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\section*{Hartogs number}

[[!redirects Hartogs's number]]

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{foundations}{}\paragraph*{{Foundations}}\label{foundations}

[[!include foundations - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{gch_implies_ac}{GCH implies AC}\dotfill \pageref*{gch_implies_ac} \linebreak
\noindent\hyperlink{equivalent_forms_of_ac}{Equivalent forms of AC}\dotfill \pageref*{equivalent_forms_of_ac} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{related_page}{Related page}\dotfill \pageref*{related_page} \linebreak
\noindent\hyperlink{related_entries}{Related entries}\dotfill \pageref*{related_entries} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

The \emph{Hartogs number} of a [[cardinal number]] $\kappa$ is the number of ways to [[well-order]] a set of cardinality at most $\kappa$.\footnote{The spelling \emph{Hartogs' number}, or even \emph{Hartog's number}, is also in use. The latter based presumably on a misapprehension since the concept was originally proposed by the German mathematician Friedrich Hartogs (1874-1943) in 1915.} 

Assuming the [[axiom of choice]], it is the smallest [[ordinal number]] whose cardinality is greater than $\kappa$ and therefore the [[successor]] of $\kappa$ as a cardinal number. But even without the axiom of choice, it makes sense and is often an effective substitute for such a successor.

\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

We will define the Hartogs number as a [[functor|functorial]] operation from [[sets]] to [[well-ordered sets]]. The operation on numbers is just a round-about way of talking about the same thing.

So let $S$ be a set. Without the axiom of choice (or more precisely, the [[well-ordering theorem]]), it may not be possible to well-order $S$ itself, but we can certainly well-order some [[subsets]] of $S$. On the other hand, if we can well-order $S$ (or a subset), then there may be many different ways to do so, even non[[isomorphism|isomorphic]] ways. So to begin with, let us form the collection of all well-ordered subsets of $S$, that is the subset of

\begin{displaymath}
\coprod_{A: \mathcal{P}S} \mathcal{P}(A \times A) ,
\end{displaymath}
where $\coprod$ indicates [[disjoint union]] and $\mathcal{P}$ indicates [[power set]], consisting of those pairs $(A,R)$ such that $R$ is a well-ordering. Then form a [[quotient set]] by identifying all well-ordered subsets that are isomorphic as well-ordered sets. This gives a set of well-order types, or [[ordinal numbers]], which can itself be well-ordered by the general theory of ordinal numbers.

The \textbf{Hartogs number} of $S$ is this well-ordered set, the set of all order types of well-ordered subsets of $S$. It is denoted $\aleph(S)$. If $\kappa$ is the cardinality of $S$, then let $\kappa^+$ be the cardinality or ordinal rank (as desired) of the Hartogs number of $S$; this is called the \textbf{Hartogs number} of $\kappa$.

There are other ways to encode $\aleph(S)$. A subset $A \subseteq S$ with a well-ordering can be specified through the collection of its initial segments or [[lower set|down-sets]], i.e., a suitable family of subsets of $S$ named by an [[element]] of $P P(S)$. Thus the set of well-orderings of subsets of $S$ can be defined as a suitable subset of $P P(S)$. Similarly, an equivalence class of such structures corresponding to a given order type is also definable as a subset of $P P(S)$, i.e., is named by a suitable element of $P P P(S)$, so $\aleph(S)$ which is the set of such equivalence classes is definable as a subset of $P P P(S)$.

All of the defining formulas in this description involve only bounded quantification, i.e., can be expressed in the language of [[ZFC|BZC]] or in [[ETCS]]. The proof that $\aleph(S)$ is well-ordered (by existence of $P$-\href{https://ncatlab.org/nlab/show/well-founded+relation#coalgebraic_formulation}{coalgebra} morphisms between representatives of order types) can be enacted in these theories without using the axiom of choice, hence the basic theory of $\aleph(S)$ can be carried out in BZ (bounded Zermelo set theory, without choice).

\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

\begin{theorem}
\label{}\hypertarget{}{}
There is no [[injection]] $\aleph(S) \to S$.

