用户: OperatorP/一篇以工地英语写就的关于0

1Iterated Ultrapower

We’ve got theories about elementary embeddings from the whole set-theoretic universe $V$ derived from an ultrafilter since that ultrafilter must be in $V$, but if we want some elemantary embeddings from arbitrary model $M$, we don’t want to take this restriction ($U\in M$) into consideration. To make things easier, we’ll work in some second-order set theory, probably GBC, but everything can also be done inside a first-order theory after extending the language of set theory with some new predicates.

Definition 1.1. A set $U\subseteq \mathcal{P}(\kappa )$ is an $M-$ultrafilter (where $M$ is a transitive set model of $ZF{C}^{-}$ or an inner model) if the following assertion holds:

1.	$(M,\in ,U)\models U$ is a normal ultrafilter over $\kappa$, and
2.	$(M,U)$ is weakly amenable: given any $\kappa -$sequence $⟨A_{\alpha }{⟩}_{\alpha \in \kappa }\in M$, the set $\{\alpha \in \kappa |A_{\alpha }\in U\}\subseteq \kappa$ also belongs to $M$.

We don’t need this $U$ to be a real ultrafilter over $\kappa$, nor $U\in M$.

It’s easy to check that there’s still an elementary embedding $j:(M,\in )\to (M^{\kappa }/U,E)$, and there’s a definable $V$ makes $j:(M,\in ,U)\to (M^{\kappa }/U,E,W)$ an embedding (though it don’t need to be elementary). We’ll denote this by $j:(M,\in ,U){\prec }^{-}(M^{\kappa }/U,E,W)$, and we left the proof of following propositions as exercises.

Proposition 1.2. Consider the well-founded (so collpased) ultrapower embedding $j:(M_0,\in ,U_0){\prec }^{-}(M_1,\in ,U_1)$,

1.	$j$ is cofinal: for any $y\in M_1$ there’s $x\in M_0$ so that $y\in j(x)$, and for any ordinal $y$ there’s ordinal $x$;
2.	$|M_0|=|M_1|$ if $M_0$ is a set;
3.	$V_{\kappa }\cap M_0=V_{\kappa }\cap N_1$ and $j$ is identical on it, $\mathcal{P}(\kappa )\cap M_0=\mathcal{P}(\kappa )\cap M_1$, and $({\kappa }^{+}{)}^{M_0}=({\kappa }^{+}{)}^{M_1}$;
4.	no matter $U_0\in M_0$ or not, $U_0\notin M_1$;
5.	$U_1$ is an $M_1-$ultrafilter over $j(\kappa )$.

The last proposition means we can make a new ultrapower embedding from $(M_1,\in ,U_1)$ to some $(M_2,\in ,U_2)$, and this process can be done to get $(M_n,\in ,U_n)$. At the limit step, since we actually got a diagram with objects $(M_n,\in ,U_n)$ and morphisms $j_{m,n}:(M_m,\in ,U_m){\prec }^{-}(M_n,\in ,U_n)(m<n)$ obtained by compositing all the ultrapower embeddings $j_{k,k+1}:(M_k,\in ,U_k){\prec }^{-}(M_{k+1},\in ,U_{k+1})$ passed by together, we can take the colimit as $(M_{\omega },E_{\omega },U_{\omega })$, and all these morphisms follows. Now the question is $E_{\alpha }$ need not to be well-founded, so maybe we can’t collapse it to $\in$ and assume $M_{\alpha }$ is transitive; but if we can, we’ll be able to make this process even further.

Definition 1.3. $(M_{\alpha },U_{\alpha },j_{\alpha ,\beta }{)}_{\alpha ,\beta <\theta }$ is an iteration of $(M_0,U_0)$ of length $\theta$ if every $M_{\alpha }$ is transitive. Moreover, if for every ordinal $\theta$ there’s iteration of $(M_0,U_0)$ of length $\theta$, we say $U$ is an iterable $M-$ultrafilter.

Let’s consider the most simple case at first.

Theorem 1.4 (Gaifman). $U$ is iterable if $U\in M$.

证明. Work inside of $M$ and we are now dealing with $U$ on a measurable, let’s prove this natural lemma firstly.

Lemma 1.5 (Factor Lemma). If we iterate $(M,U)$ $\alpha -$steps to a well-founded $(N,W)$, then the $\beta -$step iteration of $(N,W)$ is isomorphic to $\alpha +\beta -$step iteration of $(M,U)$ no matter they are well-founded or not, and the diagram (with two mmorphisms shaped $j_{\xi ,\eta }$) commutes.

证明. Put induction on $\beta$.

$\square$

Since $U\in M$, the successor step will be well-founded forever, sewe can assume for some limit $\alpha$, the $\alpha -$step iteration of $(V,\in ,U)$, namely $(M,E,W)$, is ill-founded. WLOG let $\alpha$ be the least such ordinal and pick the least $\beta$ so that there exists an infinite decreasing sequence of ‘ordinals’ in $M$ w.r.t. $E$ below $j_{0,\alpha }(\beta )$. Take such a sequence $x_0,x_1,...$ so that $x_{n+1}E{x}_n$ and $x_{0}E{j}_{0,\alpha }(\beta )$. By construction of colimit in $\mathsf{Set}$, there’s a $\gamma <\alpha$ and $\delta <j_{0,\gamma }(\beta )$ so that $x_0=j_{\gamma ,\alpha }(\delta )$. Let ${\gamma }^{\prime }\le \alpha$ be such that $\gamma +{\gamma }^{\prime }=\alpha$.

