We denote by $\frak p\le q$ the abbreviation that there is $f:\frak p\to q$ which is injective, and by $\frak p\le^\ast q$ we abbreviate that there is a surjection from $\frak q$ onto $\frak p$.


If $X$ is a set in a universe of ZF, denote by $H(X)=\min\lbrace\alpha\mid\alpha\nleq X\rbrace$ known as The Hartog number of $X$. It is clear that $H(X)\nleq X$.


Under the axiom of choice $X\leq Y\iff X\leq^\ast Y$, however without the axiom of choice the $\leq^\ast$ order can behave quite strangely. For example, it is possible that $\omega\_1\nleq2^\omega$, but it is always true that $\omega\_1\leq^\ast2^\omega$. This may occur when the continuum is a countable union of countable sets, or if we are in a Solovay-like model.


Monro showed in [1] that it is consistent to have for every $\kappa$ an infinite set $p$ such $H(p)=\omega$ and $\kappa\leq^\ast p$.


In those two instances we have examples where the axiom of choice fails, and $H(x)\leq^\ast x$. If we do not wish to add large cardinal into the mix both the examples above are such that the axiom of countable choice already fails in.


**Questions:**


1. How much choice can we have while still finding sets with the property $H(x)\leq^\ast 
x$?
2. How consistent is this property with the different type of weak choice principles (e.g. if it is consistent with DC/AC/W for some $\kappa$, would it also be consistent with BPI and its relatives)?




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**Bibliography:**


1. G.P. Monro, **[Independence results concerning Dedekind finite sets.](http://dx.doi.org/10.1017/S1446788700023521)** *Journal of the Australian Mathematical Society (Series A)* (1975), 19 : pp 35-46.



Surjective Maps onto $\aleph$-numbers
I originally posted this question in hope that someone else knew of a reference for an answer, however it seemed to me that indeed the best way is to solve this on my own. I tried to imitate Monro's proof, to a certain extent, and I believe that I have succeeded:
$\renewcommand{\Dom}{\operatorname{Dom}}\renewcommand{\Htg}{\operatorname{Htg}}$


Let $M$ be a model of ZFC, $\kappa$ is an uncountable regular. Consider the notion of forcing $(P,\le)$ where $p\in P$ is a function from a subset of $\kappa\times\kappa$ into $2$, and $|\Dom(p)|<\kappa$.


Let $M[G]$ be a generic extension, $G\_i = \lbrace\alpha<\kappa\mid\exists p\in G: p(i,\alpha)=1\rbrace$ and $H=\lbrace G\_i\mid i<\kappa\rbrace$. We give them "canonical" names, $\dot G\_i = \lbrace\langle p,\check n\rangle \mid p(i,n)=1\rbrace$ and $\dot H=\lbrace\langle 1, \dot G\_i\rangle\mid i < \kappa\rbrace$.


We now proceed to define the symmetric model. If $\pi$ is a permutation of $\kappa$ we can think of $\pi$ as an automorphism of $P$ by: $$\pi p =\lbrace \langle\pi i,\alpha,j\rangle\mid \langle i,\alpha,j\rangle\in p\rbrace$$


Let $\mathscr G$ be the permutation group of these automorphisms. Let the subgroups filter be generated by *countable* support, that is $\dot f$ is a symmetric name if and only if there is a countable set of conditions such that whenever they are fixed pointwise, $\dot f$ is fixed.


We note two things:


* For $\pi$ fixes (pointwise) countably many conditions then only $<\kappa$ many points must be fixed by $\pi$.
* For $i<\kappa$ we have $\pi\dot G\_i = \dot G\_{\pi i}$, so all the $\dot G\_i$ are symmetric and trivially hereditarily symmetric. From this follows that $\dot H$ is also hereditarily symmetric.


Let $N$ be the symmetric model, we will show the following:


1. $H\in N$, which is obvious since $\pi(\dot G\_i)=\dot G\_{\pi i}$ so $\pi(\dot H)=\dot H$ for every $\pi\in\mathscr G$.
2. $N\models DC$:


Let $S\in N$ be a non-empty set and $R\in N$ a binary relation defined on all elements of $S$. In $M[G]$ we have a function which witnesses $DC$. For every $n\in\omega$ we have $n,f(n)\in N$ therefore $\langle n,f(n)\rangle\in N$. Each ordered pair has a countable support, and the union of this countable collection is countable (in $M$) therefore supports $\dot f$ as a symmetric name, as wanted.
3. $N\models\Htg(H)=\aleph\_1$:


Assume towards contradiction that $\dot f$ is a symmetric name such that for some $p\_0\in G$ we have $p\_0\Vdash\dot f:\check\omega\_1\to H\text{ injective}$. Let $E$ be a support for $\dot f$. There exists $i<\kappa$ that for every $\alpha$ and $p\in E$ the pair $\langle i,\alpha\rangle\notin\Dom(p)$. 


Let $p\le p\_0$ be an extension of $p$ such that for some $\alpha<\omega\_1$ we have: $$p\Vdash \dot f(\check\alpha)=\dot G\_i$$
We choose now $j<\kappa$ which for every $\alpha$ and $q\in E\cup\lbrace p\rbrace$ we have $\langle j,\alpha\rangle\notin\Dom(q)$, and define $\pi$ to be induced by the 2-cycle: $(i\ j)$.


We have that $\pi$ fixes $E$ pointwise, since neither $i$ nor $j$ appear in relevant coordinates in the conditions of $E$; we have that $\pi p$ and $p$ are compatible since if $x$ is in the shared domains then $i$ and $j$ do not appear in relevant positions and so the value given to $x$ by the two conditions is similar. Lastly $\pi p\Vdash (\pi\dot f)(\pi\check\alpha)=\pi(G\_i)$, since $\pi$ fixes $\dot f$ and $\check\alpha$ we have: $$\pi p\Vdash\dot f(\check\alpha)=\dot G\_{\pi i}$$


Thus we arrive the wanted contradiction since $\pi p\cup p$ extends $p\_0$, but it contradicts $p\_0$ by forcing that $\dot f$ is not a function.
4. $N\models\kappa\leq^\ast H$:


We define from $f:H\to\kappa$ by $f(G\_i) = \min\{\beta\mid\beta\notin G\_i\}$. Since $G\_i\in N$ for all $i$ and $H\in N$ we have that indeed $f\in N$. Now we only have to show that it is onto $\kappa$, this is a simple genericity argument:


Given $\beta<\kappa$, and a condition $p\in P$ there exists $i<\kappa$ such that $\langle i,\alpha\rangle\notin\Dom(p)$ for all $\alpha$. Therefore $p\cup\lbrace\langle i,\alpha,1\rangle\mid\alpha<\beta\rbrace\cup\lbrace\langle i,\beta,0\rangle\rbrace\le p$ as wanted. By genericity there is some $i<\kappa$ such that $f(G\_i)=\beta$.




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In this model $N$ described above we have $DC$ therefore we have no infinite Dedekind finite sets, however we have a non-well orderable set which can be mapped onto an ordinal (possibly) much higher than its Hartog number. I suspect that this easily generalized to higher cardinalities as well.


