\documentclass{amsart} \usepackage[parfill]{parskip} \usepackage{graphicx} \usepackage{amssymb, latexsym} \usepackage{epstopdf} \usepackage{color} \DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png} \textwidth = 6.5 in \textheight = 9 in \oddsidemargin = 0.0 in \evensidemargin = 0.0 in \topmargin = 0.0 in \headheight = 0.0 in \headsep = 0.0 in \parskip = 0.2in \parindent = 0.0in \pagestyle{plain} \theoremstyle{plain} \newtheorem{theorem}{Theorem}[section] \newtheorem{cor}[theorem]{Corollary} \newtheorem{prop}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \theoremstyle{definition} \newtheorem*{definition}{Definition} \newtheorem*{defn}{Definition} \newtheorem{example}{Example} \theoremstyle{remark} \newtheorem*{notation}{Notation} \newtheorem*{note}{Note} \newsavebox{\savepar} \newenvironment{xbox}{\begin{lrbox}{\savepar} \begin{minipage}[c]{\textwidth}} {\end{minipage}\end{lrbox}\fcolorbox{black}{green}{\usebox{\savepar}}} \renewcommand{\baselinestretch}{1.3} \newcommand{\ord}{\text{ord}} \begin{document} \setcounter{section}{4} \section{Ordinal Numbers} \subsection{Ordinal Numbers} \hfill \framebox[\textwidth][c]{ \parbox{.9\textwidth}{ \begin{defn} Each well-ordered set $A$ is assigned a symbol in such a way that two sets are assigned the same symbol iff they are order-isomorphic. This symbol is called the \emph{ordinality} or \emph{ordinal number} of $A$ and is denoted by $\lambda =\ord(A)$. Note that by definition, \[ \ord(A) = \ord(B) \iff A \simeq B. \] In other words, an ordinal number is an equivalence class of order-isomorphic well-ordered sets. \end{defn} } } \begin{note} An ordinal number is the order type of a \emph{well}-ordered set. \end{note} \begin{example} The finite ordinal numbers are given by: $\ord(\emptyset) = 0; \; \ord(\{1,2,\dots , n\}) = n$ for any $n \in \mathbb{P}$. So the symbols $0, 1, 2, \dots$ are used to denote natural numbers, cardinal numbers \emph{and} ordinal numbers! \end{example} \begin{note} Note that since all finite, equipotent well-ordered sets are order-isomorphic, these are the only finite ordinal numbers. However, it is not the case that all infinite, equipotent well-ordered sets are order-isomorphic. \end{note} \begin{xbox} \begin{example} $\omega =\ord(\mathbb{P}) \neq \ord(\mathbb{P}^*)$. \end{example} \end{xbox} \subsection{Ordering of Ordinal Numbers} \begin{quote} \end{quote} \begin{xbox} \begin{defn} Let $\lambda$ and $\mu$ be ordinal numbers and let $A$ and $B$ be well-ordered sets such that $\lambda = \ord(A)$ and $\mu = \ord(B)$. Then $\lambda \prec \mu \iff A \lessdot B$. [Verify that this is well-defined. Then verify that this defines a quasi-order.] \end{defn} \end{xbox} \begin{xbox} \begin{example} $\omega \prec \ord(\mathbb{P}^*)$. \end{example} \end{xbox} \begin{prop} Let $\lambda = \ord(A)$ and suppose $\mu$ is an ordinal number such that $\mu \prec \lambda$. Then there exists a unique $a \in A$ such that $\mu = \ord\big(s(a)\big)$. \end{prop} \begin{xbox} \begin{proof} \end{proof} \end{xbox} \begin{theorem} The collection of ordinal numbers is well-ordered. \end{theorem} \begin{xbox} \begin{proof} Let $\mathcal{W}$ be a nonempty collection of ordinal numbers. Each ordinal number in $\mathcal{W}$ represents an order-isomorphism class of well-ordered sets. By the Axiom of Choice, we can choose a single well-ordered set from each of these classes to obtain