Fix the language of set theory $\mathcal{L}=\{\in\}$. Let $\langle M,\in\rangle$ be a set or proper class model of ZFC (e.g. $M$ could be $L$, $HOD$, $V_{\kappa}$ for some inaccessible cardinal $\kappa$, etc.) consider the complete theory of this model in the language $\mathcal{L}$ that is $T_{M}:=Th(\langle M,\in\rangle)=\{\sigma\in\mathcal{L}~|~\langle M,\in\rangle\models \sigma\}$. My questions are about the possible number of countable models of this complete theory up to isomorphism $I(T_{M},\aleph_0)$ when $M$ varies over different models of $ZFC$.  Question 1: What are $I(T_{L},\aleph_0)$ and $I(T_{HOD},\aleph_0)$, $I(T_{V_{\kappa}},\aleph_0)$ (for $\kappa$ inaccessible)?  Question 2: Is there a set or proper class model $M$ of ZFC such that $I(T_{M},\aleph_0)=\lambda$ for each cardinal $\lambda\in\{1,3,4,\cdots,\aleph_{0},\aleph_1,2^{\aleph_{0}}\}$?
121617
Assume that the meta theory allows for model theoretic techniques and handling infinite sets etc (The meta theory itself is, informally, "strong as ZFC"). Also assume that I'm studying ZFC inside this meta-theory and that I'm working with the assumption that ZFC is consistent.  Let $ZFC\subset{T}$, where $T$ is a complete theory. I would like to know the number of non-isomorphic models of $T$ for some given $\kappa\geq{\aleph_{0}}$. I feel as if this should be the maximum number possible, i.e. $2^{\kappa}$.  My idea for showing this was to use Shelah's theorem that says that unstable theories have $2^{\kappa}$ many models for any $\kappa&gt;{\aleph_{0}}$ (which establishes the result for all uncountable $\kappa$). But in order to do this I first need to show that $T$ is unstable. However I'm not sure about how to do this (or even if I'm on the right track).   Any help would be much appreciated.
81648
Appearing on the second page (under the section Digression: Size worries) of the following paper about the Yoneda Lemma:     It says that $a$ $priori$ $[\mathcal C^{op},Set](H_A,X))$ is a class. I don't understand why this is the case.  My understanding is that $a$ $priori$ each member of $[\mathcal C^{op},Set](H_A,X))$ is a class.  So it seems to me that $[\mathcal C^{op},Set](H_A,X))$ could be a collection of proper classes.  I've looked at this , but it hasn't helped me much.  Any help is appreciated -Thanks
39124
I'm studying Model Theory with the book of Chang-Keisler, and I'm trying to solve the following problem stated in Chapter 5:     The union of an elementary chain $(M_i)_{i &lt; \alpha}$ of $\alpha$-saturated models is also $\alpha$-saturated, if $\alpha$ is regular.   I'm considering the following definition:     $M$ is a $\alpha$-saturated model if for all $A \subset M$, $|A| &lt; \alpha$, every $1$-type $p$ in $A$ over $T(M) = \left \{ \varphi: M \models \varphi \right \}$ is realized in $M$, i.e. there exists $a \in M$ such that $M \models \varphi(a), \, \forall \varphi \in p.$   My attempt:  Let $(M_i)_{i &lt; \alpha}$ be an elementary chain of $\alpha$-saturated models and let $M = \bigcup_{i &lt; \alpha} M_i$. Given $A \subset M$ with $|A| &lt; \alpha$ and $p$ a $1$-type in $A$ over $T(M)$, if I prove that there exists $i &lt; \alpha$ such that $p$ is a type over $T(M_i)$, we could conclude that there exists $a$ in $M_i$, with $M_i \models \varphi(a)$ for all $\varphi \in p$. Hence $a \in M$ and $M \models \varphi(a)$, since $M_i$ is an elementary substructure of $M$.  So, how can I proof the existence of such $i &lt; \alpha$?
48412
I so stuck with a problem of set theory. But first a recursive definition:   Define $R_0=\emptyset$ If $R_\alpha$ is defined, then $R_{\alpha+1}=\mathcal{P}(R_\alpha)$ (the power set). For a limit ordinal $\gamma$, if $R_\alpha$ is defined for all $\alpha&lt;\gamma$, then define $R_\gamma=\displaystyle\bigcup_{\alpha&lt;\gamma} R_\alpha$   We define $\text{BF}=\displaystyle\bigcup_{\alpha\in\text{OR}} R_\alpha$ (the class of well founded sets).  Next, my problem     Take $A\subseteq \text{BF}$ a proper transitive class such that $(A,\in)\models\text{ZF}$. Prove that $A$ is almost universal.    I think that the exercise is false because if it was true, then we could conclude that the strongly inaccessible cardinals doesn't exists. This because if $\kappa$ is strongly inaccessible then $R_\kappa$ satisfies the hypothesis but $R_\kappa$ isn't almost universal. I really appreciate any hint or/and suggestion.   Edit: my counterexample is wrong. But, then, how can I solve the exercise?
