Proof. $(\Leftarrow ):$ By the Kunen inconsistency, $\lambda $ must be the supremum of the critical sequence $\langle \kappa _n \colon n \in \omega \rangle $ of j, where $\kappa _0$ is the critical point and $\kappa _{n + 1} = j(\kappa _n)$ . Then each $\kappa _n$ is an inaccessible cardinal and thus $2^{< \lambda } = \lambda = \beth _\lambda $ . Therefore and so . This means that , so is a definable class in . Moreover, working in , any formula $\varphi $ can be relativised to so is elementary.

$(\Rightarrow ):$ We begin by defining a standard way to code elements of by elements of . This will be done by coding ${\textrm {trcl}}(\{x\})$ by some subset of $\lambda \times \lambda $ whose Mostowski collapse is again ${\textrm {trcl}}(\{x\})$ . However, since we will be working with ${\textrm {trcl}}(\{x\})$ rather than x itself, it is necessary to do a simple, preliminary coding.

So, let . For define the relation $\operatorname {\mathrm {\hat {\in }}}$ by ${\textrm {trcl}}(\{x\}) \mathbin{\operatorname {\mathrm {\hat {\in }}}} {\textrm {trcl}}(\{y\})$ if and only if $x \in y$ and similarly for $\operatorname {\mathrm {\hat {=}}}$ . It is then clear that any first order statement $\varphi $ about is equivalent to a formula $\hat {\varphi }$ over by the obvious coding.

Now note that ${\textrm {rank}}(\lambda \times \lambda ) = \lambda $ and so any subset of $\lambda \times \lambda $ has rank at most $\lambda $ . So, for any

and bijection $f \colon |{\textrm {trcl}}(\{x\})| \rightarrow {\textrm {trcl}}(\{x\})$ let

Then the Mostowski collapse of $C_{x,f}$ , ${\textrm {coll}}(C_{x,f})$ , is ${\textrm {trcl}}(\{x\})$ . Let $\tilde {H}$ denote the definable class in

of all subsets of $\lambda \times \lambda $ which code an element of

in this way. That is $X \in \tilde {H}$ iff X is a well-founded, extensional, binary relation on $\lambda $ with a single maximal element and ${\textrm {dom}}(X) \cup {\textrm {ran}}(X)$ is a cardinal which is at most $\lambda $ .

For $Z \in \tilde {H}$ , let ${\textrm {fld}}(Z)$ be ${\textrm {dom}}(Z) \cup {\textrm {ran}}(Z)$ and define ${\textrm {max}}(Z)$ to be the unique element of ${\textrm {fld}}(Z)$ which is maximal with respect to the relation on Z. Now, for $X, Y \in \tilde {H}$ define relations $\operatorname {\mathrm {\tilde {=}}}$ and $\operatorname {\mathrm {\tilde {\in }}}$ by:

$$ \begin{align*} X \mathbin{\operatorname{\mathrm{\tilde{=}}}} Y \Longleftrightarrow \exists g \colon \lambda \rightarrow \lambda \ &\bigg( g \textit{ is a bijection} \wedge \forall \alpha, \beta \in \lambda \\ &\qquad\qquad\qquad\qquad\big( \langle \alpha, \beta \rangle \in X \leftrightarrow \langle g(\alpha), g(\beta) \rangle \in Y \big) \bigg), \end{align*} $$

$$ \begin{align*} X \mathbin{\operatorname{\mathrm{\tilde{\in}}}} Y \Longleftrightarrow \exists g \colon \lambda & \rightarrow \lambda \ \bigg( g \textit{ is injective} \wedge \langle g({\textrm{max}}(X)), {\textrm{max}}(Y) \rangle \in Y \\ & \wedge \forall \alpha, \beta \in {\textrm{fld}}(X) \ \big( \langle \alpha, \beta \rangle \in X \leftrightarrow \langle g(\alpha), g(\beta) \rangle \in Y \big) \\ & \wedge \forall \gamma \in {\textrm{fld}}(Y) \ \forall \delta \in {\textrm{fld}}(X) \\ & \phantom{\wedge \forall \gamma \in {\textrm{fld}}} \big( \langle \gamma, g(\delta) \rangle \in Y \rightarrow \exists \sigma \in {\textrm{fld}}(X) \ \sigma = g(\gamma) \big) \bigg).^{1} \end{align*} $$