By assumptions, in $V$ we have for every ${\alpha }^{\prime }\le \alpha$ and ${\beta }^{\prime }<\beta$ there’s no infinite decreasing sequence of ordinals below $j_{0,{\alpha }^{\prime }}({\beta }^{\prime })$ in the ${\alpha }^{\prime }-$step iteration of $(V,U)$. Now apply $j_{0,\gamma }$, in the $\gamma -$step iteration of $(V,U)$ which by our assumption is well-founded, we have for every ${\alpha }^{\prime }\le j_{0,\gamma }(\alpha )$ and ${\beta }^{\prime }<j_{0,\gamma }(\beta )$, in the ${\alpha }^{\prime }-$step iteration of it, which by the Factor Lemma is just the $\gamma +{\alpha }^{\prime }-$step iteration of the original $(V,U)$, there’s no infinite decreasing sequence of ordinals below $j_{\gamma ,\gamma +{\alpha }^{\prime }}({\beta }^{\prime })$. Now take ${\alpha }^{\prime }={\gamma }^{\prime }$ and ${\beta }^{\prime }=\delta$, we should say there’s no infinite decreasing sequence of ordinals below $x_0$ in $(M,E,W)$, which is obviously a contradiction since $x_1,x_2,...$ are still there.

$\square$

Now let’s consider the general case, but it’s hard if we always use the whole graph $(M_{\alpha },U_{\alpha },j_{\alpha ,\beta })$. Thanks to Kunen, we can define $j_{0,\alpha }$ directly in $(M_0,U_0)$ by the trick as follows. Let’s do this for natural numbers at first.

Theorem 1.6. Set $U^1\subseteq \mathcal{P}({\kappa }^1)\cap M$ by transforming $U$ using the canonical bijection between $\kappa$ and ${\kappa }^1$, then define recursively for every natural number $n>1$ a subset of $\mathcal{P}({\kappa }^{n+1})\cap M$ named $U^{n+1}$ by letting $X\in U^{n+1}$ iff $\{s\in {\kappa }^n|\{\xi \in \kappa |s^{⌢}⟨\xi ⟩\in X\}\in U\}\in U^n$. The follwing holds:

1.	$(M,\in ,U)\models U^n$ is a $\kappa -$complete ultrafilter on the Boolean Algebra $\mathcal{P}({\kappa }^n)$ for every $n$, and
2.	the ultrapower embedding $k_n:(M,\in ,U){\prec }^{-}(M^{{\kappa }^n}/U^n,E_n,U_n)$ equals the $n-$step iteration $j_{0,n}$ no matter they are well-founded or not.

证明. Put induction on $n$.

$\square$

To express iteration of transfinite steps, intuitionally we want some ultrafilter on ${\kappa }^{\alpha }$, but we can’t even form $U^{\omega }$ like what we’ve done for $U^n$ because there’s always an infinite undecided tails after $s$. Instead, we need another colimit of a diagram, noticing the fact below:

Theorem 1.7. If $E\subseteq F$ are finite set of ordinals, then for any $X\subseteq {\kappa }^E$, $X\in U^E$ iff ${\mathrm{in}}_{E,F}\in U^F$, where $U^E$ is $U^{|E|}$ transformed using canonical bijection between $E$ and $|E|$, ${\kappa }^E$ and ${\kappa }^{|E|}$, and power sets of them, and the inclusion map ${\mathrm{in}}_{E,F}$ is defined as ${\mathrm{in}}_{E,F}(X)=\{s\in {\kappa }^F|s\upharpoonright E\in X\}$.

证明. Induct $|F|$.

$\square$

Theorem 1.8. Noticing that the definition of ${\mathrm{in}}_{E,F}$ doesn’t need the assertion that $E,F$ are finite, we consider all $Y={\mathrm{in}}_{E,\alpha }(X)\subseteq {\kappa }^{\alpha }$ for all finite subset $E\subseteq \alpha$ and $X\subseteq {\kappa }^E$, and we say $Y$ has finite support $E$ in this case. Not only the set of all subsets with finite support of ${\kappa }^{\alpha }$ forms a Boolean Algebra $(B_{\alpha },\subseteq )$, we also have a ultrafilter on it by setting $Y\in U^{\alpha }$ iff $X\in U^E$.

证明. $U^{\alpha }$ is the colimit of all these $U^E$s.

$\square$

Theorem 1.9 (Kunen). $\mathrm{Fs}(M^{{\kappa }^{\alpha }})/U^{\alpha }$ is isomorphic to the $\alpha -$step iteration of $M$, no matter it is well-founded or not. A function $f:{\kappa }^{\alpha }\to M$ has finite support $E\in \mathrm{FinSub}(\kappa )$ iff for any $s,t\in {\kappa }^{\alpha }$ with $s\upharpoonright E=t\upharpoonright E$, $f(s)=f(t)$.

证明. Induct $\alpha$.

$\square$

Now let’s prove some equivalent forms of iterability noticed by Gaifman, Jensen and Dodd.

Theorem 1.10. TFAE.

1.	$U$ is iterable;
2.	there exists an iteration of length ${\omega }_1$;
3.	we can iterate $(M_0,\in ,U_0)$ to some $(M_{\alpha },\in ,U_{\alpha })$ where $U_{\alpha }$ is ${\omega }_1$ complete;
4.	$(M_0,\in ,U_0)$ is countably iterable: for any $j:(N,\in ,W){\prec }^{-}(M_0,\in ,U_0)$ where $(N,\in ,W)$ is countable, $(N,\in ,W)$ has an iteration of length ${\omega }_1$;
5.	any $j:(N,\in ,W){\prec }^{-}(M_0,\in ,U_0)$ gives an iterable $N-$ultrafilter $W$.

