(Standard post on cardinals, as a prerequisite for forthcoming theory model post.)

An ordinal measures a total ordering. However, it does not do a fantastic job at measuring size. For example, there is a bijection between the elements of {\omega} and {\omega+1}:

\displaystyle \begin{array}{rccccccc} \omega+1 = & \{ & \omega & 0 & 1 & 2 & \dots & \} \\ \omega = & \{ & 0 & 1 & 2 & 3 & \dots & \}. \end{array}

In fact, as you likely already know, there is even a bijection between {\omega} and {\omega^2}:

\displaystyle \begin{array}{l|cccccc} + & 0 & 1 & 2 & 3 & 4 & \dots \\ \hline 0 & 0 & 1 & 3 & 6 & 10 & \dots \\ \omega & 2 & 4 & 7 & 11 & \dots & \\ \omega \cdot 2 & 5 & 8 & 12 & \dots & & \\ \omega \cdot 3 & 9 & 13 & \dots & & & \\ \omega \cdot 4 & 14 & \dots & & & & \end{array}

So ordinals do not do a good job of keeping track of size. For this, we turn to the notion of a cardinal number.

1. Equinumerous Sets and Cardinals

Definition 1 Two sets {A} and {B} are equinumerous, written {A \approx B}, if there is a bijection between them.

Definition 2 A cardinal is an ordinal {\kappa} such that for no {\alpha < \kappa} do we have {\alpha \approx \kappa}.

Example 3 (Examples of Cardinals) Every finite number is a cardinal. Moreover, {\omega} is a cardinal. However, {\omega+1}, {\omega^2}, {\omega^{2015}} are not, because they are countable.

Example 4 ({\omega^\omega} is Countable) Even {\omega^\omega} is not a cardinal, since it is a countable union

\displaystyle \omega^\omega = \bigcup_n \omega^n

and each {\omega^n} is countable.

Ques 5 Why must an infinite cardinal be a limit ordinal?

Remark 6 There is something fishy about the definition of a cardinal: it relies on an external function {f}. That is, to verify {\kappa} is a cardinal I can’t just look at {\kappa} itself; I need to examine the entire universe {V} to make sure there does not exist a bijection {f : \kappa \rightarrow \alpha} for {\alpha < \kappa}. For now this is no issue, but later in model theory this will lead to some highly counterintuitive behavior.

2. Cardinalities

Now that we have defined a cardinal, we can discuss the size of a set by linking it to a cardinal.

Definition 7 The cardinality of a set {X} is the least ordinal {\kappa} such that {X \approx \kappa}. We denote it by {\left\lvert X \right\rvert}.

Ques 8 Why must {\left\lvert X \right\rvert} be a cardinal?

Remark 9 One needs the Well-Ordering Theorem (equivalently, Choice) in order to establish that such an ordinal {\kappa} actually exists.

Since cardinals are ordinals, it makes sense to ask whether {\kappa_1 \le \kappa_2}, and so on. Our usual intuition works well here.

Proposition 10 (Restatement of Cardinality Properties) Let {X} and {Y} be sets.

  1. {X \approx Y} if and only {\left\lvert X \right\rvert = \left\lvert Y \right\rvert}, if and only if there is a bijection between {X} and {Y}.
  2. {\left\lvert X \right\rvert \le \left\lvert Y \right\rvert} if and only if there is an injective map {X \hookrightarrow Y}.

Ques 11 Prove this.

3. Aleph Numbers

Prototypical example for this section: {\aleph_0} is {\omega}, and {\aleph_1} is the first uncountable

First, let us check that cardinals can get arbitrarily large:

Proposition 12 We have {\left\lvert X \right\rvert < \left\lvert \mathcal P(X) \right\rvert} for every set {X}.

Proof: There is an injective map {X \hookrightarrow \mathcal P(X)} but there is no injective map {\mathcal P(X) \hookrightarrow X} by Cantor’s diagonal argument. \Box

Thus we can define:

Definition 13 For a cardinal {\kappa}, we define {\kappa^+} to be the least cardinal above {\kappa}, called the successor cardinal.

This {\kappa^+} exists and has {\kappa^+ \le \left\lvert \mathcal P(\kappa) \right\rvert}.

Next, we claim that:

Exercise 14 Show that if {A} is a set of cardinals, then {\cup A} is a cardinal.

Thus by transfinite induction we obtain that:

Definition 15 For any {\alpha \in \mathrm{On}}, we define the aleph numbers as

\displaystyle \begin{aligned} \aleph_0 &= \omega \\ \aleph_{\alpha+1} &= \left( \aleph_\alpha \right)^+ \\ \aleph_{\lambda} &= \bigcup_{\alpha < \lambda} \aleph_\alpha. \end{aligned}

Thus we have the following sequence of cardinals:

\displaystyle 0 < 1 < 2 < \dots < \aleph_0 < \aleph_1 < \dots < \aleph_\omega < \aleph_{\omega+1} < \dots

By definition, {\aleph_0} is the cardinality of the natural numbers, {\aleph_1} is the first uncountable ordinal, \dots.

