# Tag Info

6

No, this is not inconsistent, at least not relative to the existence of a Mahlo cardinal. Suppose that $\kappa$ is a Mahlo cardinal, then $V_\kappa$ has a club of ordinals $\alpha$ such that $V_\alpha\prec V_\kappa$. Therefore there is a stationary set of inaccessible cardinals satisfying this. Simply enumerate these inaccessible cardinals and let $\alpha$ ...

5

Your theory is equiconsistent with what is known as the Levy scheme, or Ord is Mahlo, which is strictly weaker in consistency strength than the existence of a Mahlo cardinal. On the one hand, your theory implies Ord is Mahlo, since if $C$ is any proper class definable club (definable with parameters), then it will follow that $C$ is unbounded in the ...

4

The point is that set theory has its own definition of finite. An internal definition. It's true that FOL can't truly define "finite", and if $\sf ZFC$ is consistent then it has a model $M$ such that $\{x\mid M\models x\text{ is a finite ordinal}\}$ is an uncountable set (although $M$ thinks that this set is countable); when we work in set theory we have ...

3

In set theory this might be expressed in a variety of ways. Perhaps the simplest, most self-contained approach is to require a well-ordering of a set which is also a well-ordering if taken in the opposite direction. Finite sets have this property (by induction), and the converse is arguably then just a matter of definition. However the definition supports ...

3

No, $f$ is not surjective, as the range is the set of sequences with only finitely many non-zero entries. For example, you will never have $$f(n)=(1,2,3,4,\ldots)\ .$$

3

Let $A$ be some infinite summable set, say, $A=\{1,10,100,\dots\}$. Partition $A$ into infinitely many disjoint infinite subsets $A_1,A_2,A_3,\dots$. Let $\mathbb N\setminus A=\{b_1,b_2,b_3,\dots\}$. Then $\mathbb N$ is the union of the disjoint infinite summable sets $$A_1\cup\{b_1\},A_2\cup\{b_2\},A_3\cup\{b_3\},\dots.$$

3

First we should clarify what absoluteness between $M \subseteq N$ mean. I will assume that $M$ and $N$ are transitive sets. $\varphi$ is absolute between $M$ and $N$ means that $M \models \varphi$ if and only if $N \models \varphi$. (Note that if and only if.) From your question, if $M \models \varphi$, then $N \models \varphi$ is just upward absoluteness ...

3

Let's say that a formula $\Phi(x, y)$ is a satisfaction predicate for $\mathcal L_\in$ if is satisfies the Tarski clauses; that is, if: ($\in$) $\Phi(x_i\in x_j", a) \leftrightarrow a(i)\in a(j)$ ($=$) $\Phi(x_i = x_j", a) \leftrightarrow a(i) = a(j)$ ($\wedge$) $\Phi(\phi \wedge \psi", a) \leftrightarrow \Phi(\phi", a) \wedge \Phi(\psi", a)$ ...

2

For an easier example, try letting $X = \mathbb{R}$ and $M = \{(a, +\infty) : a \in \mathbb{R}\}$ the set of all open intervals that extend to infinity on the right. Then $M_1 = \{(a, +\infty), (-\infty, a] : a \in \mathbb{R}\}$ consists of open and closed intervals. Any intersection of intervals is again an interval (easy exercise) so $M_2$ consists only ...

2

For every natural number $n \ge 1$ define its support $\text{supp}(n)$ to be the set of prime divisors of $n$. We get a surjective map from $\{1,2,3, \ldots,\}$ to the set of finite parts of the set of prime numbers. $$\text{supp} \colon \mathbb{N}_{>0} \to \mathcal{P}_{fin}(P)$$ Each nonvoid finite subset of $P$ (the primes ) has an infinite fiber which ...

2

Here's another take on the reinterpreted question: Let $A_0 = \{a^2 : a \in \mathbb{N}\}$ be the set of all perfect squares. Let $A_1 = \{a + 1: a \in A_1\} - A_1$. And inductively, define $$A_n = \{a + n: a \in A_1\} - \bigcup_{i < n} A_i$$ So this just shifts the perfect squares, and throws away anything that was already seen. Since the squares ...

2

Existence is a semantic property, they require us to talk about a model (internally to that model, or externally to that model). Something exists in the model, if it is in the model. There's no other way to say it. We can sometimes prove that certain objects exist, even if we have no explicit means of producing them. This, for example, is how the axiom of ...

2

Actually your question follows easily from the existence of a Ramsey cardinal $\kappa$, if you only ask your sequence to be indescernible over a subset of size $<\kappa$ and $|L|<\kappa$. Recall that a Ramsey cardinal is an infinite cardinal satisfying that for each function $f:[\kappa]^{<\omega}\rightarrow\tau$, $2\leq\tau<\kappa$ there is ...

2

Then $x \cup \{ x \}$ is the set of all sets and the usual paradoxes ensue.

2

The paradoxes of set theory don't follow from the existence of a universal set. Instead they follow from the combination of naive-like separation axioms, and a universal set. To clarify, Russell's paradox showed that the Comprehension schema is inconsistent. Namely, if $\varphi(x)$ is a formula, then $\{x\mid\varphi(x)\}$ doesn't have to be a set. We then ...

2

I can't see what you mean by $H=\{a,a^2=e\}$. In any case, the notation $H=\{a\mid a^2=e\}$ means the following (here I am assuming that $H$ is supposed to be a subset of some bigger set $X$, otherwise it makes no sense): "H is the set containing all the elements $a\in X$ such that $a^2 = e$" E.g., if $X=\mathbb R$ and $e$ denotes the identity element ...

1

Say that a set $x$ is definable if there is a formula "$\phi$" in the language of set theory $\mathcal L_\in$ with only one free variable such that: for all $z$, $z\in x$ iff "$\phi$" holds of $z$. Then a simple way to read the question is: Question: Is every set definable? To ask this question, we need to say what it means for a formula in $\mathcal ... 1 Here's an answer to your reinterpreted question. Associate to each natural number$m$except$0$and$1$the pair$(p,k)$, such that$p$is the least prime appearing in the factorization of$m$,$m = p^n k$for some$n$, and$p$does not divide$k$Let$A_{p,k}$be the set of natural numbers associated to$(p,k)$. This is nonempty whenever$p$is ... 1 with the very little information you have provided, I can only guess that$X$is some element &$A$is some set and$\bar A$is complement of$A$. In this case it is true, obviously, because if$X$is in universal set but not in$A$it has to be in$\bar A$. And you should write element with small letter i.e.$x$not$X$1 If$A,B \subseteq X$and by$\overline A$you mean$X \backslash A$, then yes. For$x \in X$, then$x \notin A$if and only if$x \in \overline A$. P.S. You shouldn't write$B - A \in B \cap \overline A$but$B-A \subseteq B \cap \overline A$(or$B-A \subset B \cap \overline A$, depending on your choice of notation). The terms "being an element of" (i.e. ... 1 Sometimes the answer is yes, for example consider$T$which includes these two axioms:$V=L$. "There are no transitive models of$\sf ZFC$". Because of the first statement any countable transitive model has the form$L_\alpha$for some$\alpha<\omega_1$. But the second statement already ensures that this$\alpha$is unique, since any larger$\alpha'\$ ...

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