# Tag Info

9

To begin with, see here. I do not know of a written account where details are worked out more precisely, but the basic idea is the following: Lebesgue outer measure on $\mathbb R^n$, $\mu^*$, is defined for all subsets of $\mathbb R^n$, whether measurable or not, extends Lebesgue measure $\mu$, and satisfies monotonicity: $\mu^*(A)\le\mu^*(B)$ for ...

5

Assume the continuum hypothesis, for simplicity. Let $A\subseteq\Bbb R$ be such that both $A$ and $\Bbb R\setminus A$ meet every open interval on an uncountable set (for example, if $A$ is a Bernstein set). We order $\Bbb R$ in the order type $\omega_1+\omega_1$. Take some bijection $f$ of $\Bbb R$ with $\omega_1$ then apply it twice, to $A$ and to its ...

5

Are you sure that what you're trying to prove is true? It looks false to me, for the following reason. Let $\mathfrak c$ denote the cardinal of the continuum. It is easy (by transfinite recursion) to partition $\mathbb R$ into two sets $A$ and $B$, each of which has $\mathfrak c$ points in every nondegenerate interval. Now well-order $\mathbb R$ as ...

4

Cofinality is a very much non-increasing function, and for every regular cardinal $\kappa$ we can find a limit cardinal $\lambda$ such that $\kappa=\operatorname{cf}(\lambda)\leq\lambda$. In fact, we cannot prove the case of equality. That is, if a regular limit cardinal exists then we can prove the consistency of $\sf ZFC$, so adding to $\sf ZFC$ the ...

3

I think that you miss a certain point there. The proof is not that the ordinal $P(\alpha,\beta)\leq\max\{|\alpha|,|\beta|\}^2$, but rather that $|P(\alpha,\beta)|$ is less than that. The inequality is of cardinalities, not order types. And indeed, note that $P(\alpha,\beta)$ when both are countable, is a countable ordinal. Now we can go about our induction ...

3

Let me answer your first question, since I'm not quite sure I understand the second one at its current form. Russell's definition was given much before von Neumann introduced the term "proper class". In addition to that, von Neumann suggested we choose a representative from each equivalence class, and using the axiom of choice and the definition of ...

3

This all comes from the definition of elementary, and is a little more delicate than you might think. Model theoretically, a model $M$ is an elementary submodel of a model $N$ if $M$ is a submodel of $M$ (the universe of $M$ is a subset of that of $N$, and all interpretations of function/predicate/constant symbols are just the restrictions of those of $N$), ...

3

The second formula is an axiom of the first order logic with equality, together with the following axioms about equality: $x=x$. $x=y\rightarrow(\varphi\leftrightarrow\psi)$, where and $y$ is free for $x$ in $\varphi$ and $\psi$ is replacing some free variable $x$ by $y$ in $\varphi$. Unlike Extensionality, which is a proper axiom, these equality axioms ...

2

Yes, and that we are considering $\mathsf{ZFC}$ is not really relevant (except that for $\mathsf{ZFC}$ we know we do not have a recursive model): If $T$ is a consistent theory with an r.e. axiomatization, we can carry out the argument of Henkin's proof of the completeness theorem for $T$, and obtain a $\Delta^0_2$ model of $T$ whose universe is a ...

2

Define $f$ inductively on $\kappa$. Suppose that $f \restriction \beta$ has been appropriately defined. Set $$\begin{gather} B = \{ \alpha < \beta : \{ \alpha , \beta \} \in A \} \\ C = \{ \alpha < \beta : \{ \alpha , \beta \} \notin A \}. \end{gather}$$ Now there is a $\delta\in \kappa \setminus ( f [ B ] \cup f [ C ] )$ such that $$\begin{gather} ... 2 The second fact is a special case of the indiscernibility of identicals and is part of the definition of equality. The principle says, for all predicates \phi (u, t_1, t_2, \ldots, t_n),$$\forall x . \forall y . (x = y) \to (\phi (x, t_1, t_2, \ldots, t_n) \leftrightarrow \phi (y, t_1, t_2, \ldots, t_n)) where $t_1, t_2, \ldots, t_n$ are arbitrary ...

2

No, there is no way to define such injection for two major reasons: It is consistent with $\sf ZF$ with the axiom of choice, the usual axioms of set theory, that there is some $S\subseteq\Bbb R$ such that $\aleph_0<|S|<2^{\aleph_0}$. In that case we can easily define the partition of singletons of the elements of $S$, and the rest is just one part. ...

2

Just expanding on the comments here: Yes. When we prove theorems in model theory, we're doing just the same sort of mathematics we do when we prove theorems in group theory or real analysis. This mathematics can be formalized in ZFC, and the theorems in model theory, just like the theorems in any other branch of math, are theorems of ZFC. Hence they hold ...

1

Assuming that I correctly understood the definitions involved, the following outline should help. Let $S_0=\{0\}$. Given $S_n$ for some $n\in\omega$, let $S_{n+1}=S_n\cup g[S_n]$, and let $S_\omega=\bigcup_{n\in\omega}S_n$. Clearly $S_\omega$ is $g$-inductive, and $S_\omega\subseteq S$, so $S=S_\omega$. Let $x_0=0$. Given $x_n$ for some $n\in\omega$, let ...

Only top voted, non community-wiki answers of a minimum length are eligible