\end{theorem}
This theorem is to [[Cantor's theorem]] as [[Burali-Forti's paradox]] is to [[Russell's paradox]]. This observation alone hints how the proof goes:

\begin{proof}
As observed before, $\aleph(S)$ is well-ordered. Suppose there were an injection $i: \aleph(S) \to S$; then $i$ gives a bijection of $\aleph(S)$ onto its image $I$. By transport of structure, $I$ uniquely admits a well-ordering for which $i$ is an isomorphism of well-ordered sets. Using this well-ordering, there is an initial segment inclusion

\begin{displaymath}
j: I \hookrightarrow \aleph(S)
\end{displaymath}
which takes $x \in I$ to (the order type of) $\{y \in I: y \lt x\}$. Since there can be at most one initial segment inclusion between well-orders (cf. \href{https://ncatlab.org/nlab/show/well-founded+relation#simulations}{simulation}), $j$ must coincide with the isomorphism $i^{-1}$, which is onto. This means $I$ itself, as an order type, must be of the form $I = \{y \in I: y \lt x\}$ for some $x \in I$, which in turn implies $x \lt x$, a contradiction.

\end{proof}
According to this theorem, using the usual ordering of cardinal numbers, $\kappa^+ \nleq \kappa$. So if this $\leq$ is a [[total order]] (a statement equivalent to the [[axiom of choice]]), we can say that $\kappa^+ \gt \kappa$.

Even without choice, however, we can say this: If $\alpha$ is an ordinal number such that $|\alpha| \nleq \kappa$, then $\kappa^+ \leq \alpha$. (Notice that we've shifted our thinking of the Hartogs number from a cardinal to an ordinal.) That is, $\kappa^+$ is the smallest ordinal number whose cardinal number is not at most $\kappa$. This doesn't use any form of choice except for [[excluded middle]]; we only need choice to conclude that $|\kappa^+| \gt \kappa$.

The axiom of choice also implies the well-ordering theorem, that any set can be well-ordered. Thus with choice, $\kappa^+$ is (now as a cardinal again) the smallest cardinal number greater than $\kappa$; this explains the notation $\kappa^+$.

\hypertarget{gch_implies_ac}{}\subsubsection*{{GCH implies AC}}\label{gch_implies_ac}

In [[ZF]] (without C, the [[axiom of choice]]), a set $X$ can be well-ordered if there is an injection $i: X \to \aleph(X)$, for any subset of a well-ordered set inherits a well-order. As an amusing illustration of this principle, we will prove an old result due to Sierpiski, following \hyperlink{Gill}{Gillman}:

\begin{theorem}
\label{}\hypertarget{}{}
If the generalized [[continuum hypothesis]] holds, then the [[axiom of choice]] holds, i.e., ZF + GCH implies AC.

\end{theorem}
(Recall the GCH as formulated in ZF: for infinite sets $Y, Z$, if $|Y| \leq |Z| \leq |P(Y)|$, then $|Y| = |Z|$ or $|Z| = |P(Y)|$, letting $|Y|$ denote the cardinality as usual and where $|Y| \leq |Z|$ means there is a [[monomorphism]] $Y \to Z$.)

\begin{proof}
In fact we will show that any set $Y$ can be well-ordered, which implies AC (see the article [[Zorn's lemma]]). We start with a technical trick: any $Y$ can be embedded in a set $X$ with the property $|X| = |X + X|$ (consider for example $X = P(\mathbb{N} + Y)$), and then it suffices to well-order such $X$, since we can then pull back the well-ordering along the embedding to well-order $Y$. Iterated power sets of $X$ also have that property.

Now, the Hartogs number $\aleph(X)$ can be constructed explicitly in ZF as a subset of the triple power set $P^3(X) = P P P(X)$. We will show for $n \in \{1, 2, 3\}$ that under GCH, the hypothesis $|\aleph(X)| \leq |P^n(X)|$, true for $n = 3$, leads to one of two conclusions:

\begin{enumerate}%
\item $|X| \leq |\aleph(X)|$ (whence $X$ can be well-ordered), or


\item $|\aleph(X)| \leq |P^{n-1}(X)|$.



\end{enumerate}
If we land in case 2, then we loop around and feed that conclusion back into the hypothesis and iterate. Iterating this, we cannot descend through case 2 all the way down to $|\aleph(X)| \leq |P^0(X)| = |X|$, so after a few iterations we wind up in case 1 anyway: $X$ can be well-ordered.

