# Embedding property of weakly compact cardinals

One of the characterizations of $\kappa$ being a weakly compact cardinal is being inaccessible, and for every $\kappa$-model $M$, there is a [$\kappa$-model] $N$ and an elementary embedding $j\colon M\to N$ with critical point $\kappa$.

I am looking for a reference, or at least a proof, that this is equivalent to another property of weakly compact cardinals (which is not an embedding characterizations), for example the tree property or indescribability, or even the extension property.

However all the papers and books refer to one of two places:

1. Kanamori,
2. Cummings' Handbook article (section 16).

A diligent search through Kanamori produced nothing (the second edition, anyway). In Cummings' paper he refers to Kanamori, and only proves an implication from the embedding property to the Hauser embedding property (that we can also assume $j,M\in N$).

Does anyone knows where to find a proof for any equivalence of weak compactness? (Or if it is really folklore, what is the proof?)

It should be easy to prove some characterizations from the embedding property, but I do not quite see how to prove the embedding property from something like compactness, the tree property, indescribability, or end-extensions.

(I should also say that I scoured other standard books like Jech and the Handbook, as well newer books like Schindler's Set Theory. To non-avail.)

• In the paper Indescribable cardinals and elementary embeddings'' an equivalence of the type you asked is proved for $\Pi^m_n$-indescribable cardinals. – Mohammad Golshani Apr 26 '16 at 15:42
• Also the book Set theory an introduction to large cardinals'' by Drake gives the equivalence of weak compactness and $\Pi^1_1$-indescribability. Putting these results together gives you the required result. – Mohammad Golshani Apr 26 '16 at 15:42
• Thanks, Mohammad. But is there a direct proof anywhere? – Asaf Karagila Apr 26 '16 at 16:06
• Wait, what is the "obvious" generalization above $\Pi^1_1$? – Asaf Karagila Apr 26 '16 at 16:53
• We require the model $N$ to have one more property which is called $\Sigma^m_{n-1}$-correct. – Mohammad Golshani Apr 27 '16 at 3:26

This is a selection on from my lecture notes text, Lectures on Forcing and Large Cardinals, which I wrote long ago and which shows the main equivalences, including the ones you mention.

Theorem. If $\kappa^{<\kappa}=\kappa$, then the following are equivalent.

1. (weak compactness property) $\kappa$ is weakly compact. That is, $\kappa$ is uncountable and every $\kappa$-satisfiable theory in an $L_{\kappa,\kappa}$ language of size at most $\kappa$ is satisfiable.

2. (extension property) For every $A\newcommand\of{\subseteq}\of V_\kappa$, there is a transitive structure $W$ properly extending $V_\kappa$ and $A^*\of W$ such that $\langle V_\kappa,{\in},A\rangle\newcommand\elesub{\prec}\elesub\langle W,{\in},A^*\rangle$.

3. (tree property) $\kappa$ is inaccessible and has the tree property.

4. (filter property) If $M$ is a set containing at most $\kappa$-many subsets of $\kappa$, then there is a $\kappa$-complete nonprincipal filter $F$ measuring every set in $M$.

5. (weak embedding property) For every $A\of\kappa$ there is a transitive set $M$ of size $\kappa$ with $\kappa\in M$ and a transitive set $N$ with an embedding $j:M\to N$ with critical point $\kappa$.

6. (embedding property) For every transitive set $M$ of size $\kappa$ with $\kappa\in M$ there is a transitive set $N$ and an embedding $j:M\to N$ with critical point $\kappa$.

7. (normal embedding property) For every $\kappa$-model $M$ there is a $\kappa$-model $N$ and an embedding $j:M\to N$ with critical point $\kappa$, such that $N=\{j(f)(\kappa)\mid f\in M\}$.

8. (Hauser embedding property) For every $\kappa$-model $M$ there is a $\kappa$-model $N$ and an embedding $j:M\to N$ with critical point $\kappa$ such that $j,M\in N$.