19105
Given a collection of sets $\mathcal{C}$ and $E$ an element in the $\sigma$-algebra generated by $\mathcal{C}$, how do I show that $\exists$ a countable subcollection $\mathcal{C_0} \subset \mathcal{C}$ such that $E$ is an element of the $\sigma$-algebra, $\mathcal{A}$ generated by $\mathcal{C_0}$?    The hint says to let $H$ be the union of all $\sigma$-algebras generated by countable subsets of $\mathcal{C}$....although I don't know why.
121451
Every proof that I read seems to assume that $|L|\leq\aleph_0$. But then how do you model things like field over $\mathbb{R}$ without running out of variable symbols?  More importantly, how can I prove that $|L|\leq\aleph_0$?
79740
This means: if $V\models{\sf ZFC}$ in such a way that $2^{\aleph_0}=\aleph_2$, can $2^{\aleph_1}$ be anything it ought to be? By the way, can $2^{\aleph_0}=\aleph_2$ be considered the 'simplest' case of $\lnot{\sf CH}$ in some sense?
89287
It is well-known (forgive the pun) that the axiom of choice (which states that the product of every non-empty family of non-empty sets is non-empty) implies that the proper class of all cardinal numbers is well-ordered, at least in the presence of the ZF axioms.  Does the converse hold? If not, is the statement that the proper class of all cardinal numbers is well-ordered consistent with axioms like determinacy, which, while interesting, contradict full-blown choice?
114612
$\aleph_1$ is the cardinality of the countable ordinals. It is the least cardinal number greater than $\aleph_0$, and assuming the continuum hypothesis it's equal to $\mathfrak{c}$, the cardinality of the the real numbers.    My question is, is it possible for all $\aleph_1$ subsets of $\Bbb{R}$ to have Lebesgue measure $0$?  This is, of course, impossible assuming the continuum hypothesis, because then all sets of real numbers would have measure $0$, which is absurd.  But is $\mathsf{ZFC}$ + $\lnot\mathsf{CH}$ + "all subset of $\Bbb{R}$ of cardinality $\aleph_1$ have Lebesgue measure $0$" consistent?  If not, what if we replaced Lebesgue measure with some other measure?  It may be worth noting that, although there may be subsets of $\Bbb{R}$ with cardinality $\aleph_1$, there are no subsets of $\Bbb{R}$ which have the order-type $\omega_1$ (the order-type of the countable ordinals) under the usual ordering on $\Bbb{R}$.  Any help would be greatly appreciated.  Thank you in Advance.
7450
In ZFC with class terms a class is merely a unary predicate $\varphi$ of the language which we then write as  $$\{X\mid\varphi(X)\}$$  which only suggests that we have built a collection out of these objects. We write $Y\in\{X\mid \varphi(X)\}$ for $\varphi(Y)$. Further we say that such a class (term) $\mathcal C$ is a set if   $$\exists X\forall Y[Y\in X\leftrightarrow Y\in\mathcal C].$$  In NBG and similar set theories (which build on classes instead of sets) every set is also a class. But is this true for this ZFC approach? For me it seems that a class is just a formula. And there are sets that I cannot fully describe using such a formula, e.g. a set given to me by the axiom of choice.     Are sets and classes in ZFC (with class terms) just different concept, each one not a "sub-concept" of the other?     After Asaf's answer:  I do not feel very well when writing $\{X\mid X\in A\}$ for arbitrary sets $A$ because for me $A$ is not a symbol of the language. However, when my set $A$ is definable by a predicate $\varphi$ with   $$\mathrm{ZFC}\vdash \exists A \varphi(A) \quad\text{ and }\quad \mathrm{ZFC}\vdash\varphi(A)\wedge \varphi(B)\to A=B,$$  then I could write $\{X\mid \forall A[\varphi(A)\to X\in A]\}$, and this would feel okay. But my problem are the sets for which there is no such $\varphi$. I do not know how I can include such an abbreviating notion (like $\{X\mid X\in A\}$) into a formal proof without feeling not quite sure what I am actually doing.
