Lecture 15: Uncountable sets

\(ℝ\) is uncountable

Claim: The set of real numbers \(ℝ\) is uncountable.

Proof: in fact, we will show that the set of real numbers between 0 and 1 is uncountable; since this is a subset of \(ℝ\), the uncountability of \(ℝ\) follows immediately.

We proceed by contradiction. Suppose that \(ℝ\) were countable. Then there would exist a surjection \(f : ℕ → ℝ\). We could expand the digits of \(f\) in a table; for example, if \(f(0) = 0\), \(f(1) = 1/2\), \(f(2) = π - 3\), \(f(3) = φ - 1\), then the table would look as follows:

\(n\) \(f(n)\) digits of \(f(n)\)
0 0 \(0.00000\cdots\)
1 1/2 \(0.50000\cdots\)
2 \(π - 3\) \(0.14159\cdots\)
3 \(φ - 1\) \(0.61803\cdots\)

Given such a table, we can form a real number \(x_D\) that is not in the table by changing the \(i\)th digit of the \(i\)th number; perhaps by adding 5 (wrapping around, so that 7 + 5 = 2, for example).

In the example, \(x_D = 0.5565\cdots\).

Now \(x_D\) cannot be in the table, because \(x_D\) differs from \(f(i)\) in the \(i\)th digit. But this contradicts the fact that \(f\) is surjective, thus completing the proof.

Note: this technique is called diagonalization.

What good is countability?

Finite cardinality

Definition: If \(X\) is a set and \(n \in ℕ\), the expression "|X| = n" means \(|X| = \{1,2,\dots,n\}\).

This matches the usual notion of the number of things in the set.

Claim: If \(|A| = a\) and \(|B| = b\), and if \(A\) and \(B\) are disjoint, then \(|A \cup B| = a + b\).

Note: This can be shortened to "\(|A \cup B| = |A| + |B|\), as long as you keep in mind that this equation only makes sense if \(|A|\) and \(|B|\) are numbers (i.e. if \(A\) and \(B\) are finite)

Proof: Since \(|A| = a\), there exists a bijection \(f : A → \{1,2,\dots,a\}\). Similarly, there exists \(g : B → \{1,2,\dots,b\}\). We can build a function \(h : A \cup B → \{1,2,\dots,a+b\}\) by defining \[h(x) := \left\{\begin{array}{ll} f(x), & \text{if $x \in A$} \\ g(x), & \text{if $x \in B$} \end{array}\right.\]

To complete the proof, we need to show that \(h\) is a bijection. We left this as an exercise. While you're doing the exercise, make sure you use the fact that \(A \cup B = \emptyset\).