9. (partition property) $\kappa\to(\kappa)^2_2$.

Proof: (weak compactness implies extension property) Assume that $L_{\kappa,\kappa}$ exhibits the weak compactness property and suppose $A\of V_\kappa$. First, we argue that $\kappa$ is inaccessible. The regularity of $\kappa$ follows from our assumption that $\newcommand\ltkappa{{{<}\kappa}}\kappa^\ltkappa=\kappa$ (although even without that assumption, it follows from the weak compactness property). Suppose now that $2^\beta\geq\kappa$ for some $\beta<\kappa$. Let $L$ be the language having a constant symbol $\check\alpha$ for every $\alpha<\kappa$ and a unary predicate symbol $U$. For any $x\of\beta$, let $\sigma_x$ be the sentence $(\bigvee_{\alpha\in x}\neg U(\check\alpha))\vee(\bigvee_{\alpha\in\beta\setminus x} U(\check\alpha))$. Thus, $\sigma_x$ asserts that $U$ is different from $x$ on the set $\{\check\alpha\mid \alpha<\beta\}$. Let $S$ be the theory consisting of all $Los\sigma_x$ for $x\of\beta$. This theory has size $2^\beta$, but the language of $S$ has size only $\beta$. Since $2^\beta\geq\kappa$, the theory $S$ is $\kappa$-satisfiable, since any subtheory of size less than $\kappa$ will omit some $\sigma_x$, and then we can interpret $\check\alpha$ as $\alpha$ and $U$ as $x$ to build a model. But clearly $S$ is not satisfiable, since $U$ must pick out some pattern $x=\{\alpha<\beta\mid U(\check x)\}$ under any interpretation. So $\kappa$ is inaccessible.

Next, we show that $\kappa$ has the extension property. Let $L$ be the language with a constant symbol $\check a$ for every $a\in V_\kappa$, as well as a binary relation symbol $\check\in$, an additional constant symbol $c$ and a predicate symbol $\dot A$. Let $R$ be the first order theory $\text{Th}(\langle V_\kappa,{\in},A,a\rangle_{a\in V_\kappa})\cup\{c\neq\check a\mid a\in V_\kappa\}$, together with the infinitary assertion $\sigma=\neg\exists\vec x(\wedge x_{n+1}\in x_n)$, which asserts that $\check\in$ is well founded. This theory is $\kappa$-satisfiable, by interpreting $c$ in the structure $\langle V_\kappa,{\in},A\rangle$ to be one of the $\check a$ missing from the subtheory. By the weak compactness property, there is a model $\langle W,{\in^*},A^*\rangle$ satisfying $R$. The relation $\in^*$ on $W$ must be well founded, since the structure satisfies $\sigma$. By taking the Mostowski collapse, we may assume that $W$ is a transitive set and $\in^*$ is the $\in$ relation. Furthermore, since the first order theory of $\langle V_\kappa,\in,A\rangle$ is satisfied, it follows that $V_\kappa\of W$ and $\langle V_\kappa,{\in},A\rangle\elesub \langle W,{\in},A^*\rangle$, as desired.

(extension implies tree property) Assume $\kappa$ has the extension property. We show first that $\kappa$ is inaccessible. Regularity follows from $\kappa^\ltkappa=\kappa$ (but it also follows directly from the extension property). If $2^\beta\geq\kappa$ for some $\beta<\kappa$, then let $\vec a=\langle a_\alpha\mid \alpha<\kappa\rangle$ be a $\kappa$-sequence of distinct subsets of $\beta$. By the Extension property, there is a transitive set $W$ and $\vec a^*$ such that $\langle V_\kappa,{\in},\vec a\rangle\elesub\langle W,{\in},\vec a^*\rangle$. Let $\kappa^*=W\newcommand\intersect{\cap}\intersect\newcommand\ORD{\text{Ord}}\ORD$, and observe that $\kappa<\kappa^*$. Let $a$ be the $\kappa^{th}$ element of the sequence $\vec a^*$. Thus, $a\of\beta$ and consequently $a\in V_\kappa$. Since $W$ satisfies that $a$ appears on the sequence $\vec a^*$, it follows by elementarity that $a$ also appears on $\vec a$. Thus, $a=a_\alpha$ for some $\alpha<\kappa$, and so $a$ appears twice on $\vec a^*$, at coordinates $\alpha$ and $\kappa$, contradicting that these were distinct subsets of $\kappa$. So we have established that $\kappa$ is inaccessible.

