Processing math: 0%

Lecture 15: Uncountable sets

• Reading: MCS 8.1

• $ℝ$, $2^ℕ$ are uncountable.
• Set of (finite) strings is countable.

• Finite cardinality

$ℝ$ 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?

• the existence different sizes of infinity is pretty neat.

• diagonalization is used to prove that there are specifications with no program that implements them. One such problem is determining whether a program crashes or not. It would be nice to have a compiler that guarantees that your program never crashes. However, diagonalization can be used to show that no such program exists. We may discuss this more when we talk about computability; it is also discussed in MCS.

• this proof is also quite closely related to notions of truth and provability, which we will discuss later in the course.

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 $A \cup B = \emptyset$.