Then $\operatorname {\mathrm {\tilde {=}}}$ and $\operatorname {\mathrm {\tilde {\in }}}$ are definable in , $X \mathbin{\operatorname {\mathrm {\tilde {=}}}} Y \Leftrightarrow {\textrm {coll}}(X) = {\textrm {coll}}(Y)$ and $X \mathbin{\operatorname {\mathrm {\tilde {\in }}}} Y \Leftrightarrow {\textrm {coll}}(X) \in {\textrm {coll}}(Y)$ . Now we have that any first order statement $\hat {\varphi }$ about is equivalent to a formula $\tilde {\varphi }$ over which is defined by the following coding:

• • Replace any parameter ${\textrm {trcl}}(\{x\})$ occurring in $\hat {\varphi }$ with $C_{x, f}$ for some (any) bijection $f \colon |{\textrm {trcl}}(\{x\})| \rightarrow {\textrm {trcl}}(\{x\})$ .

• • Replace any instance of $\operatorname {\mathrm {\hat {=}}}$ with $\operatorname {\mathrm {\tilde {=}}}$ and $\operatorname {\mathrm {\hat {\in }}}$ with $\operatorname {\mathrm {\tilde {\in }}}$ .

• • Replace any unbounded quantification by the same quantifier taken over $\tilde {H}$ .

Then, by the elementarity of k,

$$ \begin{align*} X \mathbin{\operatorname{\mathrm{\tilde{=}}}} Y & \Longleftrightarrow k(X) \mathbin{\operatorname{\mathrm{\tilde{=}}}} k(Y) \\ & \textrm{ and } \\ X \mathbin{\operatorname{\mathrm{\tilde{\in}}}} Y & \Longleftrightarrow k(X) \mathbin{\operatorname{\mathrm{\tilde{\in}}}} k(Y). \end{align*} $$

Also, since is a definable class in the restriction of the embedding is still elementary.

So we can define

by setting $j(x)$ to be the unique element of ${\textrm {coll}}(k(C_{x,f}))$ of maximal rank for some bijection $f \colon | {\textrm {trcl}}(\{x\})| \rightarrow {\textrm {trcl}}(\{x\})$ . Moreover j is elementary since

⊣

Proof. First note that $\Sigma _0$ -elementarity implies $\Delta _1$ -elementarity. Now, since being an ordinal is $\Sigma _0$ definable, if $\alpha $ is an ordinal then so is $j(\alpha )$ . Next, since $\emptyset $ is definable as the unique set z such that $\forall y \in z \ (y \neq y)$ , which is a $\Sigma _0$ formula, $j(\emptyset ) = \emptyset $ . So, by induction, we have that for every ordinal $\alpha $ , $j(\alpha ) \geq \alpha $ . Now let x be a set of least rank such that $j(x) \neq x$ and let $\delta = {\textrm {rank}}(x)$ . Then for all $y \in x$ , $y = j(y) \in j(x)$ so $x \subseteq j(x)$ . Thus there must be some $z \in j(x) \setminus x$ . Now suppose that ${\textrm {rank}}(j(x)) = \delta $ , then we must have that $j(z) = z \in j(x)$ so, by elementarity, $z \in x$ which yields a contradiction. Hence, since the following is $\Delta _1$ definable, we must have that $j(\delta ) = {\textrm {rank}}(j(x))> \delta $ .⊣

Proof. Formally, this is a theorem scheme asserting that for each formula $\varphi $ there is no parameter p for which $\varphi (\cdot , \cdot , p)$ defines a non-trivial, cofinal, elementary embedding

. Using Theorem 3.2, it suffices to show that for no parameter p are we able to define a non-trivial, cofinal, $\Sigma _0$ -elementary embedding

by

$$\begin{align*}j(x) = y \Longleftrightarrow \varphi(x, y, p) \textrm{ holds.} \end{align*}$$