To show this, we need a spot of cardinal arithmetic in ZF. First, we already observed that any $P = P^n(X)$ for $n \geq 0$ has the property $|P| = |2P|$. Second, we \textbf{claim}

\begin{itemize}%
\item If $P, A$ are sets such that $|P| = |2P|$ and $|A+P| = |2^P|$, then $|2^P| \leq |A|$

\end{itemize}
whose ZF-proof we give in a moment. Granting the claim, let us continue. We have

\begin{displaymath}
|P^{n-1}(X)| \leq |\aleph(X)| + |P^{n-1}(X)| \stackrel{hypoth}{\leq} |P^n(X)| + |P^{n-1}(X)| \leq 2|P^n(X)| = |P^n(X)|
\end{displaymath}
or just

\begin{displaymath}
|P^{n-1}(X)| \leq |\aleph(X)| + |P^{n-1}(X)| \leq |P^n(X)|.
\end{displaymath}
By GCH, one of those two inequalities is an equality. If the first inequality is an equality, then we conclude $|\aleph(X)| \leq |P^{n-1}(X)|$, which lands us in case 2. If the second inequality is an equality, then the claim applies and we conclude $|P^n(X)| \leq |\aleph(X)|$, in which case $|X| \leq |P^n(X)| \leq |\aleph(X)|$ and we land in case 1.

So we are done except for the claim. Given bijections $A + P \cong 2^P$, $P + P \cong P$, let $f$ be the evident composite

\begin{displaymath}
P \stackrel{incl}{\to} A + P \cong 2^P \cong 2^{P + P} \cong 2^P \times 2^P \stackrel{\pi_1}{\to} 2^P
\end{displaymath}
where $\pi_1$ denotes first projection. Let $C$ be an element not in the image of $f$, e.g., $C = \{x \in P: x \notin f(x)\}$ ([[Cantor's theorem]]). It follows that under the bijection $A + P \cong 2^P \times 2^P$, those elements of $A + P$ that map to elements of the fiber $\pi_1^{-1}(\{C\})$ can only belong to the summand $A$. Thus it is a subset of $A$ that maps bijectively onto $\pi_1^{-1}(\{C\}) \cong 2^P$, and this proves $|2^P| \leq |A|$.

\end{proof}
\hypertarget{equivalent_forms_of_ac}{}\subsubsection*{{Equivalent forms of AC}}\label{equivalent_forms_of_ac}

Methods somewhat similar to those used to prove GCH implies AC can be used to establish equivalent forms of AC.

\begin{theorem}
\label{}\hypertarget{}{}
\textbf{(Tarski)} Over ZF (or ETCS), the axiom of choice holds iff every infinite set $Y$ can be put into bijection with its square: $Y^2 \cong Y$.

\end{theorem}
\begin{proof}
The more interesting direction is the ``if'', so we assume every infinite set can be put into bijection with its square.

It suffices to prove that every $X$ can be well-ordered. WLOG we prove this for (Dedekind) infinite $X$, since every $X$ embeds in a Dedekind infinite set (such as $X + \mathbb{N}$), and we can pull back a well-order on the latter along such an embedding to one on the former.

For infinite $X$, let $Y$ be the disjoint union $X + \aleph(X)$, and let $\phi: Y \times Y \to Y$ be a bijection. We claim that for all $x \in X$, there exist $\alpha, \beta \in \aleph(X)$ such that $\phi(x, \alpha) = \beta$. If that were not true, then there would exist $x \in X$ such that for all $\alpha \in \aleph(X)$ we have $\phi(x, \alpha) \notin \aleph(X)$, or in other words for all $\alpha \in \aleph(X)$ we have $\phi(x, \alpha) \in X$, so that $\phi(x, -)$ defines an injection $\aleph(X) \rightarrowtail X$. This is impossible.

By the claim, for each $x \in X$ there is a least pair $(\alpha_x, \beta_x) \in \aleph(X)^2$, considered in [[lexicographic order]], such that $\phi(x, \alpha_x) = \beta_x$. This assignment $x \mapsto (\alpha_x, \beta_x)$ defines an embedding $X \to \aleph(X)^2$ into a well-ordered set, so that $X$ inherits a well-order by restriction along the embedding, and we are done.