证明. $5\to 1$ and $1\to 2$ are immediate. For $2\to 3$, it’s easy to observe that we just need to verify the transitive ${\omega }_1-$step iteration $(M_{{\omega }_1},U_{{\omega }_1})$ of $(M_0,U_0)$ must let the $U_{{\omega }_1}$ be ${\omega }_1$ complete. Pick $\omega$ many $X_n\in U_{{\omega }_1}$, we claim that there is an ${\alpha }_n<{\omega }_1$ for each $X_n$ so that $\{{\kappa }_{\beta }|{\alpha }_n\le \beta <{\omega }_1\}\subseteq X$: pick the ${\alpha }_n$ be the least ordinal below ${\omega }_1$ so that there exists $Y$ such that $j_{{\alpha }_n,{\omega }_1}(Y)=X$; for any $\beta$ satisfying $\alpha \le \beta <{\omega }_1$, we have $Y_{\beta }=j_{{\alpha }_n,\beta }$ satisfying $j_{\beta ,{\omega }_1}(Y_{\beta })=X$, so $X\in U_{{\omega }_1}$ iff $Y_{\beta }\in U_{\beta }$ iff ${\kappa }_{\beta }\in j_{\beta ,\beta +1}(Y_{\beta })$ iff ${\kappa }_{\beta }\in j_{\beta ,{\omega }_1}(Y_{\beta })=X$. Now take $\alpha =\sup \{{\alpha }_n|n\in \omega \}<{\omega }_1$, we have ${\bigcap }_{n\in \omega }{X}_n\supseteq \{{\kappa }_{\beta }|\alpha \le \beta <{\omega }_1\}\ne \varnothing$, justifying that $U_{{\omega }_1}$ is ${\omega }_1$ complete.

To get $3\to 4$, since we already have $(M_0,\in ,U_0){\prec }^{-}(M_{\alpha },\in ,U_{\alpha })$, it suffice to prove that $(M_{\alpha },\in ,U_{\alpha })$ is countably iterable. Simplifying the notion, we need to prove that $(M,U)$ is countably iterable when $U$ is ${\omega }_1$ complete. Suppose there’s $k_0:(N_0,\in ,W_0){\prec }^{-}(M,\in ,U)$, we prove inductively that for every countable ordinal $\alpha$, not only the $\alpha -$step iteration of $(N_0,W_0)$ is well-founded, but also there’s $k_{\alpha }:(N_{\alpha },\in ,W_{\alpha })\to (M,\in ,U)$ making things commutes. The case $\alpha$ being a limit is obvious by the universal property of colimit, so assume this holds for every ordinal less than or equal to $\alpha$, we prove for $\alpha +1$, and this is similar to the case $\alpha =0$. For any $X\in W_0$, $k_0(X)\in U$, but $W_0$ is countable and $U$ is ${\omega }_1$ complete, we can choose one element $\eta$ in the non-empty set ${\bigcap }_{X\in W_0}{k}_0(X)$. It’s easy to verify $k_1([f{]}_{W_0}^0)=k_0(f)(\eta )$ is what we want.

To get $4\to 5$, notice that $(N,\in ,W){\prec }^{-}(M_0,\in ,U_0)$ and $(M_0,\in ,U_0)$ is countably iterable just means $(N,\in ,W)$ is countably iterable, we only need to prove countably iterable implies iterable. We prove two lemmas to represent elements in $(M_{\alpha },\in ,U_{\alpha })$ using ${\kappa }_{\gamma }$ as parameters to avoid constructing the Boolean ultrapower.

Lemma 1.11. Given $\alpha -$step iteration $(M_{\alpha },E,U_{\alpha })$ so that for every $\beta \in \alpha$ the $\beta -$step iteration is well-founded. For every $x\in M_{\alpha }$, there is a function $f\in M^{{\kappa }^n}\cap M$ and $n$ ordinals ${\gamma }_1<...<{\gamma }_n<\alpha$ such that $x=j_{0,\alpha }(f)({\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n})$.

证明. Induct $\alpha$.

$\square$

Lemma 1.12. Given $\alpha -$step iteration $(M_{\alpha },E,U_{\alpha })$ so that for every $\beta \in \alpha$ the $\beta -$step iteration is well-founded. For every ${\mathcal{L}}_{\in }$ formula $\phi (v_0,...,v_n)$ and $x\in M_0$ and ${\gamma }_1<...<{\gamma }_N<\alpha$, we have $(M_{\alpha },E)\models \phi (j_{0,\alpha }(x),{\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n})$ iff $(M,\in ,U)\models \{⟨{\xi }_1,...,{\xi }_n⟩\in {\kappa }^n|\phi (x,{\xi }_1,...,{\xi }_n)\}\in U^n$

证明. Induct $n$.

$\square$

Now assume for some $\alpha$ we have for every $\beta \in \alpha$ the $\beta -$step iteration is well-founded but $(M_{\alpha },E,U_{\alpha })$ is not. Take an infinite decreasing sequence $x_k$ w.r.t. $E$ in $M_{\alpha }$, the first lemma shows that there are $f_k\in M^{{\kappa }^{n_k}}\cap M$ and ${\gamma }_{k,1}<...<{\gamma }_{k,n_k}<\alpha$ so that $x_k=j_{0,\alpha }^M(f_k)({\kappa }_{{\gamma }_{k,1}},...,{\kappa }_{{\gamma }_{k,n_k}})$. Now take Skolem Hull of $\{f_k|k\in \omega \}$ in $(M_0,\in ,U_0)$ to get a countable well-founded elementary substructure containing every $f_k$, by collapsing we get an elementary embedding from a countable structure $e:(N,\in ,W){\prec }^{-}(M,\in ,U)$. Suppose $f_k=e(g_k)$, by the assumption that $(M,\in ,U)$ is countably iterable we can iterate to get $(N_{\delta },\in ,W_{\delta })$ for every countable ordinal $\delta$.

