\chapter{Cardinals} 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 $\omega+1$: \[ \begin{array}{rccccccc} \omega+1 = & \{ & \omega & 0 & 1 & 2 & \dots & \} \\ \omega = & \{ & 0 & 1 & 2 & 3 & \dots & \}. \end{array} \] In fact, as you likely already know, there is even a bijection between $\omega$ and $\omega^2$: \[ \begin{array}{l|cccccc} + & 0 & 1 & 2 & 3 & 4 & \dots \\ \hline 0 & 0 & 1 & 3 & 6 & 10 & \dots \\ \omega & 2 & 4 & 7 & 11 & \dots & \\ \omega \cdot 2 & 5 & 8 & 12 & \dots & & \\ \omega \cdot 3 & 9 & 13 & \dots & & & \\ \omega \cdot 4 & 14 & \dots & & & & \end{array} \] So ordinals do not do a good job of keeping track of size. For this, we turn to the notion of a cardinal number. \section{Equinumerous sets and cardinals} \begin{definition} Two sets $A$ and $B$ are \vocab{equinumerous}, written $A \approx B$, if there is a bijection between them. \end{definition} \begin{definition} A \vocab{cardinal} is an ordinal $\kappa$ such that for no $\alpha < \kappa$ do we have $\alpha \approx \kappa$. \end{definition} \begin{example}[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. \end{example} \begin{example}[$\omega^\omega$ is countable] Even $\omega^\omega$ is not a cardinal, since it is a countable union \[ \omega^\omega = \bigcup_n \omega^n \] and each $\omega^n$ is countable. \end{example} \begin{ques} Why must an infinite cardinal be a limit ordinal? \end{ques} \begin{remark} There is something fishy about the definition of a cardinal: it relies on an \emph{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 $f : \kappa \to \alpha$ for $\alpha < \kappa$. For now this is no issue, but later in model theory this will lead to some highly counterintuitive behavior. \end{remark} \section{Cardinalities} Now that we have defined a cardinal, we can discuss the size of a set by linking it to a cardinal. \begin{definition} The \vocab{cardinality} of a set $X$ is the \emph{least} ordinal $\kappa$ such that $X \approx \kappa$. We denote it by $\left\lvert X \right\rvert$. \end{definition} \begin{ques} Why must $\left\lvert X \right\rvert$ be a cardinal? \end{ques} \begin{remark} One needs the well-ordering theorem (equivalently, choice) in order to establish that such an ordinal $\kappa$ actually exists. \end{remark} Since cardinals are ordinals, it makes sense to ask whether $\kappa_1 \le \kappa_2$, and so on. Our usual intuition works well here. \begin{proposition}[Restatement of cardinality properties] Let $X$ and $Y$ be sets. \begin{enumerate}[(i)] \ii $X \approx Y$ if and only $\left\lvert X \right\rvert = \left\lvert Y \right\rvert$, if and only if there's a bijection from $X$ to $Y$. \ii $\left\lvert X \right\rvert \le \left\lvert Y \right\rvert$ if and only if there is an injective map $X \injto Y$. \end{enumerate} \end{proposition} Diligent readers are invited to try and prove this. \section{Aleph numbers} \prototype{$\aleph_0 = \omega$, and $\aleph_1$ is the first uncountable ordinal.} First, let us check that cardinals can get arbitrarily large: \begin{proposition} We have $\left\lvert X \right\rvert < \left\lvert \PP(X) \right\rvert$ for every set $X$. \end{proposition} \begin{proof} There is an injective map $X \injto \PP(X)$ but there is no injective map $\PP(X) \injto X$ by \Cref{lem:cantor_diag}. \end{proof} Thus we can define: \begin{definition} For a cardinal $\kappa$, we define $\kappa^+$ to be the least cardinal above $\kappa$, called the \vocab{successor cardinal}. \end{definition} This $\kappa^+$ exists and has $\kappa^+ \le \left\lvert \PP(\kappa) \right\rvert$. Next, we