So, seeking a contradiction, let $\sigma (p)$ be the sentence asserting that $\varphi (\cdot , \cdot , p)$ defines a $\Sigma _0$ -elementary embedding and let $\psi (p)$ asserts that $\varphi (\cdot , \cdot , p)$ defines a total function which is non-trivial, cofinal and $\Sigma _0$ -elementary. That is,

$$\begin{align*}\psi(p) \equiv \forall u \exists ! v \varphi(u, v, p) \wedge \exists t \neg \varphi(t, t, p) \wedge \forall y \exists x, z (\varphi(x, z, p) \wedge y \in z) \wedge \sigma(p). \end{align*}$$

Let $\vartheta (p, \kappa )$ assert that $\kappa $ is the critical point of j. So,

Then, by Proposition 3.1,

So denote by $\textrm {crit}_{\textit {p}}$ the (unique) $\kappa $ for which $\vartheta (p, \kappa )$ holds. Now fix p such that $\textrm {crit}_{\textit {p}}$ is as small as possible, that is such that

Then, by elementarity,

But, because the critical point of the embedding defined by $\varphi (\cdot , \cdot , s)$ must be $j(\textrm {crit}_{\textit {p}})$ , yielding a contradiction.⊣

Question 3.5. Are either of the following two statements consistent:

1. 1. There exists a non-trivial elementary embedding which is definable from parameters where ?

2. 2. There exists a non-trivial, cofinal elementary embedding which is definable from parameters where ?

Proof. Assume for a contradiction that for each $\gamma \in \mu $ the set $\{ \alpha \in S \colon f(\alpha ) = \gamma \}$ was non-stationary. Using the above comments, for each $\gamma \in \mu $ choose a club $D_\gamma $ such that for each $\alpha $ in $D_\gamma \cap S$ , $f(\alpha )\neq \gamma $ . Let

$$\begin{align*}D = \Delta_{\gamma \in \mu} D_\gamma \operatorname{\mathrm{:=}} \{ \alpha \colon \forall \beta \in \alpha \ (\alpha \in D_\beta) \}, \end{align*}$$

and note that this is club in $\mu $ . Therefore $S \cap D$ is stationary, so in particular non-empty, and for any $\alpha \in S \cap D$ and $\gamma \in \alpha $ , $f(\alpha ) \neq \gamma $ . So $f(\alpha ) \geq \alpha $ , contradicting the assumption that f was regressive.⊣

Proof. First note that for each $\alpha \in S$ there is some increasing sequence of ordinals $\langle t_n \colon n \in \omega \rangle $ cofinal in $\alpha $ . Therefore by the comments at the beginning of this section, for each $\alpha \in S$ choose an increasing sequence $\langle a^\alpha _n \colon n \in \omega \rangle $ cofinal in $\alpha $ . Then, as in the usual proof, using our first example and the regularity of $\mu $ we can fix some $n \in \omega $ such that for each $\sigma \in \mu $ , $\{ \alpha \in S \colon a^\alpha _n \geq \sigma \}$ is stationary in $\mu $ . Now define a regressive function $f \colon S \rightarrow \mu $ by $f(\alpha ) = a^\alpha _n$ . Using Fodor’s Theorem, for each $\sigma \in \mu $ fix some $S_\sigma $ stationary and $\gamma _\sigma \geq \sigma $ such that for all $\alpha \in S_\sigma $ , $f(\alpha ) = \gamma _\sigma $ . Then if $\gamma _\sigma \neq \gamma _{\sigma '}$ , $S_\sigma \cap S_{\sigma '} = \emptyset $ and, by the regularity of $\mu $ , $| \{ S_\sigma \colon \sigma \in \mu \} | = \mu ,$ which gives the required partition.⊣

Proof. Suppose for a contradiction that was a non-trivial elementary embedding with critical point $\kappa $ and let $\lambda = {\textrm {sup}} \{ j^n(\kappa ) \colon n \in \omega \}$ . Then $j(\lambda ) = \lambda $ and, since $\lambda ^+$ is definable as the least cardinal above $\lambda $ , $j(\lambda ^+) = \lambda ^+$ . Now, using Theorem 4.3, let $\langle S_\alpha : \alpha \in \kappa \rangle $ be a partition of $S^{\lambda ^+}_\omega $ into $\kappa $ many disjoint stationary sets and let $S = \{ \langle \alpha , S_\alpha \rangle \colon \alpha \in \kappa \}$ . Then $j(S) = \{ \langle \alpha , T_\alpha \rangle \colon \alpha \in j(\kappa ) \}$ and, by elementarity, $\langle T_\alpha \colon \alpha \in j(\kappa ) \rangle $ is a partition of $S^{\lambda ^+}_\omega $ into disjoint sets such that for each $\alpha $

$$\begin{align*}T_\alpha \textit{ is a stationary subset of } \lambda^+. \end{align*}$$