We just transform the sequence in $(M_{\alpha },E)$ to a sequence in $(N_{{\alpha }^{\prime }},\in )$ after this. Since $x_k=j_{0,\alpha }^M(f)({\kappa }_{{\gamma }_{k,1}},...,{\kappa }_{{\gamma }_{k,n_k}})$, we denote the ordertype of set $\{{\gamma }_{k,l}|1\le l\le n_k\}$ by ${\alpha }^{\prime }<{\omega }_1$ and the order-preserving enumerating function by $h:{\gamma }_{k,l}\mapsto {\delta }_{k,l}\in {\alpha }^{\prime }$, then take $y_k=j_{0,{\alpha }^{\prime }}^N(g_k)({\lambda }_{{\delta }_{k,1}},...,{\lambda }_{{\delta }_{k,n_k}})$, it’s easy to prove that $y_{k+1}\in y_k$ iff $x_{k+1}E{x}_k$, and this is surely a contradiction.

$\square$

2$0^{\#}$ and Elementary Embeddings

Definition 2.1. We say $0^{\#}$ exists if there exists an elementary embedding $j:L\to L$.

Theorem 2.2. $0^{\#}$ exists if there’s a measurable cardinal.

证明. Since every inner model of $L$ must be itself, we just construct the elementary embedding $j:L\to L^{\kappa }/U=L$.

$\square$

$0^{\#}$ is not the class $j$ since it’s too large, but let’s talk about it in the next section and consider the following question at first: why it’s non-trivial claiming the existence of such an elementary embedding?

Theorem 2.3. Assume $V=L$, there’s no measurable cardinal and no elemantary embedding $j:L\to L$.

证明. If there’s an elemantary embedding, then there’s a measurable. If there’s a measurable, then there’s a least measurable $\kappa$. Now $j(\kappa )>\kappa$ must also be the least measurable, leads to a contradiction.

$\square$

There are two generalization of this theorem, one makes use of this minimal critical point argument, and the other removes the assertion $V=$.

Theorem 2.4 (Suzuki). In $ZF$, there’s no definable (even with parameters) $j:V\to V$.

证明. Assume $j(x)=y$ is defined as formula $\Phi (a,x,y)$ with parameter $a$, so we have $\forall x\in M\exists y\forall z(z=y\leftrightarrow \Phi (a,x,z))$ as a assumption, surely $j\upharpoonright {\omega }_1=id$, and by an argument we learned in the first lecture, $j$ is fully elemantary iff $j$ is ${\Sigma }_1$ elementary,, which is expressible inside the first-order set theory.

To make a contradiction, since this $j$ is non-trivial, we have the following assumptions after claiming $\forall x\in M\exists y\forall z(z=y\leftrightarrow \Phi (a,x,z))$:

1.	$\exists x\in M(\neg \Phi (a,x,x))$;
2.	$\forall x_1\forall y_1\forall x_2\forall y_2((\Phi (a,x_1,y_1)\wedge \Phi (a,x_2,y_2))\to (x_1\in x_2\leftrightarrow y_1\in y_2))$;
3.	$\forall n=┌\phi ┐\in {\mathrm{Fml}}_{{\Sigma }_1}\forall x\forall y(\Phi (a,x,y)\to ({\models }_{{\Sigma }_1}(n,x)\leftrightarrow {\models }_{{\Sigma }_1}(n,y)))$.

The conjunction of these assumptions, denoted by ${\Phi }_0(a)$, claims that $\Phi (a,x,y)$ defines an embedding $j_a$, and $\mathrm{crit}(j_a)=\kappa$ can be defined as the property $\forall x\in V_{\kappa }(\Phi (a,x,x))\wedge \neg \Phi (a,\kappa ,\kappa )\wedge \kappa \in \mathbf{Ord}$. To make use of the minimal-critical-point-argument above, try ${\Phi }_1(a)$ which is ${\Phi }_0(a)\wedge \forall a^{\prime }({\Phi }_0(a^{\prime })\to \mathrm{crit}(j_a)\le \mathrm{crit}(j_{a^{\prime }}))$.

Our assumption says $\exists a({\Phi }_0(a))$, so $\exists a({\Phi }_1(a))$, and the $\kappa =\mathrm{crit}(j_a)$ now have property $\exists a({\Phi }_1(a)\wedge \kappa =\mathrm{crit}(j_a))$. Arbitrary take an $a$ here, we have $j_a(\kappa )>\kappa$ also have this property, which is a contradiction.

$\square$

Theorem 2.5 (Kunen). In $ZFC$(actually $GBC$, or $ZFC+j$ with a new functional symbol $j$), there’s no $j:V\to V$.

