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36

This is the commonplace clash between the semi-Platonic view of the laymathematician and the foundational approach for mathematics through set theory. It is often convenient, when working in "concrete" mathematics, to assume that there is a single, fixed universe of mathematics. And everyone who took a course or two in logic and set theory should be able to ...


16

In axiomatic set theory, the axioms themselves are the definition of the notion of a set: A set is whatever behaves like the axioms say sets behave. I half-agree with this. But recall that the axioms of group theory don't axiomatize the concept "element of a group." Rather, they axiomatize the concept "group." In a similar way, the axioms of ZFC don't ...


8

There is an axiomatic definition of "set", but it's actually rather uninteresting: (this space intentionally left blank) The point of things like the ZFC axioms isn't to define the notion of a set: it's to define the notion of a "universe (of sets)". Each universe can be said to give a notion of "set", but that must be understood in the same sense that ...


8

The axioms of ZFC (or any other sufficiently strong first-order formal system) cannot define the notion of "set", in the sense that you're looking for, namely that ZFC cannot pin down a unique structure that satisfies ZFC. Why so? Because ZFC cannot prove its own consistency, by Godel's incompleteness theorem, and hence ZFC cannot prove that there is a model ...


6

There are thorough answers by Asaf Karagila and user21820. I want to point out a different issue: there is more than once concept or notion of "set". Paradoxes like Russell's paradox showed that the original "naive" concept of "set" is inconsistent, and so some of the properties of these sets needed to be discarded. But there are many ways to do that. ZFC ...


6

Any witness of $\omega$-inconsistency of ZFC can be effectively transformed to a witness of $\omega$-inconsistency of ZF as follows. Suppose $\phi(x)$ is a formula such that for each numeral $n$, ZFC proves $\phi(n)$ and also $(\exists x \in \omega)\neg \phi(x)$. Let $\psi(x) = \phi^L(x)$ be obtained from $\phi(x)$ by restricting its quantifiers to $L$. Then ...


5

Of course the question is whether there's a countable $\epsilon$-cover for every $\epsilon>0$. And we really can't tell, because we don't know whether $\aleph_2>c$. The answer is no if $\aleph_2>c$ (and certainly yes if $\aleph_2\le c$): If $M$ is a metric space and there is a countable $\epsilon$-cover for every $\epsilon>0$ then $|M|\le c$. ...


4

No, your solution does not work because you appealed to the axiom of choice in choosing $z$ in order to define the injection. Instead, use the fact that $f$ is a bijection to conclude that $f(\{(x,y)\mid y\in b\})\cap b$ creates a partition of $b$ indexed by $a$, and therefore defines a surjection from $b$ onto $a$.


3

Update: I'm marking this answer as community wiki so as not to "leech reputation." OP here. I want to include some information I've gathered from answers and comments to this question (as well as several others) which I think most directly address what I was asking about. I think the most succinct answer to my question is goblin's, where they write that ...


3

I'll just take a moment to set up some notations. Let $T$ be an arbitrary tree, with ordering $\sqsubseteq$. Given $t \in T$, $\operatorname{ht}_T (t) = \operatorname{ot} \{ s \in T : s \sqsubset t \}$ is the height of $t$ in $T$. For an ordinal $\alpha$, $T_\alpha = \{ t \in T : \operatorname{ht}_T (t) = \alpha \}$ is the $\alpha$th level of $T$. For $t ...


3

Your argument about (1) is correct, and it is really the reason that it cannot contain a club. If $\kappa>\omega$ is regular, and $D\subseteq\kappa$ is a club, then it must have limit points of every cofinality below $\kappa$ (alternatively, simply note that $\{\alpha<\omega_2\mid\operatorname{cf}(\alpha)=\omega_1\}$ is stationary, and therefore the ...


3

The axiom of infinity fails in $V_\omega$. Even though each of the natural numbers is itself in $V_\omega$, the set of all of them is not itself a member of $V_\omega$. It only shows up as an element of $V_{\omega+1}$. In $V_{\omega+1}$ the MK axiom of infinity still fails. You do have $\omega$ as an element of $V_{\omega+1}$, but it is not a "set" (as ...