34114
Is it possible in ZFC that you have a cyclic containment of sets, e.g., a inclusion like $A \in B$ and $B \in A$?  I never took set theory classes, I am just curious.
109014
In predicate logic, to define a language, we consider the following $3$ sets:  (1) $V$, the set of variables;  (2) $F$, the sets of function symbols;  (3) $P$, the set of predicate symbols with their arities.   Now I am concidering about determine the cardinality of a language over a set with arbitary cardinality, $κ$(Note that $κ$ can be either finite, countablely infinite or uncountablely infinite ).  May I please ask:  If the cardinality of $V$, $\#V = κ$, what conditions on $V$ , $F$ and $P$ will give us a language over $V$, $F$ and $P$ with cardinality $κ$?  And may I please ask if it is the fact that if $\kappa$ is finite, then to get a language with cardinality $\kappa$ we need $(V+F+P)^{&lt;\omega}=(\kappa+F+P)^{&lt;\omega}=\kappa$, so we can never have a language of cardinality $\kappa$. I think it is something weird here. Am I wrong with something?  I had googled it but I have get nothing from websites, now I am confused. Could someone please help? Thanks so much!  EDIT: Latter I am told that in my course, the cardinality of a language refers to the set of expression instead of this of the set of strings. So now I need to consider further about when a string makes sense.  EDIT':Now I am considering that if we can find out a general way to determine the cardinality of expression, which may related to combinatorics. See this question:  Any help would be appreciate.
45993
How to show that the class of all sets of a particular cardinality ,say $h$ is not a set.   My argument:  I assume that I've shown the following lemma.  Lemma: If $X$ is an infinite set of cardinality $h$ , $a\ne x$ for every $x\in X$ then $\{a\}\cup X$ is of cardinality $h$ too i.e. adding new element to an infinite set doesn't change cardinality.  Let $h$ be an infinite cardinal number. Let $T_h=\{x:\text{x is a set of cardinality h}\}$. Suppose $T_h$ is a set.   Let $a$ be any set and consider any $t\in T_h$. Let $R=t\cup \{a\}$. Using the above lemma, $R\in T_h$.   So, $a\in R\subset \bigcup T_h$. As $T_h$ is a set then $\bigcup T_h$ is also a set.  As $a$ was arbitrary set then we have, $a\in\bigcup T_h$ for every set $a$. Hence, the set $\bigcup T_h$ contains all sets which meas that the collection of all sets is a set, contradiction.    If $h$ is non-zero finite cardinal number. Define $T_h$ the same way as above. Let $a$ be any set, Let $A_1=a, A_{n+1}=A_n\times a$. Consider $A=\{A_1,A_2,...,A_h\}$. So, $A\in T_k$. Hence, $a=A_1\in \bigcup T_h=K$. So K is a set containing all sets $a$ since $a$ was arbitary,contradiction.  My main question is, Is there any flaws in my argument? If yes, point it out please.   If the argument is valid, Is there any standard proof to show the same fact?
114082
ZFC works as a foundation because it can prove many sentences that are "translations" of theorems from "standard" mathematics into the language of ZFC.  But there's a subtlety. When we say, "ZFC can found most of mathematics," what do we really mean?  Do we mean   ZFC proves most theorems (suitably translated into the language of sets) in the mathematics literature A metatheory that makes sense of "consistency" + the assumption that ZFC is consistent can be used to prove most theorems (suitably translated into the language of sets) in the mathematics literature A metatheory that makes sense of "models" + the assumption that ZFC has a model can be used to prove most theorems (suitably translated into the language of sets) in the mathematics literature A metatheory that makes sense of "models" + the assumption that ZFC has a standard model can be used to prove most theorems (suitably translated into the language of sets) in the mathematics literature Something else?
