Lecture 5: cardinality and countability

reading: MCS 8 intro/8.1
cardinality definitions: $|X| \leq |Y|$ , $|X| \geq |Y|$ , $|X| = |Y|$ , countable
countable examples

Definitions

For now, we will not use the symbol $|X|$ by itself. It does not mean "the number of elements of the set", although our definitions will be consistent with this meaning. Until told otherwise, you may only use the symbol $|X|$ as part of the phrases " $|X| \leq |Y|$ ", " $|X| \geq |Y|$ ", or " $|X| = |Y|$ ". This will prevent you from accidentally using reasoning that only applies to finite sets.

Definitions: Let $X$ and $Y$ be sets.

$|X| \geq |Y|$ means "there exists a surjection $f : X → Y$ .
$|X| \leq |Y|$ means "there exists an injection $f : X → Y$ .
$|X| = |Y|$ means "there exists a bijection $f : X → Y$ .

MCS uses the notation "X surj Y" (respectively "inj" or "bij"), but this can be confusing: surjectivity is a property of functions, not sets. For example, just because you find one non-injective function from $X$ to $Y$ does not mean that $|X| \nleq |Y|$ .

Properties of cardinality

It is not obvious that this use of $\leq$ and $\geq$ is justified. There are many things to prove, most of which are easy. We proved one of them:

Claim: (Cantor-Schroder-Bernstein) $|X| \geq |Y|$ then $|Y| \leq |X|$ .

Proof: Suppose $|X| \geq |Y|$ . Then by definition, there exists a surjection $f : X → Y$ . We showed last time that $f$ must have a right inverse $g : Y → X$ (which means $f \circ g = id$ ). This means $g$ has a left inverse (namely $f$ ), so $g$ must be injective. Therefore there is an injection from $Y$ to $X$ , so $|Y| \leq |X|$ as required.

Other properties you could check for practice:

if $|X| \leq |Y|$ then $|Y| \geq |X|$
if $|X| \leq |Y|$ and $|Y| \leq |Z|$ then $|X| \leq |Z|$
if $|X| = |Y|$ then $|X| \leq |Y|$ and $|X| \geq |Y|$
$|X| = |X|$
if $|X| = |Y|$ then $|Y| = |X|$ .

There is one property that is true, but the proof is very non-obvious:

Claim: if $|X| \leq |Y|$ and $|Y| \leq |X|$ then $|Y| = |X|$ .

The proof is beyond the scope of the course, but it is worth starting it to see why it is hard. If you are curious, here is a proof from a previous semester. You are not responsible for knowing this proof, but it only uses techniques that you should be comfortable with so you should be able to read it.

Note: You may use these properties without proof, unless we ask you to prove them.

Countability

Informally, a set $X$ is countable if you can put it in a (potentially infinite) list. This can be formalized by saying that there is a "first element", a "second element", and so on, and each element is the "nth" element for some $n$ . In other words, there should exist a surjection $f : \mathbb{N} → X$ . Even more concisely:

Definition: $X$ is countable if $|\mathbb{N}| \geq |X|$ .

Equivalently, $X$ is countable if

there exists a surjection $f : \mathbb{N} → X$
there exists an injection $f : X → \mathbb{N}$
$|X| \leq |\mathbb{N}|$ .

If $|X| = |\mathbb{N}|$ then we say $X$ is countably infinite.

Examples

the set $\mathbb{N}$ is countable; the identity function is a surjection
the set $X = \mathbb{N} \cup \{-1\}$ is countable; let $f : \mathbb{N} → X$ be given by $f(n) = n - 1$ . You can check that $f$ is surjective
the set of integers $\mathbb{Z}$ is countable. Let $f : \mathbb{N} → \mathbb{Z}$ be given by $f(n) ::= -n/2$ if $n$ is even and $(n+1)/2$ if $n$ is odd. We must show $f$ is surjective, i.e. for all $i \in \mathbb{Z}$ , there exists $n \in \mathbb{N}$ such that $f(n) = i$ . To do so, choose an arbitrary $i$ . If $i \gt 0$ , let $n = 2i - 1$ . We see that $n$ is odd, so $f(n) = (n+1)/2 = i$ , and thus $i$ is in the image of $f$ . If $i \leq 0$ on the other hand, we can choose $n = -2i$ . Then $n$ is even, so $f(n) = -(-2i)/2 = i$ , and again, we see that $i$ is in the image of $f$ . In either case, $i$ is in the image of $f$ , so $f$ must be surjective.
Next time we will show that $\mathbb{Q}$ is countable, but $\mathbb{R}$ is uncountable.