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One dollar short.

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edited Jun 18 2012 at 23:51

Asaf Karagila
7,448●1●15●44

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:

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

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.
$N\models\Htg(H)=\aleph_1$:

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.
$N\models\kappa\leq^\ast H$:

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 $f(G_i)=\beta$.

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.

Corrected some notation, and made explicit why G_i and H are in N.

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edited Jun 18 2012 at 23:41

Asaf Karagila
7,448●1●15●44

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:

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

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.
$N\models\Htg(H)=\aleph_1$:

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 $\pi p\perp p$ 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.
$N\models\kappa\leq^\ast H$:

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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answered Jan 9 2012 at 0:53

Asaf Karagila
7,448●1●15●44

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 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 $\pi\dot G_i = \dot G_{\pi i}$.

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

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

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.
$N\models\Htg(H)=\aleph_1$:

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 $\pi p\perp p$ 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.
$N\models\kappa\leq^\ast H$:

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 $f(G_i)=\beta$.