2212
My question is about a proof of Theorem 4.12 given in the text Introduction to Set Theory by Hrbacek and Jech. The authors first prove Theorems 4.9, 4.10, and 4.11. For reference, I quote these theorems below.   Let $V$ be the class of all sets, $\operatorname{Ord}$ be the class of all ordinals, and $G:V\to V$ be a class function.  Theorem 4.9: Transfinite Recursion Theorem  There exists a class function $F:\operatorname{Ord}\to V$ such that $F(\alpha)=G(F\restriction \alpha)$ for all $\alpha\in\operatorname{Ord}$.  Theorem 4.10: Transfinite Recursion Theorem, Parametric Version  There exists a class function $F:\operatorname{Ord}\to V$ such that $F(z,\alpha)=G(z,F_z\restriction \alpha)$ for all $\alpha\in\operatorname{Ord}$ and $z\in V$ where $F_z\restriction \alpha:=\{\langle\beta,F(z,\beta)\rangle\mid\beta&lt;\alpha\}$.  Theorem 4.11:  Let $G_1,G_2,G_3$ be class functions from $V$ to $V$. There exists a class function $F:\operatorname{Ord}\to V$ such that (1) $F(0)=G_1(\emptyset)$ (2) $F(\alpha+1)=G_2(F(\alpha))$ for all $\alpha\in\operatorname{Ord}$ (3) $F(\alpha)=G_3(F\restriction\alpha)$ for all limit $\alpha\neq 0$   Then they say A parametric version of Theorem 4.11 is straightforward and we leave it to the reader. From my attempt, the parametric version of Theorem 4.11 is as follows (Honestly, I'm not sure if my attempt is correct or not)  Let $G_1,G_2,G_3$ be class functions from $V$ to $V$. There exists a class function $F$ such that, for all $z\in V$ $F(z,0)=G_1(z,\emptyset)$ $F(z,\alpha+1)=G_2(z,F_z(\alpha))$ for all $\alpha\in\operatorname{Ord}$ $F(z,\alpha)=G_3(z,F_z\restriction\alpha)$ for all limit $\alpha\neq 0$  Next they make use of this parametric version of Theorem 4.11 in the proof of Theorem 4.12. Theorem 4.12:  For any set $a$, there is a unique infinite sequence $(f_n\mid n\in \Bbb N)$ such that (1) $f_0=a$ (2) $f_{n+1}=G(f_n,n)$ for all $n\in \Bbb N$  The proof for Theorem 4.12 given in the text is as follows:  Let $G:V\to V$ be a class function. We want to find, for every set $a$, a sequence $(f_n\mid n\in \Bbb N)$ such that $f_0=a$ and $f_{n+1}=G(f_n,n)$ for all $n\in \Bbb N$. By the parametric version of the Transfinite Recursion Theorem 4.11, there is class function $F$ such that $F(0)=a$ and $F(n+1)=G(F(n),n)$ for all $n\in \Bbb N$. Now we apply the Axiom of Replacement: There exists a sequence $(f_n\mid n\in \Bbb N)$ that is equal to $F\restriction \omega$ and the Theorem follows.   I don't understand how the parametric version of Theorem 4.11 is used to derive the class function $F$ in the proof of Theorem 4.12. Can someone please help me fill in the blanks?
49565
I was reading David Marker’s Model Theory and came upon the following problem in Chapter 3.  Setting  Let $ \mathcal{M} $ be a saturated $ \mathcal{L} $-structure.   A definable subset $ X \subseteq M $ is said to be strongly minimal if and only if $ X $ is defined by an $ \mathcal{L}_M $-formula $ \phi(v) $ and every subset of $ X^{\mathcal{M}} $ definable in $ \mathcal{L}_{\mathcal{M}} $ is either finite or co-finite. $ {\text{acl}^{X}}(A) \stackrel{\text{df}}{=} X \cap \text{acl}(A) $ is the set of all $ b \in X $ such that $ b $ is algebraic over $ A $, i.e., there is an $ \mathcal{L}_{A} $-formula $ \psi $ such that $ \mathcal{M} \models \psi(b) $ and $ |\{ c \mid \mathcal{M} \models \psi(b) \}| &lt; \infty $. $ A \subseteq X $ is called independent if and only if $ a \notin \text{acl}^{X}(A \setminus \{ a \}) $ for all $ a \in X $.   Problem  I would like to show that if $ b \in {\text{acl}^{X}}(A \cup \{ c \}) $ and $ b \notin {\text{acl}^{X}}(A) $, then $ c \in {\text{acl}^{X}}(A \cup \{ b \}) $, and that for any $ A \subseteq X $, there is an independent $ B \subseteq A $ so that $ A = {\text{acl}^{X}}(B) $.
25222
On the top of my head, I cannot think of any fields of cardinality more than that of the reals. (It is known that the process of algebraic closure does not increase the cardinality of an infinite field.)  What is the simplest way to give an example of a field (and an ordered field) of a specific cardinality $\alpha$?  I see there is the "Field" of surreal numbers, but it is a proper class rather than a set (and hence do not have a cardinality as such). However, there seems to be some modified construction which gives proper fields with the cardinality of some strongly inaccessible cardinal.