We claim the aleph numbers constitute all the cardinals:

Lemma 16 (Aleph Numbers Constitute All Infinite Cardinals) If {\kappa} is a cardinal then either {\kappa} is finite (i.e. {\kappa \in \omega}) or {\kappa = \aleph_\alpha} for some {\alpha \in \mathrm{On}}.

Proof: Assume {\kappa} is infinite, and take {\alpha} minimal with {\aleph_\alpha \ge \kappa}. Suppose for contradiction that we have {\aleph_\alpha > \kappa}. We may assume {\alpha > 0}, since the case {\alpha = 0} is trivial.

If {\alpha = \overline{\alpha} + 1} is a successor, then

\displaystyle \aleph_{\overline{\alpha}} < \kappa < \aleph_{\alpha} = (\aleph_{\overline{\alpha}})^+

which contradicts the fact the definition of the successor cardinal. If {\alpha = \lambda} is a limit ordinal, then {\aleph_\lambda} is the supremum {\bigcup_{\gamma < \lambda} \aleph_\gamma}. So there must be some {\gamma < \lambda} has {\aleph_\gamma > \kappa}, which contradicts the minimality of {\alpha}. \Box

Definition 17 An infinite cardinal which is not a successor cardinal is called a limit cardinal. It is exactly those cardinals of the form {\aleph_\lambda}, for {\lambda} a limit ordinal, plus {\aleph_0}.

4. Cardinal Arithmetic

Prototypical example for this section: {\aleph_0 \cdot \aleph_0 = \aleph_0 + \aleph_0 = \aleph_0}

Recall the way we set up ordinal arithmetic. Note that in particular, {\omega + \omega > \omega} and {\omega^2 > \omega}. Since cardinals count size, this property is undesirable, and we want to have

\displaystyle \begin{aligned} \aleph_0 + \aleph_0 &= \aleph_0 \\ \aleph_0 \cdot \aleph_0 &= \aleph_0 \end{aligned}

because {\omega + \omega} and {\omega \cdot \omega} are countable. In the case of cardinals, we simply “ignore order”.

The definition of cardinal arithmetic is as expected:

Definition 18 (Cardinal Arithmetic) Given cardinals {\kappa} and {\mu}, define

\displaystyle \kappa + \mu \overset{\mathrm{def}}{=} \left\lvert \left( \left\{ 0 \right\} \times \kappa \right) \cup \left( \left\{ 1 \right\} \times \mu \right) \right\rvert


\displaystyle k \cdot \mu \overset{\mathrm{def}}{=} \left\lvert \mu \times \kappa \right\rvert .

Ques 19 Check this agrees with what you learned in pre-school for finite cardinals.

This is a slight abuse of notation since we are using the same symbols as for ordinal arithmetic, even though the results are different ({\omega \cdot \omega = \omega^2} but {\aleph_0 \cdot \aleph_0 = \aleph_0}). In general, I’ll make it abundantly clear whether I am talking about cardinal arithmetic or ordinal arithmetic. To help combat this confusion, we use separate symbols for ordinals and cardinals. Specifically, {\omega} will always refer to {\{0,1,\dots\}} viewed as an ordinal; {\aleph_0} will always refer to the same set viewed as a cardinal. More generally,

Definition 20 Let {\omega_\alpha = \aleph_\alpha} viewed as an ordinal.

However, as we’ve seen already we have that {\aleph_0 \cdot \aleph_0 = \aleph_0}. In fact, this holds even more generally:

Theorem 21 (Infinite Cardinals Squared) Let {\kappa} be an infinite cardinal. Then {\kappa \cdot \kappa = \kappa}.

Proof: Obviously {\kappa \cdot \kappa \ge \kappa}, so we want to show {\kappa \cdot \kappa \le \kappa}.

The idea is to repeat the same proof that we had for {\aleph_0 \cdot \aleph_0 = \aleph_0}, so we re-iterate it here. We took the “square” of elements of {\aleph_0}, and then re-ordered it according to the diagonal:

\displaystyle \begin{array}{l|cccccc} & 0 & 1 & 2 & 3 & 4 & \dots \\ \hline 0 & 0 & 1 & 3 & 6 & 10 & \dots \\ 1 & 2 & 4 & 7 & 11 & \dots & \\ 2 & 5 & 8 & 12 & \dots & & \\ 3 & 9 & 13 & \dots & & & \\ 4 & 14 & \dots & & & & \end{array}

Let’s copy this idea for a general {\kappa}.