Now suppose that $T$ is a $\kappa$-tree. By replacing $T$ with an isomorphic copy, if necessary, we may assume $T\of\kappa^\ltkappa\of V_\kappa$. By the Extension property, there is a transitive set $W$ and a subset $T^*\of W$ with $\langle V_\kappa,{\in},T\rangle\elesub\langle W,{\in},T^*\rangle$. It follows that $T^*$ is a tree of height $\kappa^*=W\intersect\ORD$. Furthermore, for any $\alpha<\kappa$, the $\alpha^{th}$ level of $T$, which is an element of $V_\kappa$, is by elementarity also the $\alpha^{th}$ level of $T^*$. Thus, $T^*$ is an end-extension of $T$. Let $q\in T^*$ be any node on the $\kappa^{th}$ level of $T^*$, and consider the set of predecessors $b=\{p\in T^*\mid p<_{T^*} q\}$. The elements of $b$ form a linearly ordered subset of $T^*$ on the levels below $\kappa$. Thus, $b$ is a $\kappa$-branch through $T$. So $\kappa$ has the tree property.

(tree property implies filter property) Suppose that $\kappa$ is inaccessible and has the tree property. Suppose that $\{A_\alpha\mid \alpha<\kappa\}$ is a collection of $\kappa$ many subsets of $\kappa$. By enlarging the collection if necessary, let us assume that all singletons $\{\xi\}$, for $\xi<\kappa$, appear on the list. For each $s\in 2^\beta$, where $\beta<\kappa$, let $A_s$ be the result of intersecting all $A_\alpha$ or the complement $\kappa\setminus A_\alpha$, chosen according to the values of $s(\alpha)$. That is, $A_s=(\newcommand\Intersect{\bigcap}\intersect\{A_\alpha\mid s(\alpha)=1\})\intersect(\Intersect\{\kappa\setminus A_\alpha\mid s(\alpha)=0\})$. Let $T=\{s\in 2^\ltkappa\mid |A_s|=\kappa\}$. This is clearly a tree. For any $\beta<\kappa$, we may define for any $\gamma<\kappa$ a sequence $s_\gamma\in 2^\beta$ so that $s_\gamma(\alpha)=1\iff\gamma\in A_\alpha$ for all $\alpha<\beta$. In particular, $\gamma\in A_{s_\gamma}$, and so $\kappa=\newcommand\Union{\bigcup}\Union\{A_s\mid s\in 2^\beta\}$. Since $2^\beta<\kappa$, it must be that some $A_s$ has size $\kappa$, and so $T$ has nodes on the $\beta^{th}$ level. Thus, $T$ is a $\kappa$-tree. By the tree property, there is a $\kappa$-branch $b\in[T]$. Let $F$ be the filter generated by the $A_{b\newcommand\restrict{\upharpoonright}\restrict\beta}$ for $\beta<\kappa$. Thus, $X\in F$ if and only if $A_{b\restrict\beta}\of X$ for some $\beta<\kappa$. Since $b$ cannot choose to add a singleton $\{\xi\}$, since this does not have size $\kappa$, it must choose the complement, and so the filter $F$ is not principal. Since the sequence of $A_{b\restrict\beta}$ is descending and $\kappa$ is regular, the filter $F$ is $\kappa$-complete. And since either $A_\alpha\in F$ or $\kappa\setminus A_\alpha\in F$ explicitly at stage $\alpha$, depending on whether $b(\alpha)=1$ or $0$, the filter $F$ decides every set in our original family. So $\kappa$ has the filter property.

