Book of Proof - Amazon S3

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The set A must be uncountable, as follows. For each b ∈ B, let a b be an
element of A for which f (a b ) = b. (Such an element must exist because f is
surjective.) Now form the set U = {a b : b ∈ B}. Then the function f : U → B is
bijective, by construction. Then since B is uncountable, so is U. Therefore U is
an uncountable subset of A, so A is uncountable by Theorem 13.9.
3. Prove or disprove: If A is uncountable, then |A| = |R|.
This is false. Let A = P(R). Then A is uncountable, and by Theorem 13.7,
|R| < |P(R)| = |A|.
5. Prove or disprove: The set {0,1} × R is uncountable.
This is true. To see why, first note that the function f : R → {0} × R defined as
f (x) = (0, x) is a bijection. Thus |R| = |{0} × R|, and since R is uncountable, so is
{0} × R. Then {0} × R is an uncountable subset of the set {0,1} × R, so {0,1} × R is
uncountable by Theorem 13.9.
7. Prove or disprove: If A ⊆ B and A is countably infinite and B is uncountable,
then B − A is uncountable.
This is true. To see why, suppose to the contrary that B− A is countably infinite.
Then B = A ∪ (B − A) is a union of countably infinite sets, and thus countable,
by Theorem 13.6. This contradicts the fact that B is uncountable.
Exercises for Section 13.4
1. Show that if A ⊆ B and there is an injection g : B → A, then |A| = |B|.
Just note that the map f : A → B defined as f (x) = x is an injection. Now apply
the Cantor-Bernstein-Schröeder theorem.
3. Let F be the set of all functions N → { 0,1}. Show that |R| = |F |.
Because |R| = |P(N)|, it suffices to show that |F | = |P(N)|. To do this, we will
exhibit a bijection f : F → P(N). Define f as follows. Given a function ϕ ∈ F ,
let f (ϕ) = {n ∈ N : ϕ(n) = 1}. To see that f is injective, suppose f (ϕ) = f (θ). Then
{n ∈ N : ϕ(n) = 1} = {n ∈ N : θ(n) = 1}. Put X = {n ∈ N : ϕ(n) = 1}. Now we see that if
n ∈ X, then ϕ(n) = 1 = θ(n). And if n ∈ N − X, then ϕ(n) = 0 = θ(n). Consequently
ϕ(n) = θ(n) for any n ∈ N, so ϕ = θ. Thus f is injective. To see that f is surjective,
take any X ∈ P(N). Consider the function ϕ ∈ F for which ϕ(n) = 1 if n ∈ X and
ϕ(n) = 0 if n ∉ X. Then f (ϕ) = X, so f is surjective.
5. Consider the subset B = { (x, y) : x 2 + y 2 ≤ 1 } ⊆ R 2 . Show that |B| = |R 2 |.
This will follow from the Cantor-Bernstein-Schröeder theorem provided that
we can find injections f : B → R 2 and g : R 2 → B. The function f : B → R 2 defined
as f (x, y) = (x, y) is clearly injective. For g : R 2 → B, consider the function
(
x 2 + y 2
g(x, y) =
x 2 + y 2 + 1 x, x 2 + y 2 )
x 2 + y 2 + 1 y .
Verify that this is an injective function g : R 2 → B.