Page 339 - HOW TO PROVE IT: A Structured Approach, Second Edition

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The Cantor–Schr¨ oder–Bernstein Theorem 325
g is one-to-one, f (a) = b. Since a ∈ A n−1 , a ∈ X,so h(a) = f (a) = b. Thus,
b ∈ Ran(h).
−1
Case 2. g(b) ∈ Y. Then h(g(b)) = g (g(b)) = b,so b ∈ Ran(h).
In both cases we have b ∈ Ran(h), so h is onto.

The Cantor–Schr¨oder–Bernstein theorem is often useful for showing that
sets are equinumerous. For example, in exercise 3 of Section 7.1 you were
asked to show that (0, 1] ∼ (0, 1), where

(0, 1] ={x ∈ R | 0 < x ≤ 1}
and
(0, 1) ={x ∈ R | 0 < x < 1}.
It is surprisingly difficult to find a one-to-one correspondence between these
two sets, but it is easy to show that they are equinumerous using the Cantor–
Schr¨oder–Bernstein theorem. Of course, (0, 1) ⊆ (0, 1], so clearly (0, 1)
(0, 1]. For the other direction, define f :(0, 1] → (0, 1) by the formula
x
f (x) = .
2
It is easy to check that this function is one-to-one (although it is not onto), so
(0, 1] (0, 1). Thus, by the Cantor–Schr¨oder–Bernstein theorem, (0, 1] ∼
(0, 1). For more on this example see exercise 9.
Our next theorem gives a more surprising consequence of the Cantor–
Schr¨oder–Bernstein theorem.

Theorem 7.3.3. R ∼ P (Z ).
+
It is quite difficult to prove Theorem 7.3.3 directly by giving an example
of a one-to-one, onto function from R to P (Z ). In our proof we’ll use the
+
Cantor–Schr¨oder–Bernstein theorem and the following lemma.

Lemma 7.3.4. Suppose x and y are real numbers and x < y. Then there is a
rational number q such that x < q < y.
Proof. Let k be a positive integer larger than 1 . Then 1 < y − x.We
y−x k
will show that there is a fraction with denominator k that is between x
and y.
j
Let m and n be integers such that m < x < n, and let S ={ j ∈ N | m + >
k
k(n−m)
x}. Note that m + = n > x, and therefore k(n − m) ∈ S. Thus S = ∅,
k
so by the well-ordering principle it has a smallest element. Let j be the smallest
0
element of S. Note also that m + = m < x,so0 /∈ S, and therefore j > 0.
k