We proceed by transfinite induction on {\kappa}. The base case is {\kappa = \aleph_0}, done above. For the inductive step, first we put the “diagonal” ordering {<_{\text{diag}}} on {\kappa \times \kappa} as follows: for {(\alpha_1, \beta_1)} and {(\alpha_1, \beta_2)} in {\kappa \times \kappa} we declare {(\alpha_1, \beta_1) <_{\text{diag}} (\alpha_2, \beta_2)} if

  • {\max \left\{ \alpha_1, \beta_1 \right\} < \max \left\{ \alpha_2, \beta_2 \right\}} (they are on different diagonals), or
  • {\max \left\{ \alpha_1, \beta_1 \right\} = \max \left\{ \alpha_2, \beta_2 \right\}} and {\alpha_1 < \alpha_2} (same diagonal).

Then {<_{\text{diag}}} is a well-ordering of {\kappa \times \kappa}, so we know it is in order-preserving bijection with some ordinal {\gamma}. Our goal is to show that {\left\lvert \gamma \right\rvert \le \kappa}. To do so, it suffices to prove that for any {\overline{\gamma} \in \gamma}, we have {\left\lvert \overline{\gamma} \right\rvert < \kappa}.

Suppose {\overline{\gamma}} corresponds to the point {(\alpha, \beta) \in \kappa} under this bijection. If {\alpha} and {\beta} are both finite, then certainly {\overline{\gamma}} is finite too. Otherwise, let {\overline{\kappa} = \max \{\alpha, \beta\} < \kappa}; then the number of points below {\overline{\gamma}} is at most

\displaystyle \left\lvert \alpha \right\rvert \cdot \left\lvert \beta \right\rvert \le \overline{\kappa} \cdot \overline{\kappa} = \overline{\kappa}

by the inductive hypothesis. So {\left\lvert \overline{\gamma} \right\rvert \le \overline{\kappa} < \kappa} as desired. \Box

From this it follows that cardinal addition and multiplication is really boring:

Theorem 22 (Infinite Cardinal Arithmetic is Trivial) Given cardinals {\kappa} and {\mu}, one of which is infinite, we have

\displaystyle \kappa \cdot \mu = \kappa + \mu = \max\left( \kappa, \mu \right).

Proof: The point is that both of these are less than the square of the maximum. Writing out the details:

\displaystyle \begin{aligned} \max \left( \kappa, \mu \right) &\le \kappa + \mu \\ &\le \kappa \cdot \mu \\ &\le \max \left( \kappa, \mu \right) \cdot \max \left( \kappa, \mu \right) \\ &= \max\left( \kappa, \mu \right). \end{aligned}


5. Cardinal Exponentiation

Prototypical example for this section: {2^\kappa = \left\lvert \mathcal P(\kappa) \right\rvert}.

Definition 23 Suppose {\kappa} and {\lambda} are cardinals. Then

\displaystyle \kappa^\lambda \overset{\mathrm{def}}{=} \left\lvert \mathscr F(\lambda, \kappa) \right\rvert.

Here {\mathscr F(A,B)} is the set of functions from {A} to {B}.

As before, we are using the same notation for both cardinal and ordinal arithmetic. Sorry!

In particular, {2^\kappa = \left\lvert \mathcal P(\kappa) \right\rvert > \kappa}, and so from now on we can use the notation {2^\kappa} freely. (Note that this is totally different from ordinal arithmetic; there we had {2^\omega = \bigcup_{n\in\omega} 2^n = \omega}. In cardinal arithmetic {2^{\aleph_0} > \aleph_0}.)

I have unfortunately not told you what {2^{\aleph_0}} equals. A natural conjecture is that {2^{\aleph_0} = \aleph_1}; this is called the Continuum Hypothesis. It turns out to that this is undecidable — it is not possible to prove or disprove this from the {\mathsf{ZFC}} axioms.

6. Cofinality

Prototypical example for this section: {\aleph_0}, {\aleph_1}, \dots are all regular, but {\aleph_\omega} has cofinality {\omega}.

Definition 24 Let {\lambda} be a limit ordinal, and {\alpha} another ordinal. A map {f : \alpha \rightarrow \lambda} of ordinals is called cofinal if for every {\overline{\lambda} < \lambda}, there is some {\overline{\alpha} \in \alpha} such that {f(\overline{\alpha}) \ge \overline{\lambda}}. In other words, the map reaches arbitrarily high into {\lambda}.

Example 25 (Example of a Cofinal Map) {\empty}

  1. The map {\omega \rightarrow \omega^\omega} by {n \mapsto \omega^n} is cofinal.
  2. For any ordinal {\alpha}, the identity map {\alpha \rightarrow \alpha} is cofinal.

Definition 26 Let {\lambda} be a limit ordinal. The cofinality of {\lambda}, denoted {\text{cof }(\lambda)}, is the smallest ordinal {\alpha} such that there is a cofinal map {\alpha \rightarrow \lambda}.