a nonempty collection $\mathcal{A}$ of pairwise non-order-isomorphic well-ordered sets. ..... \end{proof} \end{xbox} \begin{xbox} \begin{cor}[Law of Trichotomy for Ordinal Numbers] \end{cor} \end{xbox} \begin{cor} Every ordinal number has an immediate successor. However, $\lambda = \ord(A)$ has an immediate predecessor if and only if $A$ has a last element. \end{cor} \begin{xbox} \begin{proof} \end{proof} \end{xbox} \begin{defn} A \emph{limit ordinal} is a nonzero ordinal number with no immediate predecessors; equivalently, a limit ordinal is the ordinal number of a nonempty well-ordered set with no last element. \end{defn} \begin{xbox} \begin{example} $\omega ; \: \ord(\mathbb{P}^*)$ \end{example} \end{xbox} \begin{note} For any ordinal number $\lambda$, the initial segment $s(\lambda)$ is the set of all ordinal numbers that strictly precede $\lambda$. As a subset of a well-ordered collection, $s(\lambda)$ is itself a well-ordered set and thus has its own ordinal number, $\ord\big(s(\lambda)\big)$. \end{note} \begin{xbox} \begin{example} $s(\omega) = \mathbb{N}$, so $\ord(s(\omega)) = \omega$. (To show that $s(\omega) = \mathbb{N}$, you must show two set inclusions. Refer to the `ordinal' definition of the natural numbers given in Example 1. To show $\ord(s(\omega)) = \omega$, you will need to show that $\mathbb{N} \simeq \mathbb{P}$.) \end{example} \end{xbox} \begin{theorem} For every ordinal number $\lambda$, $ord(s(\lambda) )= \lambda$. \end{theorem} \begin{xbox} \begin{proof} Let $\lambda = \ord(A)$, where $A$ is a well-ordered set and define $f:A \to s(\lambda)$ by %$f(a) = \ord\big(s_A(a)\big)$ for all $a \in A$. ... \end{proof} \end{xbox} \begin{note} From now on, when we consider $0< n \in \mathbb{N}$ as an ordinal number, we will view it as \[ n =\ord(s_{\mathbb{N}} (n)) = \ord(\{0, 1, 2, \dots, n-1\}) . \] This is subtly different from the definition given in Example 1. \end{note} \subsection{Ordinal Number Arithmetic} \hfill \begin{xbox} \begin{defn} Let $\lambda$ and $\mu$ be ordinal numbers and let $A$ and $B$ be \emph{disjoint} well-ordered sets such that $\lambda = \ord(A)$ and $\mu = \ord(B)$. Then $\lambda + \mu= \ord( \{A : B\})$, where $\{A : B \}$ is $A \cup B$ together with the concatenation order. \end{defn} \end{xbox} \begin{xbox} \begin{example} $\ord(\mathbb{P}^*) = \omega + \omega$ \end{example} \end{xbox} \begin{xbox} \begin{example} For any $n \in \mathbb{P}, \, n + \omega = \omega \prec \omega + n$, so ordinal number addition is \emph{not} commutative. \end{example} \end{xbox} \begin{prop} \hfill \begin{enumerate} \item Ordinal number addition is associative. \item The ordinal number 0 is an additive identity; that is, $\lambda + 0 = 0 + \lambda = \lambda$ for any ordinal number $\lambda$. \item For any ordinal number $\lambda$, $\lambda + 1$ is the immediate successor of $\lambda$. \end{enumerate} \end{prop} \begin{xbox} \begin{proof} \end{proof} \end{xbox} \begin{xbox} QUESTION: Determine whether the following statements are true for all ordinal numbers $\lambda, \, \mu$ and $\eta$, where $\lambda \neq 0$: \begin{enumerate} \item$\mu \prec \lambda + \mu$; \item $\mu \prec \mu + \lambda$; \item if $\mu \prec \eta$, then $\lambda +\mu \prec \lambda + \eta$; \item If $\mu \prec \eta$, then $\mu + \lambda \prec \eta +\lambda $. \end{enumerate} \end{xbox} \begin{xbox} \begin{defn} Let $\lambda$ and $\mu$ be ordinal numbers and let $A$ and $B$ be well-ordered sets such that $\lambda = \ord(A)$ and $\mu = \ord(B)$. Then $\lambda \cdot \mu= \ord( A \times B)$, where $A \times B$ is given the \emph{reverse} lexicographic order, defined by \[ (a, b) \prec (a^\prime, b^\prime) \iff b \prec b^\prime \text{ or } [b = b^\prime \text{ and } a \prec a^\prime]. \] \end{defn} \end{xbox} \begin{note} Unless otherwise stated, the Cartesian product of two well-ordered sets is assumed to be ordered reverse lexicographically. \end{note} \begin{xbox} \begin{example} Ordinal number multiplication is \emph{not} commutative because $\omega = 2 \cdot \omega \prec \omega \cdot 2 = \omega + \omega$. [Prove this carefully!] More generally, for any $1 < n \in \mathbb{P}$, $\omega = n \cdot \omega \prec \omega \cdot n $. [Explain how your careful proof can be generalized.] \end{example} \end{xbox} \begin{prop} \hfill \begin{enumerate} \item Ordinal number multiplication is associative. \item The ordinal number 1 is a multiplicative identity; that is, $\lambda \cdot 1 = 1 \cdot \lambda = \lambda$ for any ordinal number $\lambda$. \end{enumerate} \end{prop} \begin{xbox} \begin{proof} \end{proof} \end{xbox} \begin{prop} The \emph{left} distributive law of multiplication over addition holds for ordinal numbers; that is, $\lambda \cdot (\mu + \eta) = \lambda \cdot \mu + \lambda \cdot \eta$, for all ordinal numbers $\lambda, \, \mu$ and $\eta$. However, the \emph{right} distributive law does not in general hold. \end{prop} \begin{xbox} \begin{proof} \end{proof} \end{xbox} \begin{cor} Let $\lambda$ be an ordinal number. Then for all $n \in \mathbb{P}$, $\lambda \cdot n = \lambda + \lambda + \dots + \lambda$ ($n$ times). \end{cor} \begin{xbox} \begin{proof} (Hint. Use mathematical induction.) \end{proof} \end{xbox} \begin{defn} Let $\lambda$ be an ordinal number. Then for all $n \in \mathbb{P}$, we define $\lambda^n = \lambda \cdot \lambda \cdot \dots \cdot \lambda$ ($n$ times). \end{defn} \newpage We can write down many of the first few ordinal numbers in order. \begin{itemize} \item First we have the finite ordinal numbers: $0, 1,2, 3, \dots$. \item Then come the first limit ordinal number and its immediate successors: $\omega, \omega +1, \omega +2, \dots$ \item The next ordinal number is $\ord(\{0, 1, 2, \dots : \omega, \omega +1, \omega +2, \dots \}) = \omega + \omega =\omega \cdot 2$. \item Then come: $\omega \cdot 2 + 1, \omega \cdot 2 + 2, \dots : \omega \cdot 3, \omega \cdot 3 +1, \dots : \omega \cdot 4, \dots : \omega \cdot 5, \dots $ \item For each $n \in \mathbb{P}$, $s(\omega \cdot n)$ is the set of all ordinal numbers that strictly precede $\omega \cdot n$. The set $A = \bigcup \{ s(\omega \cdot n) \mid n \in \mathbb{P} \}$ is the set of all ordinal numbers that strictly precede $\omega \cdot n$ for some $n \in \mathbb{P}$. This subset of the collection of ordinal numbers is well-ordered, by Proposition 4.1(2). Since $A$ has no last element, $\ord(A)$ will be a limit ordinal. \begin{xbox} Then $ \ord(A) = \ord\big(\{0, 1, \dots : \omega, \dots : \omega \cdot 2, \dots : \omega \cdot n, \dots \}\big) = \omega \cdot \omega = \omega ^2. $ \end{xbox} \item We continue: $\omega^2 +1, \dots : \omega^2 \cdot 2, \dots : \omega^2 \cdot 3, \dots ; \omega^2 \cdot n, \dots : \omega^2 \cdot \omega = \omega ^3$. \item Then we have the finite powers of $\omega: \; \omega^3, \omega^3 +1, \dots, \omega^4, \dots , \omega ^5, \dots \omega^n, \dots $ \item Let $B= \bigcup \{ s(\omega^n) \mid n \in \mathbb{P} \}$ be the set of all ordinal numbers that precede an ordinal number of the form $\omega^n$, for some $n \in \mathbb{P}$. This subset of the collection of ordinal numbers is well-ordered, by Proposition 4.1(2). Note that $B$ has no last element and so $\ord(B)$ is a limit ordinal. We define \[ \ord(B)=\ord \big( \{ 0, 1, \dots ; \omega, \dots ; \omega \cdot 2, \dots ; \omega^2, \cdots ; \omega^n, \dots \} \big) = \omega^\omega. \] \item Next we have $\omega^\omega +1, \dots, \omega^\omega \cdot 2, \dots, (\omega^\omega)^\omega, \dots, \big((\omega^\omega)^\omega \big)^\omega, \dots$ \end{itemize} \begin{xbox} \begin{note} Each of the infinite ordinal numbers above is the ordinal number of a countably infinite set! \end{note} \end{xbox} \subsection{Well-Ordering Theorem} \begin{quote} \end{quote} \framebox[\textwidth][c]{ \parbox{.9\textwidth}{ \vspace{8pt} \textbf{WELL-ORDERING THEOREM.} \emph{Every set can be well-ordered.} \vspace{8pt} } } \begin{note} We give the proof first proposed by Ernest Zermelo in 1904, based on the Axiom of Choice. Zermelo also showed that the Axiom of Choice can be proved assuming the Well-Ordering Theorem to be true. Here's how: Let $\{A_i \, | i \in I \}$ be a nonempty collection of nonempty sets. If every $A_i$ can be well-ordered, then we can define $f: \{A_i \, | \, i \in I \} \to \bigcup \{A_i \, | \, i \in I \}$ by stipulating that $f(A_i)$ is the first element of $A_i$. Therefore, \textbf{the Well-Ordering Theorem and the Axiom of Choice are equivalent!} \end{note} \begin{proof} Let $X$ be a set. If $X$ is countable, then any function establishing countability can be used to define a well-ordering on $X$, so we assume $X$ is uncountable. Let $\wp^*(X)$ denote the set of nonempty subsets of $X$. Now, if we were \emph{given} an arbitrary well-ordering on $X$, then as above, we could \emph{define} a choice function $f$ on $\wp^*(X)$ by stipulating that $f$ `chooses' the first element of every nonempty subset of $X$. In particular, the first element of $X \backslash s_X(x)$ is $x$, so we would have $f(X \backslash s_X(x)) = x$. What we will show is that \emph{given} an arbitrary choice function $f$, we can \emph{define} a well-ordering on $X$ in such a way that $f$ satisfies this particular property. So we begin by assuming that $f$ is an arbitrary choice function on $\wp^*(X)$; such a function exists by the Axiom of Choice. Repeating a trick used in the proof of Theorem 2.4, we let $x_0=f(X)$, $x_1=f(X\backslash\{x_0\})$, $x_2 = f(X\backslash \{x_0, x_1\})$ and more generally, $x_n = f(X\backslash \{x_0, \dots, x_{n-1}\})$ for all $n \in \mathbb{P}$. Then the denumerable subset $N = \{x_0, x_1, \dots, x_n, \dots\} \subset X$ can be well-ordered by $x_0 \prec x_1 \prec \dots \prec x_n \prec \dots$. If we let $s_N(x_n)$ denote the initial segment of $x_n$ within the well-ordered set $N$, then \[ f\big(X\backslash s_N(x_n)\big) = f\big(X\backslash \{x_0, \dots, x_{n-1}\} \big) = x_n. \] By construction, we have arranged that $f$ satisfies the particular property, at least on complements of initial segments of $N$. \begin{defn} A subset $M \subseteq X$ is \emph{normal} iff $M$ has a well-ordering such that for all $m \in M$, \[f\big(X\backslash s_M(m)\big) = m. \] \end{defn} Our ultimate goal is to show that $X$ is a normal subset of itself; we will get there by proving a series of claims. We have shown that $X$ has at least one normal subset, $N$. Next we show that any two normal subsets can intersect in only a very restricted way. \emph{Claim 1.