(filter property implies weak embedding property) Assume the filter property and suppose $A\of\kappa$. By the Löwenheim-Skolem theorem and using $\kappa^\ltkappa=\kappa$, there is a transitive set $M\elesub H_{\kappa^+}$ with $A\in M$ and $M^\ltkappa\of M$. By the filter property, there is a $\kappa$-complete nonprincipal filter $F$ deciding every element of $P(\kappa)^M$. Consider the ultrapower $M^\kappa/F$, where we use only functions $f:\kappa\to M$ with $f\in M$. The relations $f=_F g$ and $f\in_F g$ are still equivalence relations, even though $F$ may not be an ultrafilter, because the corresponding sets $\{\alpha\mid f(\alpha)=g(\alpha)\}$ and $\{\alpha\mid f(\alpha)\in g(\alpha)\}$ are in $M$ if $f$ and $g$ are. Since $F$ is $\kappa$-complete and $M$ is closed under $\omega$-sequences, it follows that $\in_F$ is well founded. Thus, the ultrapower $N=\newcommand\Ult{\text{Ult}}\Ult(M,F)$ is a transitive set. The proof that the canonical embedding $j:M\to N$ is elementary proceeds as in Łos' Theorem, by induction on the complexity of the formula, appealing to the Axiom of Choice in $M$ in the existential case. Similarly, using $\kappa$-completeness, one shows for $\alpha<\kappa$ that if $[f]_F\in[c_\alpha]_F$, then $f=_F c_\beta$ for some $\beta<\alpha$, and consequently $j(\alpha)=\alpha$; meanwhile, $\kappa\leq [id]_F<j(\kappa)$, so the critical point of $j$ is $\kappa$. So $j:M\to N$ is as desired.

(weak embedding implies embedding property) Assume the weak embedding property and suppose $M$ is a transitive set of size $\kappa$. Since $M\in H_{\kappa^+}$, we may code $M$ with a set $A\of\kappa$, and then find a transitive set $\bar M$ with $M\in \bar M$ and an embedding $j:\bar M\to \bar N$ with critical point $\kappa$. The restriction $j\restrict M:M\to j(M)$ is as desired.

(embedding implies normal embedding property) Assume that $\kappa$ has the embedding property, and suppose that $M$ is a $\kappa$-model. By the embedding property there is an embedding $j:M\to N$ with critical point $\kappa$. Let $X=\{j(f)(\kappa)\mid f\in M\}$ be the seed hull of $\kappa$ via $j$. It follows that $X\elesub N$, and so if $\pi:X\cong N_0$ is the Mostowski collapse, we obtain the induced factor embedding, where $k=\pi^{-1}$ and $j_0=\pi\circ j$.

Note that $N_0=\{\pi(j(f)(\kappa))\mid f\in M\}=\{j_0(f)(\kappa)\mid f\in M\}$. If $\vec x=\langle x_\alpha\mid \alpha<\beta\rangle\in N_0^\ltkappa$, then there are functions $f_\alpha\in M$ with $x_\alpha=j_0(f_\alpha)(\kappa)$. Since $M$ is a $\kappa$-model, it follows that $\langle f_\alpha\mid \alpha<\beta\rangle\in M$. Thus, $\langle j_0(f_\alpha)\mid \alpha<\beta\rangle\in N_0$, and so by evaluation these functions at $\kappa$, it follows that $\vec x=\langle j_0(f)(\kappa)\mid \alpha<\beta\rangle\in N_0$, and so $N_0$ is a $\kappa$-model as desired.