证明. This theorem is known as Kunen’s Inconsistency, and it remains unknown whether we can avoid using AC here or not; it’s one of the hardest and central questions in the theory of large cardinals. The proof here belongs to Woodin. Assume there is $j$ and set $\kappa =\mathrm{crit}(j)$, we find a contradiction by trying to split stationary subsets of ${\lambda }^{+}$ where $\lambda =j_{0,\omega }(\kappa )$ because it’s easy to see $j(\lambda )=\lambda$ and $j({\lambda }^{+})={\lambda }^{+}$. Suppose $S:\kappa \to \mathcal{P}({\lambda }^{+})$ split the stationary set $\{\xi \in {\lambda }^{+}|\mathrm{cf}(\xi )=\omega \}$ into $\kappa <{\lambda }^{+}$ parts by Solovay’s famous theorem, we have $j(S):j(\kappa )\to \mathcal{P}({\lambda }^{+})$ also a partition and $j(S)(\kappa )$ must be a stationary subset of ${\lambda }^{+}$ with every element having cofinality $\omega$. Since the club filter on ${\lambda }^{+}$ is ${\lambda }^{+}$ complete, there must be some $\alpha <\kappa$ making $j(S)(\kappa )\cap S(\alpha )$ stationary, but the set $C=\{\xi \in {\lambda }^{+}|j(\xi )=\xi \wedge \mathrm{cf}(\xi )=\omega \}$ is $\omega -$closed unbounded, so we can choose an element $\xi \in j(S)(\kappa )\cap S(\alpha )\cap C$, but then $\xi =j(\xi )\in j(S(\alpha ))=j(S)(\alpha )$, and $j(S)(\kappa )\cap j(S)(\alpha )$ is non-empty, contradicting that $j(S)$ gives a partition.

$\square$

Let’s back to $0^{\#}$. The $U$ on measurable don’t need to be an $L-$ultrafilter because it do not need to be weakly amenable, but we want this $j:L\to L$ to be an ultrapower embedding so we can analysis it more clearly. Kunen proves that we are lucky.

Theorem 2.6 (Kunen). TFAE.

1.	there’s $j:L\to L$;
2.	there’s $j:L_{\alpha }\to L_{\beta }$ with $L_{\alpha }\models ZF{C}^{-}$ and $\mathrm{crit}(j)<|\alpha |$;
3.	there’s an $L-$ultrafilter with its ultrapower embedding well-founded;
4.	there’s an iterable $L-$ultrafilter.

证明. $1\to 2$ is trivial: set $\kappa =\mathrm{crit}(j)$ and take an arbitrary $\alpha$ so that $L_{\alpha }\models ZF{C}^{-}$ and $\kappa <|\alpha |$, $j\upharpoonright L_{\alpha }:L_{\alpha }\prec j(L_{\alpha })=L_{\beta }$ is what we want. As for $2\to 3$ we just check $U=\{X\in \mathcal{P}(X)\cap L|\kappa \in j(X)\}$ makes a well-founded ultrapower embedding of $L$ because it’s easy to verify this gives an $L-$ultrafilter. Assume there is a sequence $f_n$ and $X_n=\{\xi \in \kappa |f_{n+1}(\xi )\in f_n(\xi )\}\in U$, lets’ take some cardinal $\gamma$ so that $\kappa$ and every $f_n$ is in $L_{\gamma }$, and then take Skolem Hull and collapse it to get an elementary embedding from a countable structure $e:L_{\delta }\to L_{\gamma }$ where $\delta <{\kappa }^{+}\le \alpha$ and $g_n\in L_{\delta }$ making $e(g_n)=f_n$. Now since for $\xi \in \kappa$ we have $e(\xi )=\xi$, $g_{n+1}(\xi )\in g_n(\xi )$ iff $f_{n+1}(\xi )\in f_n(\xi )$, so $X_n\in U$ makes $j(g_n)(\kappa )$ an infinite decreasing sequence in $L_{\beta }$, leads to a contradiction.

$4\to 1$ is also trivial, so the most important part is to prove $3\to 4$, that is an well-founded $L-$ultrafilter must be iterable. Let’s try to prove the easier successor step: assume $(M_{\alpha },\in ,U_{\alpha })=(L,\in ,\alpha )$ is already well-founded, by lemma we have for every $x\in M_{\alpha }$ a function $f\in L^{{\kappa }^n}\cap L$ and ordinals ${\gamma }_1<...<{\gamma }_n<\alpha$ so that $x=j_{0,\alpha }(f)({\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n})$. To get $\alpha +1$, notice that $M_1=L^{\kappa }/U=L$, so we can assume $f=[g{]}_U^0$ where for all $\xi <\kappa$ we have $g(\xi )$ is a function on ${\xi }^n$. Now define $\bar{f}:L^{{\kappa }^n}\to L^{\kappa }$ by setting $\bar{f}({\xi }_1,...,{\xi }_n)(\xi )=g(\xi )({\xi }_1,...,{\xi }_n)$(let it be $0$ if RHS remains undefined). Just check $e:(M_{\alpha },\in )\to (M_{\alpha +1},E)$ where $e(j_{0,\alpha }(f)({\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n}))=[j_{0,\alpha }(\bar{f})({\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n}){]}_{U_{\alpha }}^0$ is an isomorphism witnessing well-foundedness of $E$.

The limit step is harder. To make the colimit well-founded, we try to reduct the colimit into some coproduct. Firstly, recall that there’s a ${\Sigma }_1-{\mathcal{L}}_{\in }$ formula $\Phi$ defining the ${\Sigma }_1$ Skolem Function $h$ in models of $ZF{C}^{-}+V=L$ without parameters, which means for every coding $n\in \omega$ of a ${\Sigma }_1$ formula $\phi$ and parameters $x_1,...,x_{v(n)}\in L_{\alpha }$ with the arity function $n\mapsto v(n)$ obviously a ${\Delta }_0$ definable function in set theory, $y=h(n,x_1,...,x_{v(n)})$ iff $L_{\alpha }\models \Phi (n,y,x_1,...,x_{v(n)})$, and $y=\varnothing$ if the outmost quantifier of $\phi$ is not an $\exists$, $L_{\alpha }\models \psi (y,x_1,...,x_{v_n})\leftrightarrow L_{\alpha }\models \phi (x_1,...,x_{v_n})$ if $\phi$ has the form $\exists y(\psi (y,x_1,...,x_n))$. Now for $M\models ZF{C}^{-}+V=L$ and a subset $X\subseteq M$, we denote the ${\Sigma }_1$ Skolem Hull of $X$ in $M$ by $h_M(X)=\{h(n,x_1,...,x_{v(n)})|n\in \omega ,x_1,...,x_{v(n)}\in X\}$, and now $h_M(X)$ is a ${\Sigma }_1$ substructure of $M$. Since $V=L$ is a ${\Pi }_2$ assertion, by replacing the use of $V=WF$ by $V=L$ in the proof of cofinal ${\Sigma }_1$ elementary embedding of something into a model of $ZF$ is fully elementary, we can prove that $h_M(X)$ is an elementary substructure if $X$ is cofinal.