Also, we have that for each $\alpha \in \kappa $ , $j(S_\alpha ) = T_\alpha $ . We claim that there is some $\beta \in \kappa $ such that $T_\kappa \cap S_\beta $ is stationary. For suppose not, then by our comments on choosing from set many classes, for each $\alpha $ we can fix a club $C_\alpha $ such that $T_\kappa \cap S_\alpha \cap C_\alpha = \emptyset $ . Letting $C = \bigcap _{\alpha \in \kappa } C_\alpha $ we must have that

$$\begin{align*}\emptyset = T_\kappa \cap C \cap \bigcup_{\alpha \in \kappa} S_\alpha = T_\kappa \cap C, \end{align*}$$

contradicting the assumption that $T_\kappa $ was stationary. So fix $\beta $ such that $T_\kappa \cap S_\beta $ is stationary. Now, let

$$\begin{align*}U = \{ \gamma \in \lambda^+ \colon \gamma = j(\gamma) \}, \end{align*}$$

and note that U is closed under sequences of length $\omega $ . Therefore there exists some $\sigma \in U \cap T_\kappa \cap S_\beta $ . But then $\sigma = j(\sigma ) \in j(S_\beta ) = T_\beta $ , contradicting the assumption that the $T_\alpha $ were disjoint. Hence no such embedding can exist.⊣

Proof. Suppose for a contradiction that was a non-trivial, cofinal, $\Sigma _0$ -elementary embedding with critical point $\kappa $ and let $\lambda = {\textrm {sup}} \{ j^n(\kappa ) \colon n \in \omega \}$ . Note that, by an instance of Replacement in the language expanded to include the predicate j, . Now there are two cases:

• • Case 1: $\lambda ^+$ exists.

• • Case 2: For all , there is an injection $f \colon x \rightarrow \lambda $ .

Case 1: Using Theorem 3.2, any such embedding must be fully elementary. Therefore, this is just a special case of Theorem 5.2.

Case 2: First note that by elementarity, since so is for each $n \in \omega $ and therefore . Note also that and, by the Well-Ordering Principle, for each there is a bijection

$$\begin{align*}f \colon | {\textrm{trcl}}(\{x\}) | \rightarrow {\textrm{trcl}}(\{x\}). \end{align*}$$

Moreover, since there is an injection of x into $\lambda $ , we must have that $| {\textrm {trcl}}(\{x\}) | \leq \lambda $ for each . Now let

$$\begin{align*}C_{x, f} \operatorname{\mathrm{:=}} \{ \langle \alpha, \beta \rangle \in \lambda \times \lambda \colon f(\alpha) \in f(\beta) \}. \end{align*}$$

Then

and therefore so is its Mostowski collapse, with ${\textrm {coll}}(C_{x, f}) = {\textrm {trcl}}(\{x\})$ . This means that for any x and bijection $f \colon | {\textrm {trcl}}(\{x\}) | \rightarrow {\textrm {trcl}}(\{x\})$ ,

That is, j is completely determined by its construction up to

. Now, let

and note that, since

, so is

Therefore, by defining $\varphi (\cdot , \cdot , i, \lambda )$ as

we have that $\varphi (x, y, i, \lambda )$ holds if and only if $j(x) = y$ so j is definable from the parameters i and $\lambda $ , both of which lie in V, contradicting Theorem 3.4.⊣