claim that: \begin{exercise} Show that if $A$ is a set of cardinals, then $\cup A$ is a cardinal. \end{exercise} Thus by transfinite induction we obtain that: \begin{definition} For any $\alpha \in \On$, we define the \vocab{aleph numbers} as \begin{align*} \aleph_0 &= \omega \\ \aleph_{\alpha+1} &= \left( \aleph_\alpha \right)^+ \\ \aleph_{\lambda} &= \bigcup_{\alpha < \lambda} \aleph_\alpha. \end{align*} \end{definition} Thus we have the sequence of cardinals \[ 0 < 1 < 2 < \dots < \aleph_0 < \aleph_1 < \dots < \aleph_\omega < \aleph_{\omega+1} < \dots. \] 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: \begin{lemma}[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 $\alpha \in \On$. \end{lemma} \begin{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 $\alpha = \ol\alpha + 1$ is a successor, then \[ \aleph_{\ol\alpha} < \kappa < \aleph_{\alpha} = (\aleph_{\ol\alpha})^+ \] which contradicts 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 $\gamma < \lambda$ with $\aleph_\gamma > \kappa$, which contradicts the minimality of $\alpha$. \end{proof} \begin{definition} An infinite cardinal which is not a successor cardinal is called a \vocab{limit cardinal}. It is exactly those cardinals of the form $\aleph_\lambda$, for $\lambda$ a limit ordinal, plus $\aleph_0$. \end{definition} \section{Cardinal arithmetic} \prototype{$\aleph_0 \cdot \aleph_0 = \aleph_0 + \aleph_0 = \aleph_0$} 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 \begin{align*} \aleph_0 + \aleph_0 &= \aleph_0 \\ \aleph_0 \cdot \aleph_0 &= \aleph_0 \end{align*} 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: \begin{definition}[Cardinal arithmetic] Given cardinals $\kappa$ and $\mu$, define \[ \kappa + \mu \defeq \left\lvert \left( \left\{ 0 \right\} \times \kappa \right) \cup \left( \left\{ 1 \right\} \times \mu \right) \right\rvert \] and \[ \kappa \cdot \mu \defeq \left\lvert \mu \times \kappa \right\rvert . \] \end{definition} \begin{ques} Check this agrees with what you learned in pre-school for finite cardinals. \end{ques} \begin{abuse} 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. \end{abuse} 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, \begin{definition} Let $\omega_\alpha = \aleph_\alpha$ viewed as an ordinal. \end{definition} However, as we've seen already we have that $\aleph_0 \cdot \aleph_0 = \aleph_0$. In fact, this holds even more generally: \begin{theorem}[Infinite cardinals squared] Let $\kappa$ be an infinite cardinal. Then $\kappa \cdot \kappa = \kappa$. \end{theorem} \begin{proof} Obviously $\kappa \cdot \kappa \ge \kappa$, so we want to show $\kappa \cdot \kappa \le \kappa$. The idea is to try 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 \emph{re-ordered} it according to the diagonal: \[ \begin{array}{l|cccccc} & 0 & 1 & 2 & 3 & 4 & \dots \\ \hline 0 & 0 & 1 & 3 & 6 & 10 & \dots \\ 1 & 2 & 4 & 7 & 11 & \dots & \\ 2 & 5 & 8 & 12 & \dots & & \\ 3 & 9 & 13 & \dots & & & \\ 4 & 14 & \dots & & & & \end{array} \] We'd like to copy this idea for a general $\kappa$; however, since addition is less well-behaved for infinite ordinals it will be more convenient to use $\max\{\alpha,\beta\}$ rather than $\alpha+\beta$. Specifically, we put the ordering $<_{\text{max}}$ on $\kappa \times \kappa$ as follows: for $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$ in $\kappa \times \kappa$ we declare $(\alpha_1, \beta_1) <_{\text{max}} (\alpha_2, \beta_2)$ if \begin{itemize} \ii $\max \left\{ \alpha_1, \beta_1 \right\} < \max \left\{ \alpha_2, \beta_2 \right\}$ or \ii $\max \left\{ \alpha_1, \beta_1 \right\} = \max \left\{ \alpha_2, \beta_2 \right\}$ and $(\alpha_1, \beta_1)$ is lexicographically earlier than $(\alpha_2, \beta_2)$. \end{itemize} This alternate ordering (which deliberately avoids referring to the addition) looks like: \[ \begin{array}{l|cccccc} & 0 & 1 & 2 & 3 & 4 & \dots \\ \hline 0 & 0 & 1 & 4 & 9 & 16 & \dots \\ 1 & 2 & 3 & 5 & 10 & 17 & \dots \\ 2 & 6 & 7 & 8 & 11 & 18 & \dots \\ 3 & 12 & 13 & 14 & 15 & 19 & \dots \\ 4 & 20 & 21 & 22 & 23 & 24 & \dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array} \] Now we proceed by transfinite induction on $\kappa$. The base case is $\kappa = \aleph_0$, done above. Now, $<_{\text{max}}$ 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 $\ol\gamma \in \gamma$, we have $\left\lvert \ol\gamma \right\rvert < \kappa$. Suppose $\ol\gamma$ corresponds to the point $(\alpha, \beta) \in \kappa \times \kappa$ under this bijection. If $\alpha$ and $\beta$ are both finite then certainly $\ol\gamma$ is finite too. Otherwise, let $\ol\kappa = \max \{\alpha, \beta\} < \kappa$; then the number of points below $\ol\gamma$ is at most \[ \left\lvert \alpha \right\rvert \cdot \left\lvert \beta \right\rvert \le \ol\kappa \cdot \ol\kappa = \ol\kappa \] by the inductive hypothesis. So $\left\lvert \ol\gamma \right\rvert \le \ol\kappa < \kappa$ as desired. \end{proof} From this it follows that cardinal addition and multiplication is really boring: \begin{theorem}[Infinite cardinal arithmetic is trivial] Given cardinals $\kappa$ and $\mu$, one of which is infinite, we have \[ \kappa \cdot \mu = \kappa + \mu = \max\left\{ \kappa, \mu \right\}.\] \end{theorem} \begin{proof} The point is that both of these are less than the square of the maximum. Writing out the details: \begin{align*} \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\}. \qedhere \end{align*} \end{proof} \section{Cardinal exponentiation} \prototype{$2^\kappa = \left\lvert \PP(\kappa) \right\rvert$.} \begin{definition} Suppose $\kappa$ and $\lambda$ are cardinals. Then \[ \kappa^\lambda \defeq \left\lvert \mathscr F(\lambda, \kappa) \right\rvert. \] Here $\mathscr F(A,B)$ is the set of functions from $A$ to $B$. \end{definition} \begin{abuse} As before, we are using the same notation for both cardinal and ordinal arithmetic. Sorry! \end{abuse} In particular, $2^\kappa = \left\lvert \PP(\kappa) \right\rvert > \kappa$, 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 \vocab{Continuum Hypothesis}. It turns out that this is \emph{undecidable} -- it is not possible to prove or disprove this from the $\ZFC$ axioms. \section{Cofinality} \prototype{$\aleph_0$, $\aleph_1$, \dots\ are all regular, but $\aleph_\omega$ has cofinality $\omega$.} \begin{definition} Let $\lambda$ be an ordinal (usually a limit ordinal), and $\alpha$ another ordinal. A map $f : \alpha \to \lambda$ of ordinals is called \vocab{cofinal} if for every $\ol\lambda < \lambda$, there is some $\ol\alpha \in \alpha$ such that $f(\ol\alpha) \ge \ol\lambda$. In other words, the map reaches arbitrarily high into $\lambda$. \end{definition} \begin{example} [Example of a cofinal map] \listhack \begin{enumerate}[(a)] \ii The map $\omega \to \omega^\omega$ by $n \mapsto \omega^n$ is cofinal. \ii For any ordinal $\alpha$, the identity map $\alpha \to \alpha$ is cofinal. \end{enumerate} \end{example} \begin{definition} Let $\lambda$ be a limit ordinal. The \vocab{cofinality} of $\lambda$, denoted $\cof(\lambda)$, is the smallest ordinal $\alpha$ such that there is a cofinal