Back to proof. Assume $\alpha$ is a limit ordinal and every $\beta <\alpha$ makes $M_{\beta }$ from $M_0=L$ equals $L$, we must show $M_{\alpha }=L$ holds. Consider $H=\{x\in L|\forall \beta <\alpha (j_{0,\beta }(x)=x)\}$, it’s easy to prove that $\kappa \subseteq H$ and $H$ is (cofinally) a proper class in $Ord$ (and in $L$). Now we define $H_{\beta }=h_L(H\cup \{{\kappa }_{\gamma }|\gamma \in \beta \})$, $N_{\beta }=L$ its collapse, and $e_{\beta }:N_{\beta }\to L$ the resulting elementary embedding, also define $H_{\alpha }=h_L(H\cup \{{\kappa }_{\beta }|\beta \in \alpha \})$ and $N_{\alpha },e_{\alpha }$ in the same manner. We claim that $e_{\gamma ,\beta }=e_{\beta }^{-1}\circ e_{\gamma }:L\to L(\gamma \in \beta )$ equals $j_{\gamma ,\beta }$, and the claim implies $e_{\beta ,\alpha }=j_{\beta ,\alpha }$ and $M_{\alpha }=N_{\alpha }=L$ since $H_{\alpha }={\bigcup }_{\beta \in \alpha }{H}_{\beta }$.

Firstly, we prove that ${\kappa }_{\beta }\subseteq H_{\beta }$, which implies that $e_{\beta }\upharpoonright {\kappa }_{\beta }=id$. For $\xi \in {\kappa }_{\beta }$, assume $\xi =j_{0,\beta }(f)({\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n})$ where $f\in L^{{\kappa }^n}\cap L$ and ${\gamma }_1<...<{\gamma }_n<\beta$. Notice that $H_0=H=h_L(H)$ since $j_{0,\beta }(h(n,x_1,...,x_{v(n)}))=h(j_{0,\beta }(n),j_{0,\beta }(x_1),...,j_{0,\beta }(x_{v(n)}))=h(n,x_1,...,x_{v(n)})$, $e_0$ is an elementary embedding from $L$ into $H$, $e_0(f)$ has domain $e_0(\kappa {)}^n$, but $\kappa \subseteq H$ implies $e_0$ is constant on it, so we also have $f=e_0(f)\upharpoonright {\kappa }^n$, and $j_{0,\beta }(f)\upharpoonright {\kappa }^n=f$ gives $\xi =e_0(f)({\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n})\in H_{\beta }$ at last.

Secondly, we prove that for any $X\in \mathcal{P}({\kappa }_{\gamma })\cap L$, $j_{\gamma ,\beta }(X)=e_{\gamma ,\beta }(X)\cap {\kappa }_{\beta }$. $e_{\gamma }\upharpoonright {\kappa }_{\gamma }=id$ gives $X=e_{\gamma }(X)\cap {\kappa }_{\gamma }$, and $e_{\beta }\upharpoonright {\kappa }_{\beta }=id$(with ${\kappa }_{\beta }\supseteq {\kappa }_{\gamma }$) gives $e_{\gamma }(X)\cap {\kappa }_{\gamma }=e_{\gamma ,\beta }(X)\cap {\kappa }_{\gamma }$. Notice that $j_{\gamma ,\beta }$ fixes $H$ by definition, and fixes ${\kappa }_{\delta }(\delta <\gamma )$ cause $j_{\gamma }$ and $j_{\beta }$ both do so, we have $j_{\gamma ,\beta }$ fixes $H_{\gamma }$, and here comes $j_{\gamma ,\beta }(X)=j_{\gamma ,\beta }(e_{\gamma }(X)\cap {\kappa }_{\gamma })=e_{\gamma }(X)\cap {\kappa }_{\beta }=e_{\gamma ,\beta }(X)\cap {\kappa }_{\beta }$.

Take $X={\kappa }_{\gamma }\subseteq {\kappa }_{\beta }$, the results gives $e_{\gamma ,\beta }({\kappa }_{\gamma })={\kappa }_{\beta }$, and take $X=\xi <{\kappa }_{\gamma }\subseteq {\kappa }_{\beta }$ we have $e_{\gamma ,\beta }(\xi )=\xi$, so $\mathrm{crit}(e_{\gamma ,\beta })={\kappa }_{\gamma }$. Lastly, we prove $j_{\gamma ,\beta }=e_{\gamma ,\beta }$ by put induction on $\beta$. The limit step comes immediately from definition of colimit, so we concentrate on the successor step, and the only thing we should prove is $j_{\beta ,\beta +1}=e_{\beta ,\beta +1}$. Construct $k:M_{\beta +1}\to N_{\beta +1}$ by sending $j_{\beta ,\beta +1}(f)({\kappa }_{\beta })$ to $e_{\beta ,\beta +1}(f)({\kappa }_{\beta })$(where $f\in L^{{\kappa }_{\beta }}\cap L$), we just need to prove $k$ is an isomorphism: $M_{\beta }=L=N_{\beta }$ gives $k=id$ then, and $j_{\beta ,\beta +1}(x)=j_{\beta ,\beta +1}(c_x)({\kappa }_{\beta })=k(...)=e_{\beta ,\beta +1}(c_x)({\kappa }_{\beta })=e_{\beta ,\beta +1}(x)$ for $x\in L$.