Proof. Let $\varphi $ and $\psi $ be formulae and u be a set such that for some y, $\psi (y, u)$ and for every $\alpha \in \delta $ and every $\alpha $ -sequence $s = \langle x_\beta \colon \beta \in \alpha \rangle $ satisfying $\psi (x_\beta , u)$ for each $\beta $ , there is a z satisfying $\psi (z, u)$ and $\varphi (\alpha , s, z, u)$ . We define a new formula $\vartheta $ extending $\varphi $ to apply to any $\alpha $ -sequence, s, for $\alpha \in \mu $ by

$$ \begin{align*} &\vartheta(\alpha, s, z, u, \delta) \equiv \big( \alpha < \delta \wedge \forall \beta \in \alpha \ \psi(s(\beta), u) \wedge \varphi(\alpha, s, z, u) \big) \\[3pt] &\phantom{\big( \alpha < \delta \wedge \forall \beta \in \alpha \ \psi(s(\beta), u) \wedge \varphi(\alpha, s, z, u) \big)} \ \ \vee \big( \alpha \geq \delta \wedge \psi(z, u) \big). \end{align*} $$

Then for any function f with domain $\mu $ witnessing this instance of the -Scheme, $f \operatorname {\mathrm {\upharpoonright }} \delta $ witnesses that holds for $\psi $ and $\varphi $ .⊣

Proof. Let $\mathcal {C} = \{ x \colon \psi (x) \}$ be a proper class. We shall in fact prove the equivalent statement that for any $\nu \leq \mu $ there is a subset b of $\mathcal {C}$ and a bijection between b and $\nu $ . Suppose for a contradiction that this were not the case and let $\gamma $ be the least ordinal for which no such subset of size $\gamma $ exists. It is obvious that $\gamma $ must be an infinite cardinal. Let $\varphi (\alpha , s, y)$ be the statement that s is a sequence of length $\alpha $ and ${\textrm {ran}}(s) \cup \{y\}$ is a subset of $\mathcal {C}$ and $y \not \in {\textrm {ran}}(s)$ . Then, by assumption, for every $\alpha \in \gamma $ there is a sequence of elements of $\mathcal {C}$ of length $\alpha $ . Also, since $\mathcal {C}$ is a proper class, if s is an $\alpha $ length sequence from $\mathcal {C}$ then there is some $y \in \mathcal {C}$ which is not in s so the hypothesis of is satisfied. Therefore, by , there is a function f with domain $\gamma $ and whose range gives a subset of $\mathcal {C}$ of cardinality $\gamma $ , giving us our desired contradiction.⊣

Proof. This is proven by induction on $\alpha \in \kappa + 1$ . Clearly limit cases follow by an instance of Collection so it suffices to prove that for $\alpha \in \kappa $ , if then so is . First note that j fixes every set of rank less than $\kappa $ so is the identity. Now suppose for sake of a contradiction that was a proper class. Then, by Corollary 6.8, we could fix a set and a surjection

$$\begin{align*}h \colon b \twoheadrightarrow \kappa. \end{align*}$$

So, by elementarity, there is a surjection

$$\begin{align*}j(h) \colon j(b) \twoheadrightarrow j(\kappa) \end{align*}$$

in M. However, since , $j(b)$ is also a subset of and for any $x \in j(b)$ , $j(x) = x$ . Therefore,

$$\begin{align*}x \in j(b) \Longleftrightarrow j(x) \in j(b) \Longleftrightarrow x \in b, \end{align*}$$

and hence $b = j(b)$ . Then, for any $x \in b$ ,

$$\begin{align*}j(h)(x) = j(h)(j(x)) = j(h(x)) = h(x), \end{align*}$$

so $j(h) = h$ . But this then contradicts the assumption that $j(h)$ was a surjection onto $j(\kappa ).$ Hence must be a set in V as required.⊣

Question 6.11. Is the existence of a non-trivial, cofinal, $\Sigma _0$ -elementary embedding such that inconsistent?

Question 6.12. Suppose that , and is a non-trivial elementary embedding. Is ? Is ?Footnote ⁵

Proof. Given a proper class $\mathcal {C}$ , define

Then S must be unbounded in the ordinals so, given an ordinal $\alpha $ , we can take the first $\alpha $ many elements of S, $\{ \gamma _\beta : \beta \in \alpha \}$ . Then

$$ \begin{align*} f(x) = \begin{cases} \beta, & \textit{if} \ {\textrm{rank}}(x) = \gamma_\beta, \\ 0, &\textit{otherwise}, \end{cases} \end{align*} $$

defines a surjection of $\mathcal {C}$ onto $\alpha $ .⊣