3

There are syntactic interpretations of PA into ZF-I and of ZF-I into PA. For the purposes of equiconsistency, it does not matter whether these are inverses of each other, an issue that is discussed by Kaye and Wong in this paper. They are worried about something stronger than mere equiconsistency. I will assume we are talking about interpretations in their ...


3

There's no paradox here. If there are no sets that contain themselves (such as in modern ZFC set theory), then $\{x\mid x\in x\}$ is simply the empty set -- and we know the empty set exists! If there are sets that contain themselves (but not too many of them), then we can still assert that there is a set that consists of all of them. That set may or may ...


2

Let $\kappa$ be a Mahlo cardinal in $V$, i.e. let $\kappa$ be inaccessible such that $S:= \{ \alpha \in \kappa \mid \alpha \text{ is regular} \}$ is stationary in $\kappa$. First, note that $L \models \kappa \text{ is inaccessible}$. Indeed, if $L \models \kappa \text{ is not a cardinal}$, then there is some $\mu < \kappa$ and some $f \in L$ such that $L ...


2

HINT: It seems easier to show the contrapositive of the forward implication. Suppose that $C$ is a club disjoint from $\{\min N_\alpha:\alpha<\kappa\}$. Show that for each $\alpha<\kappa$ the set $C\cap\min N_\alpha$ has a maximum element $\gamma_\alpha$. Define $f:\bigcup_{\alpha<\kappa}N_\alpha\to\kappa$ by $$\large ...


2

Take a ordinal $\alpha_0$. Let $G(\alpha_0)$ be the group generated by $\alpha_0$. Then $|G(\alpha_0)|<\kappa$. Define $\alpha_1$ to be any ordinal larger than all element of $G(\alpha_0)$ repeat this construction with $\alpha_1$. This gives $\alpha_n$ increasing. Your desired ordinal subgroup is then $\alpha=\sup\alpha_n$.


1

HINT: Suppose that $\alpha$ is a subgroup of $\kappa$. Let $G_0$ be the closure of $\alpha+1$ under the group operations. Let $\beta_0=\sup G_0$, and let $G_1$ be the closure of $\beta_0+1$ under the group operations. In general, given a bounded subgroup $G_n$ of $\kappa$, let $\beta_n=\sup G_n$, and let $G_{n+1}$ be the closure of $\beta_n+1$ under the ...


1

I think the informal argument that the cumulative hierarchy is a model of ZFC is meant to appeal to some kind of direct perception of the cumulative hierarchy, much like the informal argument that $\mathbb{N}$ is a model of PA relies on some kind of direct perception of $\mathbb{N}$. I don't expect the informal argument to be formalizable in weak theories ...


1

No, you don't have a problem. Nobody told you that doing these things will enlarge your set. At some point, your sequence is going to stabilize. Namely, there will be some $\alpha$ such that $C_\alpha=C_\beta$ for all $\beta>\alpha$. Try and see what happens if $f=\operatorname{id}$. This understanding extends to any situation when $C_0$ is already ...


1

If every set of cardinal numbers is well-ordered by cardinality, then in particular any two cardinals are comparable by cardinality (i.e., trichotomy holds). To see that trichotomy implies the well-ordering theorem: Let $X$ be any set that we want to well-order, and let $\alpha$ be its Hartogs number, which is the first ordinal that cannot be injected into ...


1

Exercise III 6.17: For $f \in \mathcal F$ let $S_{f} := \{ \alpha < \omega_{1} \mid f(\alpha) < g(\alpha) \}$. First consider the case that $g \colon \omega_{1} \to \lambda$ satisfies $g(\alpha) = \omega_{\alpha}$ for every $\alpha < \omega_{1}$. By Fodor, we may, for each $f \in F$, choose some stationary $T_{f} \subseteq S_{f}$ and some cardinal ...


1

HINT: For the first part, if $c_\beta\ne\bigwedge_{\alpha<\beta}c_\alpha$, then $c_\beta<\bigwedge_{\alpha<\beta}c_\alpha$, and $C\cup\left\{\bigwedge_{\alpha<\beta}c_\alpha\right\}$ is a strictly larger descending chain with no minimum element. For the second part you must prove two things: that $A$ is an antichain, and that it is maximal. For ...



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