(normal embedding implies Hauser embedding property) Suppose that $\kappa$ has the normal embedding property and suppose that $M$ is a $\kappa$-model. Since $M$ has size $\kappa$, there is a relation $E$ on $\kappa$ such that $\langle M,{\in}\rangle\cong\langle \kappa,E\rangle$. Thus, $M$ must be the Mostowski collapse of the relation $E$. By the Löwenheim-Skolem theorem theorem, there is a $\kappa$-model $\bar M$ with $E\in \bar M$. By the normal embedding property there is an embedding $j:\bar M\to \bar N$ with critical point $\kappa$. Since $j(E)\intersect\kappa\times\kappa=E$, it follows that $E\in\bar N$, and since the Mostowski collapse of $E$ is unique, that $M$ is in both $\bar M$ and $\bar N$. By elementarity, if $x\in M$ is coded by $\alpha$ with respect to $E$, then $j(x)$ is coded by $j(\alpha)=\alpha$ with respect to $j(E)$. The model $N$ can therefore reconstruct $j\restrict M$ from $E$ and $j(E)$. Since $\bar M$ can see that $M^\ltkappa\of M$, it follows that $\bar N\models j(M)^{< j(\kappa)}\of j(M)$. In particular, $M$ and $j\restrict M$, which have size $\kappa$ in $N$, must be in $j(M)$. Thus, $j\restrict M:M\to j(M)$ has the desired Hauser embedding property.

(embedding properties all equivalent) The Hauser embedding property immediately implies the weak embedding property, since any $A\of\kappa$ can be placed into a $\kappa$-model $M$.

(embedding implies weak compactness property) Assume that $\kappa$ has the embedding properties. We show first that $\kappa$ is inaccessible. Regularity follows from $\kappa^\ltkappa=\kappa$ (although it also follows from the weak embedding property). If $2^\beta\geq\kappa$ for some $\beta<\kappa$, then there is a $\kappa$-sequence $\vec a=\langle a_\alpha\mid \alpha<\kappa\rangle$ of distinct subsets of $\beta$. We may code $\vec a$ with a subset of $\kappa$, and therefore find a $\kappa$-model $M$ and an embedding $j:M\to N$ with $\vec a\in M$. If $a=j(\vec a)(\kappa)$ be the $\kappa^{th}$ element of $j(\vec a)$, then by $M^\ltkappa\of M$, it follows that $a\in M$. Since $a\of\beta<\kappa$, it follows that $j(a)=a$. Since $j(a)$ appears on $j(\vec a)$, it follows by elementarity that $a$ appears on $\vec a$. So $a=a_\alpha$ for some $\alpha<\kappa$. But since $j(\vec a)(\alpha)=j(a_\alpha)=a_\alpha$, this means that $a$ appears at least twice on the sequence $j(\vec a)$, at coordinates $\alpha$ and $\kappa$, contradicting the assumption that the sets were distinct. So $\kappa$ is inaccessible.

Lastly, we verify the weak compactness property. Suppose that $T$ is a $\kappa$-satisfiable theory in an $L_{\kappa,\kappa}$ language $L$ of size at most $\kappa$. Since $\kappa$ is inaccessible, it follows that $T$ has size at most $\kappa$. We may assume that the symbols of $L$ are built from ordinals below $\kappa$, and so $T$ may be coded with a subset of $\kappa$. Thus, by the $\kappa$-model embedding property, there is a $\kappa$-model $M$ with $T,L\in M$ and an embedding $j:M\to N$ with critical point $\kappa$ into a transitive set $N$. Since $V_\kappa\of M$, the model $M$ has all the subtheories of $T$ of size less than $\kappa$, and the models that exist for them (for this conclusion, one needs an infinitary version of the upward Löwenheim-Skolem theorem, which can be proved by an infinitary analogue of the classical proof). Thus, $M$ satisfies that $T$ is $\kappa$-satisfiable. By elementarity, it follows that $N$ satisfies that $j(T)$ is $j(\kappa)$-satisfiable. But $j$ fixes every element of $T$ individually, so $T=j(T)\intersect V_\kappa\in N$, and so $T$ is a subtheory of $j(T)$ of size less than $j(\kappa)$. Thus, $N$ has a model $\cal A$ of $T$. Furthermore, the satisfaction relation is absolute to transitive sets, so $\cal A$ really is a model of $T$, as desired.