map $\alpha \to \lambda$. \end{definition} \begin{ques} Why must $\alpha$ be an infinite cardinal? \end{ques} 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 \vocab{regular}. \begin{example}[$\aleph_0$ is regular] $\cof(\aleph_0) = \aleph_0$, because no finite subset of $\aleph_ 0 = \omega$ can reach arbitrarily high. \end{example} \begin{example}[$\aleph_1$ is regular] $\cof(\aleph_1) = \aleph_1$. Indeed, assume for contradiction that some countable set of ordinals $A = \{ \alpha_0, \alpha_1, \dots \} \subseteq \aleph_1$ reaches arbitrarily high inside $\aleph_1$. Then $\Lambda = \cup A$ is a \emph{countable} ordinal, because it is a countable union of countable ordinals. In other words $\Lambda \in \aleph_1$. But $\Lambda$ is an upper bound for $A$, contradiction. \end{example} On the other hand, there \emph{are} cardinals which are not regular; since these are the ``rare'' cases we call them \vocab{singular}. \begin{example}[$\aleph_\omega$ 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 $\cof(\aleph_\omega) = \aleph_0$. \end{example} We now confirm a suspicion you may have: \begin{theorem} [Successor cardinals are regular] If $\kappa = \ol\kappa^+$ is a successor cardinal, then it is regular. \end{theorem} \begin{proof} We copy the proof that $\aleph_1$ was regular. Assume for contradiction that for some $\mu \le \ol\kappa$, there are $\mu$ sets reaching arbitrarily high in $\kappa$ as a cardinal. Observe that each of these sets must have cardinality at most $\ol\kappa$. 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 \[ \#\text{sets} \cdot \#\text{elms} \le \mu \cdot \ol\kappa = \ol\kappa \] and hence $\left\lvert \Lambda \right\rvert \le \ol\kappa < \kappa$. \end{proof} \section{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. \begin{example}[Beth fixed point] Consider the monstrous cardinal \[ \kappa = \aleph_{\aleph_{\aleph_{\ddots}}}. \] This might look frighteningly huge, as $\kappa = \aleph_\kappa$, but its cofinality is $\omega$ as it is the limit of the sequence \[ \aleph_0, \aleph_{\aleph_0}, \aleph_{\aleph_{\aleph_0}}, \dots \] \end{example} More generally, one can in fact prove that \[ \cof(\aleph_\lambda) = \cof(\lambda). \] 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 \vocab{weakly inaccessible}. Such cardinals are so large that it is impossible to prove or disprove their existence in $\ZFC$. It is the first of many so-called ``large cardinals''. An infinite cardinal $\kappa$ is a strong limit cardinal if \[ \forall \ol\kappa < \kappa \quad 2^{\ol\kappa} < \kappa \] for any cardinal $\ol\kappa$. For example, $\aleph_0$ is a strong limit cardinal. \begin{ques} Why must strong limit cardinals actually be limit cardinals? (This is offensively easy.) \end{ques} A regular strong limit cardinal other than $\aleph_0$ is called \vocab{strongly inaccessible}. \section\problemhead \begin{problem} Compute $\left\lvert V_\omega \right\rvert$. \begin{hint} $\sup_{k \in \omega} \left\lvert V_k \right\rvert$. \end{hint} \end{problem} \begin{problem} Prove that for any limit ordinal $\alpha$, $\cof(\alpha)$ is a \emph{regular} cardinal. \begin{hint} Rearrange the cofinal maps to be nondecreasing. \end{hint} \end{problem} \begin{sproblem} [Strongly inaccessible cardinals] \label{prob:strongly_inaccessible} Show that for any strongly inaccessible $\kappa$, we have $\left\lvert V_\kappa \right\rvert = \kappa$. \end{sproblem} \begin{problem} [K\"onig's theorem] Show that \[ \kappa^{\cof(\kappa)} > \kappa \] for every infinite cardinal $\kappa$. \end{problem}