\end{proof}
A very similar method establishes the following claim.

\begin{prop}
\label{}\hypertarget{}{}
Over ZF, AC is equivalent to the statement that every inhabited set admits a group structure.

\end{prop}
\begin{proof}
Every inhabited finite set admits a cyclic group structure, and under AC any infinite set can be put in bijection with the collection of its finite subsets, which forms a group under symmetric difference.

Conversely, suppose $Y = X + \aleph(X)$ admits a group structure. Then for every $x \in X$, there exist $\alpha, \beta \in \aleph(X)$ such that $x\alpha = \beta$. For if that were not true, then there exists $x \in X$ such that for all $\alpha \in \aleph(X)$, we have $x \alpha \notin \aleph(X)$, i.e., for all $\alpha \in \aleph(X)$ we have $x\alpha \in X$, which implies left multiplication by $x$ defines an injection $\aleph(X) \rightarrowtail X$, contradiction.

So for each $x \in X$, there exists a least $(\alpha_x, \beta_x) \in \aleph(X)^2$ in lexicographic order such that $x \alpha_x = \beta_x$. This again defines an embedding $x \mapsto (\alpha_x, \beta_x)$ into a well-ordered set $\aleph(X)^2$, and we are done.

\end{proof}
\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

For $n$ a [[natural number]] regarded as the cardinal number of a [[finite set]], $n^+$ is the usual [[successor]] $n + 1$. This result uses excluded middle; else we get the \emph{plump} successor of $n$, which may be rather larger.

For $\aleph_0$ the cardinality of the set of all [[natural number]]s, the Hartogs number $\aleph_0^+ = \omega_1$ is the smallest [[uncountable set|uncountable]] ordinal. Assuming the axiom of choice ([[countable choice]] and excluded middle are enough), we have $\aleph_0^+ = \aleph_1$ as a cardinal.

In general, we get a sequence $\omega_\alpha$ of infinite cardinalities of well-orderable sets; assuming excluded middle, every infinite well-orderable cardinality shows up in this sequence. Assuming the axiom of choice, every infinite cardinal shows up, and we have $|\omega_\alpha| = \aleph_\alpha$. (Actually, there's no real need to begin with infinite cardinals; if we started with $\omega_0 = 0$ instead of $\omega_0 = \mathbf{N}$ and $\aleph_0 = 0$ instead of $\aleph_0 = |\mathbf{N}|$, then absolutely \emph{every} cardinality or well-orderable cardinality would appear.)

\hypertarget{related_page}{}\subsection*{{Related page}}\label{related_page}

\begin{itemize}%
\item \href{https://en.wikipedia.org/wiki/Hartogs_number}{wikipedia entry}

\end{itemize}
\hypertarget{related_entries}{}\subsection*{{Related entries}}\label{related_entries}

\begin{itemize}%
\item [[Lindenbaum number]]


\item [[well-ordering theorem]]


\item [[axiom of choice]]


\item [[diagonal argument]]



\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

\begin{itemize}%
\item Friedrich Hartogs, \emph{\"U{}ber das Problem der Wohlordnung} , Math. Ann. \textbf{76} no.4 (1915) pp.438-443. (\href{http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002266105}{gdz})


\item [[Radu Diaconescu]], \emph{On Comparability in a Topos} , Proc. AMS \textbf{98} no.3 (1986) pp.389-393. (\href{http://www.ams.org/journals/proc/1986-098-03/S0002-9939-1986-0857927-9/S0002-9939-1986-0857927-9.pdf}{pdf})


\item Leonard Gillman, \emph{Two Classical Surprises Concerning the Axiom of Choice and the Continuum Hypothesis}, Amer. Math. Monthly 109 (June-July 2002), 544-553. (\href{https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Gillman544-553.pdf}{pdf})



\end{itemize}
The last section of Gillman's paper is based on Sierpiski's 1947 proof which may be found \href{http://matwbn.icm.edu.pl/ksiazki/fm/fm34/fm3411.pdf}{here}.

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