Let’s show $k$ is a well-defined injection: $j_{\beta ,\beta +1}(f)({\kappa }_{\beta })=j_{\beta ,\beta +1}(g)({\kappa }_{\beta })$ iff ${\kappa }_{\beta }\in j_{\beta ,\beta +1}(X)$ where $X=\{\xi \in {\kappa }_{\beta }|f(\xi )=g(\xi )\}$, but $X\subseteq {\kappa }_{\beta }$ gives $j_{\beta ,\beta +1}(X)=e_{\beta ,\beta +1}(X)\cap {\kappa }_{\beta +1}$, so ${\kappa }_{\beta }\in {\kappa }_{\beta +1}$ gives ${\kappa }_{\beta }\in j_{\beta ,\beta +1}(X)$ iff ${\kappa }_{\beta }\in e_{\beta ,\beta +1}(X)$, and ${\kappa }_{\beta }\in e_{\beta ,\beta +1}(X)$ iff $e_{\beta ,\beta +1}(f)({\kappa }_{\beta })=e_{\beta ,\beta +1}(g)({\kappa }_{\beta })$ is trivial. Replace $=$ by $\in$, we have $k$ is an embedding.

To show $k$ is surjective, suppose $y\in N_{\beta +1}$, we have $e_{\beta +1}(y)\in H_{\beta +1}$, so there’s $\phi$ and $x_1,...,x_m\in H$ and ${\gamma }_1<...<{\gamma }_n<\beta$ making $e_{\beta +1}(y)=h(┌\phi ┐,x_1,...,x_m,{\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n},{\kappa }_{\beta })$. Define $$\begin{gathered} f:{\kappa }_{\beta }\to L,\xi \mapsto h(┌\phi ┐,x_1,...,x_m, \\ {\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n},\xi ) \end{gathered}$$, if we can show $e_{\beta }(f)({\kappa }_{\beta })=e_{\beta +1}(y)$, then $e_{\beta ,\beta +1}(f)({\kappa }_{\beta })=e_{\beta +1}^{-1}(e_{\beta }(f)({\kappa }_{\beta }))=e_{\beta +1}^{-1}(e_{\beta +1}(y))=y$. Since $h$ is defined from $\Phi$, the definition of $f$ gives $\forall \xi <{\kappa }_{\beta }(\Phi (┌\phi ┐,f(\xi ),x_1,...,x_m,{\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n},\xi ))$, so $\forall \xi <{\kappa }_{\beta +1}(\Phi (┌\phi ┐),e_{\beta ,\beta +1}(f)(\xi ),x_1,...,.x_m,{\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n},\xi )$, by taking $\xi ={\kappa }_{\beta }<{\kappa }_{\beta +1}$ we get $\Phi (┌\phi ┐,e_{\beta ,\beta +1}(f)({\kappa }_{\beta }),x_1,...,x_m,{\kappa }_{{\gamma }_1},...,{\kappa }_{{\gamma }_n},{\kappa }_{\beta })$, so we only need to show $e_{\beta ,\beta +1}(f)\upharpoonright {\kappa }_{\beta +1}=e_{\beta }(f)\upharpoonleft {\kappa }_{\beta +1}$, which immediately comes from $e_{\beta +1}\upharpoonleft {\kappa }_{\beta +1}=id$.

$\square$

The next natural question is to find what large cardinal hypothesis can we derive from the existence of $0^{\#}$. To prove results, we need a more elegant and modern approach to $0^{\#}$ that is to consider it as a special countable model $(L_{\alpha },\in ,U)$, which will be called a mouse. It can give us a more accurate picture of set of ordinals in $L$.

Definition 2.7. $(L_{\alpha },\in ,U)$ will be called a ($0-$)premouse if the following holds:

1.	$(L_{\alpha },\in )\models ZF{C}^{-}+$there’s a largest cardinal $\kappa$;
2.	$(L_{\alpha },\in ,U)\models U$ is a normal $\kappa$ complete nonprincipal ultrafilter on $\kappa$;
3.	$U$ is amenable: for every $x\in L_{\alpha }$, $x\cap U\in L_{\alpha }$.

We say it’s a mouse if this $U$ is iterable.

Here comes the most important lemma about all kinds of this small stuffs.

Theorem 2.8 (The Comparison Lemma). Given two mouses $(M,\in ,U)$ and $(N,\in ,W)$, either there’s ordinal $\alpha$ so that the $\alpha -$step iteration of $(M,\in ,U)$ gives $(M_{\alpha },\in ,U_{\alpha })=(N,\in ,W)$ or the converse holds.

证明.

$\square$

Now we want a ‘minimal’ mouse in sense of this lemma, and the trick is to take an appropriate Skolem Hull.

Theorem 2.9. If there exists a mouse, there’s exactly one mouse such that it is the Skolem Hull of $\omega$ in itself, and every mouse is an iteration of it.

证明.

$\square$

Now lets’ define $0^{\#}$ as this special mouse. We want to prove that our definition of ‘$0^{\#}$ exists’ above coincides with this definition, which can also be considered as strengthening weakly amenable to fully amenable.

Theorem 2.10 (Kunen). An elemantary embedding $j:L\to L$ exists iff there is a mouse.

证明.