} If $M$ and $K$ are two normal subsets of $X$, then either $M=K$ or one is an initial segment of the other. \emph{Proof of Claim 1.} By Theorem 4.15, either $M$ and $K$ are order-isomorphic or one is order-isomorphic to an initial segment of the other. Without loss of generality, let $g:M \to K$ be an order-preserving injection; the range $g(M)$ is either all of $K$ or an initial segment of $K$. Let $\widehat{M} = \{ m \in M \, | \, g(m) \neq m \}$. If $\widehat{M} = \emptyset$, then we are done, so suppose $\widehat{M}$ is nonempty. Since $M$ is well-ordered, $\widehat{M}$ must have a first element $m_0$; note that by construction, $g$ must act as the identity on all elements of $s_M(m_0)$. In fact, $s_M(m_0) = s_K\big(g(m_0)\big)$. By definition of normal subset, \[ m_0 = f\big(X\backslash s_M(m_0)\big) = f\big(X\backslash s_K(g(m_0))\big) = g(m_0); \] this is a contradiction which proves Claim 1. Next, let $Y$ be the union of all normal subsets of $X$. By Claim 1, these normal subsets are nested as initial segments of one another. Hence, given any distinct $y, z \in Y$, there exists a normal subset $M$ containing both. Within $M$, $y \prec_M z$ or $z \prec_M y$; moreover, if $K$ is any other normal subset containing both $y$ and $z$, the $\prec_K$ relationship between them would be the same. \emph{Claim 2.} This defines a well-ordering on $Y$. \emph{Proof of Claim 2.} It is easy to verify that this relation defines a total order on $Y$. It remains to show that every nonempty subset of $Y$ has a first element. Assume $\emptyset \neq Z \subseteq Y$. By definition of $Y$, there exists a normal subset $M$ such that $Z \cap M \neq \emptyset$. Since $Z \cap M$ is a nonempty subset of the well-orderd set $M$, it must have a first element, $z_0$. We will show that $z_0$ is the first element of $Z$ as a whole. Suppose not. Then there exists $w \in Z$ such that $w \prec z_0$. Now, $w$ must be an element of a normal subset $K$ and so either $K = M$ or one is an initial segment of the other. In both cases, $w \in Z \cap M$, a contradiction. This finishes the proof of Claim 2. \emph{Claim 3.} The set $Y$ is a normal subset of $X$. \emph{Proof of Claim 3.} Let $y \in Y$; we must show that $f\big( X \backslash s_Y(y)\big) = y$. Now, $y \in M$ for some normal subset $M$ and by definition, $f\big( X \backslash s_M(y)\big) = y$, so it suffices to show that $ s_Y(y)=s_M(y)$. \emph{Claim 4.} In fact, $Y = X$. \emph{Proof of Claim 4.} Suppose $Y$ is a proper subset of $X$; then $X \backslash Y \in \wp^*(X)$. Let $x = f(X \backslash Y)$; note that by definition of a choice function, $x \in X \backslash Y$. Let $\widehat{Y} = Y \cup \{x\}$, with the concatenation order; that is, $\widehat{Y} = \{ Y : \{x\} \}$. By Proposition 4.2, $\widehat{Y}$ is well-ordered. First we show that $\widehat{Y}$ is a normal subset of $X$. If $y \in Y$, then note that $s_Y(y) =s _{\widehat{Y}}(y)$. Since $Y$ is a normal subset of $X$, $f\big( X \backslash s_{\widehat{Y}}(y)\big) =f\big( X \backslash s_Y(y)\big) = y$. Next, note that $s_{\widehat{Y}}(x) = Y$, so $f\big( X \backslash s_{\widehat{Y}}(x)\big)=f\big( X \backslash Y) = x$. Now, $Y$ was defined as the union of all normal subsets of $X$. We have just shown that if $Y$ is a proper subset of $X$, then there exists a normal subset $\widehat{Y}$ of $X$ that is clearly not in $Y$. This is a contradiction. Hence, $Y = X$, which finishes the proof of Claim 4. We have achieved our goal; we have shown that $X$ is a normal subset of itself, which in particular means that $X$ has a well-ordering. \end{proof} \begin{note} The Well-Ordering Theorem implies that $\mathbb{R}$ \emph{can} be well-ordered. Clearly, any well-ordering will not coincide with the usual ordering on $\mathbb{R}$, so how can we actually find a well-ordering? Unfortunately, the Well-Ordering Theorem is not very helpful on this score. The proof suggests that we begin with a specific choice function on $\wp^*(\mathbb{R})$; the Axiom of Choice asserts that a choice function exists, but is not very helpful when it comes to constructing one. This lack of determinism has left some dissatisfied with both the Axiom of Choice and the Well-Ordering Theorem. \end{note} \subsection{Relationship between Cardinal and Ordinal Numbers} \hfill \begin{xbox} \begin{defn} Let $\lambda = \ord(A)$, where $A$ is a well-ordered set. Then the \emph{cardinal number of} $\lambda$ is $ | \lambda| = |A|$. \end{defn} \end{xbox} % \begin{example} %As shown in section 5.3, $|\omega| = |\omega \cdot 2| = \aleph_0$. In fact, if $\lambda$ is any ordinal number in the list at the end of section 5.2, $|\lambda| = \aleph_0$. % \end{example} \begin{prop} For any cardinal number $\alpha$, there exists an ordinal number $\lambda$ such that $\alpha = |\lambda|$. However, this ordinal number may not be not unique. \end{prop} \begin{xbox} \begin{proof} (\emph{Hint.} Use the Well-Ordering Theorem.) \end{proof} \end{xbox} \begin{theorem} Let $\alpha$ and $\beta$ be cardinal numbers, and let $\lambda$ and $\mu$ be ordinal numbers such that $\alpha =|\lambda|$ and $\beta = |\mu|$. Then: \begin{enumerate} \item if $\lambda = \mu$, then $\alpha = \beta$, but the converse is false; \item if $\alpha < \beta$, then $\lambda \prec \mu$; \item If $\lambda \prec \mu$, then $\alpha \leq \beta$, but it is not necessarily true that $\alpha < \beta$. \end{enumerate} \end{theorem} \begin{xbox} \begin{proof} Let $A$ and $B$ be well-ordered sets such $\lambda = \ord(A),\alpha = |A|, \mu= \ord(B)$ and $\beta = |B|$...... \end{proof} \end{xbox} \begin{theorem}[Law of Trichotomy for Cardinal Numbers] For any two cardinal numbers $\alpha$ and $\beta$, exactly one of the following is true: $\alpha < \beta$, $\alpha = \beta$ or $\alpha > \beta$. \end{theorem} \begin{xbox} \begin{proof} \end{proof} \end{xbox} \begin{theorem} The collection of cardinal numbers is well-ordered (under the ordering of Chapter 2). \end{theorem} \begin{xbox} \begin{proof} \end{proof} \end{xbox} \begin{example} Every infinite cardinal number $\alpha$ satisfies $\aleph_0 \leq \alpha$. In other words, $\aleph_0$ is the first element in the nonempty subcollection of all infinite cardinal numbers. \end{example} \framebox[\textwidth][c]{ \parbox{.9\textwidth}{ \vspace{8pt} \textbf{Aleph Notation.