(embedding implies partition property) Most other accounts derive the partition property from the tree property, but the embedding point of view, ever efficacious, seems to simplify things. Assume that $\kappa$ has the embedding properties and suppose $F:[\kappa]^2\to 2$. Since $F\in H_{\kappa^+}$, we may find a $\kappa$-model $M_0$ with $F\in M_0$. By the normal embedding property there is a $\kappa$-model $N_0$ and an embedding $j_0:M_0\to N_0$ with critical point $\kappa$. Since the entire embedding $j_0:M_0\to N_0$ has hereditary size $\kappa$, we may find a $\kappa$-model $M$ with $M_0$, $N_0$ and $j_0$ all having size $\kappa$ in $M$. By the Hauser embedding property there is an embedding $j:M\to N$ such that $j,M\in N$. Applying $j$ to $j_0:M_0\to N_0$ yields an embedding $h=j(j_0):j(M_0)\to j(N_0)$ and a factor diagram. Let $j^*=h\circ j\restrict M_0$ be the composition of the diagonal. This diagram commutes because $j_0(x)=y$ implies $j(j_0)(j(x))=j(y)$. Let $\mu_0=\{X\of\kappa\mid X\in M_0\And\kappa\in j_0(X)\}$ be the $M_0$-normal filter induced by $j_0$. Since $\mu_0$ has size $\kappa$ in $M$, it follows from $P(\kappa)^M\of N$ that $\mu_0\in N$. Using $j\in N$, it follows also that $j\newcommand\image{''}\image\mu_0\in N$. Since $\mu_0\of M_0$, it follows that $j\image\mu_0\of j(M_0)$. Since $M$ knows that $M_0^\ltkappa\of M_0$, it follows that $N\models j(M_0)^\kappa\of j(M_0)$, and consequently $j\image\mu_0\in j(M_0)$. This is a collection of $\kappa$ many elements of $j(\mu_0)$ in $j(M_0)$, and so by the $j(\kappa)$-completeness of $j(\mu_0)$ over $j(M_0)$, it follows that $I=\Intersect j\image\mu_0\in j(\mu_0)$. Fix any $\delta_0\in I$ and observe that $X\in\mu_0\iff\delta_0\in j(X)$ for $X\in M_0$. Let $\delta_1=j(\kappa)$, and apply $j$ to the definition of $\mu_0$ to see that $Y\in j(\mu_0)\iff \delta_1\in h(Y)$ for all $Y\in j(M_0)$.

Suppose $j^*(F)(\delta_0,\delta_1)=i$. We will build an increasing monochromatic sequence $\langle \gamma_\alpha\mid \alpha<\kappa\rangle$ such that (i) $\alpha<\alpha'\implies F(\gamma_\alpha,\gamma_{\alpha'})=i$; and (ii) $\{\xi\mid F(\gamma_\alpha,\xi)=i\}\in\mu_0$ for every $\alpha<\kappa$. Suppose that $\langle \gamma_\alpha\mid \alpha<\beta\rangle$ is defined and we would like to define $\gamma_\beta$. Consider the set of possible values for $\gamma_\beta$, namely $X=\{\gamma<\kappa\mid \forall\alpha<\beta\,\,F(\gamma_\alpha,\gamma)=i\text{ and }\{\xi\mid F(\gamma,\xi)=i\}\in\mu_0\}$. It suffices to show that $X$ is not empty. By (ii) for any $\alpha<\beta$, we know that $j(F)(\gamma_\alpha,\delta_0)=i$, since $\delta_0$ is a seed for $\mu_0$ via $j$. Similarly, $\{\xi\mid j(F)(\delta_0,\xi)=i\}\in j(\mu_0)$, since $\delta_1$ is in $h$ of this set, in light of $j^*(F)(\delta_0,\delta_1)=i$. Thus, $\delta_0\in j(X)$. It follows that $X$ is unbounded in $\kappa$, and so we may choose $\gamma_\beta\in X$ and continue the construction. So $F$ admits a monochromatic set of size $\kappa$.

