ Z = B \ W , so
b0 ∈ W = f (X ). Let x0 ∈ X be a witness, i.e. assume f (x0 ) = b0
and let m0 be such that x0 ∈ Am0 (a∃).
Thus y0 = g (b0 ) = g (f (x0 )) ∈ g (f (Am0 )) = Am0 +1 ⊆ X , so y0 is
in both X and Y which is a contradiction since these sets are
disjoint (c∗). So our assumption was false and Claim 3 follows.
CBS Theorem
J. Larson UF
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, g (Z ) = Y .
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 5.2.3.
Since g ∩ (Z × Y ) is one-to-one and onto, its inverse is a function,
(g ∩ (Z × Y ))−1 = g −1 ∩ (Y × Z ) : Y → Z , and it is one-to-one
and onto, by Theorem 5.3.4.
CBS Theorem
J. Larson UF
Claim 5: (f ∩ (X × W )) ∪ (g −1 ∩ (Y × Z )) : A → B is one-to-one and
onto.
Proof.
By Claims 2 and 4, the functions f ∩ (X × W ) and g −1 ∩ (Y × Z )
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 −1 ∩ (Y × Z ))
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.
CBS Theorem
J. Larson UF
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 5.1.4.
CBS Theorem
J. Larson UF
Closing of the Proof:
By Claim 6, h : A → B is one-to-one and onto, so it witnesses that
A ∼ B (p∃).
We assumed A B and B A and proved A ∼ B, so the
implication and the theorem follow (c→).
CBS Theorem
J. Larson UF
CBS Theorem
J. Larson UF