Ques 27 Why must {\alpha} be an infinite cardinal?

Usually, we are interested in taking the cofinality of a cardinal {\kappa}.

Pictorially, you can imagine standing at the bottom of the universe and looking up the chain of ordinals to {\kappa}. You have a machine gun and are firing bullets upwards, and you want to get arbitrarily high but less than {\kappa}. The cofinality is then the number of bullets you need to do this.

We now observe that “most” of the time, the cofinality of a cardinal is itself. Such a cardinal is called regular.

Example 28 ({\aleph_0} is Regular) {\text{cof }(\aleph_0) = \aleph_0}, because no finite subset of {\omega} can reach arbitrarily high.

Example 29 ({\aleph_1} is Regular) {\text{cof }(\aleph_1) = \aleph_1}. Indeed, assume for contradiction that some countable set of ordinals {A = \{ \alpha_0, \alpha_1, \dots \} \subseteq \omega_1} reaches arbitrarily high inside {\omega_1}. Then {\Lambda = \cup A} is a countable ordinal, because it is a countable union of countable ordinals. In other words {\Lambda \in \omega_1}. But {\Lambda} is an upper bound for {A}, contradiction.

On the other hand, there are cardinals which are not regular; since these are the “rare” cases we call them singular.

Example 30 ({\aleph_\omega} is Not Regular) Notice that {\aleph_0 < \aleph_1 < \aleph_2 < \dots} reaches arbitrarily high in {\aleph_\omega}, despite only having {\aleph_0} terms. It follows that {\text{cof }(\aleph_\omega) = \aleph_0}.

We now confirm a suspicion you may have:

Theorem 31 (Successor Cardinals Are Regular) If {\kappa = \overline{\kappa}^+} is a successor cardinal, then it is regular.

Proof: We copy the proof that {\aleph_1} was regular.

Assume for contradiction that for some {\mu \le \overline{\kappa}}, there are {\mu} sets reaching arbitrarily high in {\kappa} as a cardinal. Observe that each of these sets must have cardinality at most {\overline{\kappa}}. We take the union of all {\mu} sets, which gives an ordinal {\Lambda} serving as an upper bound.

The number of elements in the union is at most

\displaystyle \#\text{sets} \cdot \#\text{elms} \le \mu \cdot \overline{\kappa} = \overline{\kappa}

and hence {\left\lvert \Lambda \right\rvert \le \overline{\kappa} < \kappa}. \Box

7. Inaccessible Cardinals

So, what about limit cardinals? It seems to be that most of them are singular: if {\aleph_\lambda \ne \aleph_0} is a limit ordinal, then the sequence {\{\aleph_\alpha\}_{\alpha \in \lambda}} (of length {\lambda}) is certainly cofinal.

Example 32 (Beth Fixed Point) Consider the monstrous cardinal

\displaystyle \kappa = \aleph_{\aleph_{\aleph_{\ddots}}}.

This might look frighteningly huge, as {\kappa = \aleph_\kappa}, but its cofinality is {\omega} as it is the limit of the sequence

\displaystyle \aleph_0, \aleph_{\aleph_0}, \aleph_{\aleph_{\aleph_0}}, \dots

More generally, one can in fact prove that

\displaystyle \text{cof }(\aleph_\lambda) = \text{cof }(\lambda).

But it is actually conceivable that {\lambda} is so large that {\left\lvert \lambda \right\rvert = \left\lvert \aleph_\lambda \right\rvert}.

A regular limit cardinal other than {\aleph_0} has a special name: it is weakly inaccessible. Such cardinals are so large that it is impossible to prove or disprove their existence in {\mathsf{ZFC}}. It is the first of many so-called “large cardinals”.

An infinite cardinal {\kappa} is a strong limit cardinal if

\displaystyle \forall \overline{\kappa} < \kappa \quad 2^{\overline{\kappa}} < \kappa

for any cardinal {\overline{\kappa}}. For example, {\aleph_0} is a strong limit cardinal.

Ques 33 Why must strong limit cardinals actually be limit cardinals? (This is offensively easy.)

A regular strong limit cardinal other than {\aleph_0} is called strongly inaccessible.

8. Exercises

Problem 1 Compute {\left\lvert V_\omega \right\rvert}.

Problem 2 Prove that for any limit ordinal {\alpha}, {\text{cof }(\alpha)} is a regular cardinal.

Sproblem 3 (Strongly Inaccessible Cardinals) Show that for any strongly inaccessible {\kappa}, we have {\left\lvert V_\kappa \right\rvert = \kappa}.

Problem 4 (Konig’s Theorem) Show that

\displaystyle \kappa^{\text{cof }(\kappa)} > \kappa

for every infinite cardinal {\kappa}.

(This post is a draft of a chapter from my Napkin project.)


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s