(partition implies tree property) Assume that $\kappa\to(\kappa)^2_2$. We first show that $\kappa$ is inaccessible. Regularity follows from $\kappa^\ltkappa=\kappa$ (but even without this hypothesis, it follows from the partition property). If $2^\beta\geq\kappa$ for some $\beta<\kappa$, then let $\langle a_\alpha\mid \alpha<\kappa\rangle$ be a $\kappa$-sequence of distinct subsets of $\beta$. Let $F(\alpha,\beta)=0$ if $a_\alpha$ precedes $a_\beta$ in the lexical ordering of $2^\beta$, and otherwise $1$. It is not difficult to prove that there is no monotone sequence of length $\beta^+$ in the lexical order on $2^\beta$. Consequently, there can be no monochromatic set for $F$ of size $\kappa$. So $\kappa$ is inaccessible.

Now suppose that $T$ is a $\kappa$-tree. We may assume that the underlying set of $T$ is exactly $\kappa$. For any two nodes $\alpha$ and $\beta$ in $T$, let us say that $\beta$ is to the {\df right} of $\alpha$ in $T$, if $\alpha\perp_T\beta$ and $\alpha'<\beta'$, where $\alpha'\leq_T\alpha$ and $\beta'\leq_T\beta$ are least in $T$ such that $\alpha'\perp_T\beta'$. Define $F(\alpha,\beta)=0$ if $\beta$ is above or to the right of $\alpha$, and otherwise $1$. Suppose that $H\of\kappa$ is a monochromatic set of size $\kappa$. Suppose first that the monochromatic value is $0$. Thus, whenever $\alpha<\beta$ in $H$, then $\beta$ is above or to the right of $\alpha$ in $T$. By applying the partition property once more, we may further assume that only one of these answers arises. If any $\alpha<\beta$ in $H$ has $\alpha<_T\beta$, then the elements of $H$ are linearly ordered, and $T$ has a $\kappa$-branch. Otherwise, we assume that whenever $\alpha<\beta$ in $H$, then $\beta$ is to the right of $\alpha$ in $T$. We may therefore follow the right-most'' path through (the $T$-predecessors of elements of) $H$. Specifically, using the fact that the levels of $T$ have size less than $\kappa$ and $\kappa$ is regular, there is on each level $\xi$ a right-most node $\zeta$ that occurs as a $T$-predecessor of all sufficiently large elements of $H$. The set of such nodes is clearly linearly ordered, and therefore forms a $\kappa$-branch through $T$. The final case, when the monochromatic value of $F$ on $H$ is $1$, is similar, except that we follow the left-most branch through $H$ to provide a $\kappa$-branch through $T$. QED

I believe that it was in these lecture notes that the terms $\kappa$-model and the "Hauser property" were introduced, since I remember inventing that terminology. (I knew Kai Hauser from the time I was an undergraduate at Caltech, where he was a graduate student.)

Finally, let me mention that Sean Cox, Brent Cody, Thomas Johnstone and myself have a paper in preparation called, "The weakly compact embedding property," which is all about the property when you drop the inaccessibility requirement. There are some very interesting things to say, and I'l post a link when the article is available. But meanwhile, I refer you to the slides for my talk The weakly compact embedding property.

• I really think this answer was possibly the most detailed that it could have been; how was this not accepted? – Keith Millar Oct 21 '17 at 23:56

Let me take the following as a definition of weakly compact cardinal and show its equivalence to the embedding property:

Definition. $\kappa$ is weakly compact if for any collection $X$ of $\kappa$-many subsets of $\kappa,$ there exists a non-principal $\kappa$-complete filter on $\kappa$ deciding every element of $X$.

Theorem. The following are equivalent:

(1) $\kappa$ is weakly compact.

(2) $\kappa$ has the embedding property described in the question.