$\square$

3$0^{\#}$ and EM blueprint

In many set theory textbooks, the first definition of $0^{\#}$ is the unique well-founded remarkable Ehrenfeucht Mostowski blueprint(if exists), which can be easily verified as a ${\Pi }_2^1$ definable subset of $\omega$(or more confusingly, a real number). Firstly let’s replace $U$ by a sequence of indiscernibles, which can be produced quickly from our iterated ultrapowers.

Theorem 3.1 (Silver). TFAE.

1.	$0^{\#}$ exists;
2.	there’s a proper class of indiscernible ordinals for $L$.

证明.

$\square$

Now the question is how many indiscernibles actually do we need since we know a class elementary $j:L\to L$ can actually be recovered from a countable one on mouse $0^{\#}$. To derive some results, we should take a closer look at how indiscernibles will influence our models.

Definition 3.2. Extend the language ${\mathcal{L}}_{\in }$ with $\omega$ many constant symbols $c_k$. We denote the theory $\mathrm{Th}(L_{\alpha },\in ,{\alpha }_k)$ as an E.M. blueprint for any ${\alpha }_0<...<{\alpha }_k<...<\alpha$.

Theorem 3.3. For every EM blueprint $T$ and infinite ordinal $\alpha$, there exists up to isomorphism exactly one model $M(T,\alpha )$ of $T\upharpoonright {\mathcal{L}}_{\in }$ so that there’s an $\alpha -$sequence of indiscernible ‘ordinals’ $I\subseteq M(T,\alpha )$ and every $\omega -$sequence ${\alpha }_0<...<{\alpha }_k<...$ in $I$ makes $(M,E,{\alpha }_0,...,{\alpha }_k,...)$ a model of $T$; moreover, $M$ is the Skolem Hull of $I$ in itself.

证明.

$\square$

Notice that this model $(M(T,\alpha ),E)$ may not be well-founded, surely we want a theorem to tell us when it’s well-founded for every $\alpha$.

Theorem 3.4. For every EM blueprint $T$, for every $\alpha$ the model $M(T,\alpha )$ is well-founded iff it holds for every countable $\alpha$.

证明.

$\square$

But we still want properties of this $I\equiv \alpha$ be similar to the class of indiscernibles in $L$ we construct earlier. The most obvious choice is to say $I$ is CLUB, but we want a more powerful assertion.

Definition 3.5. If the set $I$ is unbounded in $Or{d}^M$ for some $(M,I)$, we’ll say that $(M,I)$ is unbounded.

Theorem 3.6. For every EM blueprint $T$, TFAE:

1.	for every $\alpha$, $M(T,\alpha )$ is unbounded;
2.	for some $\alpha$, $M(T,\alpha )$ is unbounded;
3.	for every Skolem Term $t(v_0,...,v_n)$, the sentence ‘$t(c_0,...,c_n)<c_{n+1}$ if $t(c_0,...,c_n)$ is an ordinal’ is in $T$.

we say $T$ is unbounded if these holds.

证明.

$\square$

Definition 3.7. If every ordinal $x\in M$ less than the ${\omega }^{th}$ element $i_{\omega }$ of $I$ is in the Skolem Hull of $\{i_n|n\in \omega \}\subseteq I$, we say $(M,I)$ is remarkable.

Theorem 3.8. For every unbounded EM blueprint $T$, TFAE:

1.	for every $\alpha$, $M(T,\alpha )$ is remarkable;
2.	for some $\alpha$, $M(T,\alpha )$ is remarkable;
3.	for every Skolem Term $t(v_0,...,v_{m+n})$, the sentence ‘$t(c_0,...,c_{m+n})=t(c_0,...,c_{m-1},c_{m+n+1},...,c_{m+2n+1})$ if $t(c_0,...,c_{m+n})$ is an ordinal less than $c_m$’ is in $T$.

we say $T$ is remarkable if these holds.

证明.

$\square$

Theorem 3.9. $I$ is closed in $M$ if $(M,I)$ is remarkable.

Now we define $0^{\#}$ in this way.

Theorem 3.10. TFAE.

1.	$0^{\#}$ exists;
2.	there’re uncountably many indiscernibles for some $L_{\alpha }$;
3.	there’s exactly one well-founded remarkable EM blueprint, which will be called $0^{\#}$.

证明.

$\square$

4$0^{\#}$ and The Covering Lemma

The goal of this section is to prove the following ‘lemma’.

Theorem 4.1 (Jensen). $0^{\#}$ does not exist if and only if for any set of ordinals $x$, there’s a set of ordinals $y\in L$ so that $y\supseteq x$ (so $y$ covers $x$) and $|y|=|x|+{\aleph }_1$.

Sometimes people will denote this result by ‘Jensen’s Dichotomy Theorem’, because it gives us two ways to go: either large enough cardinal exists and there’s a set of ordinals that can’t be covered so $L$ is very far from the real universe $V$, or there’s only large cardinals compatible with $L$ and $V=L$ is almost correct. Many corollaries of this lemma justifies this intuition.

Theorem 4.2. $0^{\#}$ exists iff ${\aleph }_{\omega }$ is regular in $L$.

Definition 4.3. A class inner model $M$ satisfies the weak covering property if it computes every successor of singulars correctly: for any singular $\kappa \in V$, ${\kappa }^{+}=({\kappa }^{+}{)}^M$.

Theorem 4.4. $L$ satisfies the WCP if $0^{\#}$ doesn’t exist. In fact a stronger form of $SCH$ holds in $V$: for any singular $\kappa$, ${\kappa }^{\mathrm{cf}(\kappa )}={\kappa }^{+}\cdot 2^{\mathrm{cf}(\kappa )}$.

Some remarks on first-order definability