} Recall that in a well-ordered collection, every element (except the last one, if there is a last one) has an immediate successor. In particular, this is true of the subcollection of all infinite cardinal numbers. We let $\aleph_1$ denote the immediate successor of $\aleph_0$, and we let $\aleph_2$ denote the immediate successor of $\aleph_1$. More generally, for any $n \in \mathbb{N}$, we let $\aleph_{n+1}$ denote the immediate successor of $\aleph_n$. More generally still, let $\alpha$ be any infinite cardinal number and let $s_\infty (\alpha)$ denote the set of all infinite cardinal numbers strictly less than $\alpha$. Then $s_\infty (\alpha)$ is a well-ordered set and so we can let $\lambda = \ord \big(s_\infty (\alpha)\big)$. Thus $\alpha$ is the ``$\lambda$-th'' infinite cardinal number; we write $\alpha =\aleph_{\lambda}$. } } \begin{example} Let $\alpha$ denote the first cardinal number that is greater than $\aleph_n$ for all $n \in \mathbb{N}$. Then $s_\infty(\alpha) = \{ \aleph_n \, | \, n \in \mathbb{N} \} \simeq \mathbb{N}$. Hence $\ord \big(s_\infty (\alpha)\big)=\ord(\mathbb{N})=\omega$, so we have $\alpha = \aleph_\omega$. Note that this is a limit element in the well-ordered collection of infinite cardinal numbers. \end{example} Do any of these fantastically large cardinal numbers actually `exist'? In other words, are there sets having such large cardinalities? Cantor's Theorem states that given any set $A$, its power set $\wp(A)$ has strictly larger cardinality - but how much larger? Recall that the Continuum Hypothesis conjectures there is no cardinal number strictly between $\aleph_0$ and $\mathbf{c}$. We also saw that $\mathbf{c} = 2^{\aleph_0}$. Combining this with the notation above, we can reformulate the Continuum Hypothesis as stating that $\aleph_1= 2^{\aleph_0}$. \framebox[\textwidth][c]{ \parbox{.9\textwidth}{ \vspace{8pt} \textbf{GENERALIZED CONTINUUM HYPOTHESIS.} For any ordinal number $\lambda, \, \aleph_{\lambda+1} = 2^{\aleph_\lambda}$. } } The Generalized Continuum Hypothesis conjectures that the power set of an infinite set has the `next greatest' cardinality. Thus $|\wp(\mathbb{P})| = \aleph_1, |\wp(\wp(\mathbb{P}))| = \aleph_2, |\wp(\wp(\wp(\mathbb{P})))| = \aleph_3,$ etc. Assuming that the Generalized Continuum Hypothesis is true - and this is a big assumption! - we have a way of generating infinite cardinal numbers that are immediate successors of other infinite cardinal numbers. What about limit infinite cardinal numbers, such as $\aleph_\omega$? For each $n \in \mathbb{N}$, let $A_n$ be a set of cardinality $\aleph_n$. Without loss of generality, we can assume these sets are pairwise disjoint. Then let $A_\omega = \bigcup\{A_n \, | \, n \in \mathbb{N}\}$, with the concatenation order; that is, $A_\omega = \{A_0 : A_1 : A_2 : \dots \}$. By Proposition 4.2, this is a well-ordered set. Since there is clearly an injection $A_n \to A_\omega$, by definition, $\aleph_n \leq |A_\omega|$ for all $n \in \mathbb{N}$. Now let $\Omega$ be the set of all infinite cardinal numbers that are greater than or equal to $\aleph_n$ for all $n \in \mathbb{N}$. Since $|A_\omega| \in \Omega$, we know $\Omega \neq \emptyset$. Since the infinite cardinal numbers form a well-ordered collection and $\Omega$ is a nonempty subset of this collection, $\Omega$ has a first element, which will by definition be $\aleph_\omega$. (Note that this proof does not assert that $|A_\omega|=\aleph_\omega$; it only indirectly asserts the existence of a set of cardinality exactly $\aleph_\omega$.) \end{document}