Proof. $(1) \to (2).$ Take a $\kappa$-model $M$. Let $X=P(\kappa) \cap M.$ Let $F$ be the filter produced by $(1)$ which decides every element of $X$. Take $M^\kappa/F.$ Note that this is well-founded (as $M$ is closed under $\omega$-sequences and $F$ is countably complete), so let $N$ be its transitive collapse and consider the resulting $j: M \to N.$ This witnesses $(2)$.

$(2) \to (1)$. Assume $X$ is given. Take $\kappa$-model $M$ with $X \in M, X \subset M.$ Let $j: M \to N$ witness $(2).$ Define $F$ by $F= \{ A \subset \kappa: A \in M, \kappa \in j(A) \}$. Then $F$ witnesses $(1)$ with respect to $X$.

Let me directly show $\Pi^1_1$-indescribability implies the embedding property (following Hauser's proof). Thus let $\kappa$ be $\Pi^1_1$-indescribable but assume it does not satisfy the embedding property as witnessed by the $\kappa$-model $M$. Let $E \subset \kappa \times \kappa$ code $(M, \in)$ such that if $\pi$ is the transitive collapse map, then $\pi(0)=\kappa.$ Let $F=\pi \restriction \kappa$ and $T=\{(n, \bar{\xi}): n$ is a (code of a) first order formula that holds in $(\kappa, E)$ under the assignment $\bar{\xi} \}$.

Let $\Phi(\kappa, E, F, T)$ be the formula:

$\Phi(\kappa, E, F, T)$ $(\kappa, E)$ is well founded and extensional $F=\pi \restriction \kappa$, $\pi(0)=\kappa$ and $T$ is the first order theory of $(\kappa, E)$.

Consider the $\Pi^1_1$-sentence which is satisfied over $V_\kappa:$

$\forall M [M, \kappa$-model of size $\kappa$ and $\kappa, E, F ,T \in M \implies$ $M \models \Phi(\kappa, E, F, T)$ but there is no $N, j$ with $|N|=\kappa$ and $j$ from transitive collapse of $(\kappa, E)$ into $N$ with critical point $\kappa].$

By $\Pi^1_1$-indescribability, there is inaccessible $\lambda < \kappa$ such that $(\lambda, E \cap \lambda \times \lambda)$ is well founded and extensional, $F \cap \lambda \times \lambda=$ (transitive collapse map)$^{-1} \restriction \lambda$ which sends $0$ to $\lambda$, and $T \cap \omega \times \lambda^{<\omega}$ is the theory of $(\lambda, E \cap \lambda \times \lambda)$.

Let $(M^*, \in)$ be the transitive collapse of $(\lambda, E \cap \lambda \times \lambda)$. Then we have $j^*: M^* \to M$ with $crit(j^*)=\lambda$ and $j^*(\lambda)=\kappa.$ Let $X \prec M$ with $X$ a $\lambda$-model and $j^*[M^*] \cup \{ \lambda\} \subset X$. Let $k: X \to N$ be the transitive collapse map and $j=k \circ j^*: M^* \to N.$ But then $j$ and $N$ witness that the above sentence fails in $V_\lambda$ a contradiction.

• That's helpful, but actually that is probably the least helpful equivalent here. But it is helpful! – Asaf Karagila Apr 27 '16 at 5:06
• Your welcome. Maybe the best thing is to look at Hauser paper I mentioned above and simplify his argument for the case of weakly compact cardinals. I gave the above as it was the simplest way (for me) to give such a characterization. – Mohammad Golshani Apr 27 '16 at 5:07
• Yes, that is probably what's going to happen later tonight. I've just arrived after two weeks away, and a lot of things need attention. – Asaf Karagila Apr 27 '16 at 5:09
• Another natural question that can be added is who (and where) proved the equivalence between weak compactness and embedding property. I think the extension property is due to Keisler. – Mohammad Golshani Apr 27 '16 at 5:11
• I referred specifically to compactness, and three other definitions, in my edit. – Asaf Karagila Apr 27 '16 at 5:15