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8

After working out a little bit of math, I suddenly remembered that this was asked, in some variation, on MathOverflow. Thankfully, Nate Eldredge posted an answer to this question which had just the right reference. Hardin, Christopher S.; Taylor, Alan D. An introduction to infinite hat problems. Math. Intelligencer 30 (2008), no. 4, 20–25. MR2501394. ...


7

According to the book Albrecht Pietsch: History of Banach Spaces and Linear Operators (Birkhäuser, 2007, DOI: 10.1007/978-0-8176-4596-0) this is an open problem. (Or at least it was at the time of publishing the book.) I quote from p.586: We list two more consequences of the axiom of choice, which have already been discussed in 1.2.2 and 1.5.10, ...


6

The following should be a comment, because it doesn't attempt to say who first wrote down the ZFC axioms and called them ZFC, but it's way too long. I'd better begin by quoting, from the historical appendix in Peter Freyd's book "Abelian Categories": "The origin of concepts, even for a scholar, is very difficult to trace. For a nonscholar such as me, it is ...


5

What you want is the usual argument that the bounding number $\mathfrak{b}$ is uncountable. Given a countable family $\mathscr{F}=\{f_n:n\in\omega\}\subseteq{}^\omega\omega$, define $$f:\omega\to\omega:n\mapsto 1+\max_{k\le n}f_k(n)\;;$$ then for each $n\in\omega$ we have $f(k)>f_n(k)$ for all $k\ge n$. Using this, it’s easy to construct recursively a ...


4

Since my comments on the comment section were getting a little big, I decided to expand the issue on first-order logic into an answer of its own. But first, let me preface this with a caveat, due to a paper by Gregory Moore ("Historians and Philosophers of Logic: Are They Compatible?"): it seems that establishing priority claims in cases such as this is ...


4

Reading in Kanamori's introduction to the Handbook of Set Theory, the text seems to support Robert Israel's answer. On p.10 it says that von Neumann wrote in his Ph.D. dissertation an axiomatization of set theory, however it is not clear to me (from this text, and from other texts that I have seen this mentioned) whether or not his was the basis for the ...


4

Here are some relevant quotes from Fraenkel, Bar-Hillel, Levy's "Foundations of set theory": "Zermelo's vague notion of a definite statement did not live up to the standard of rigor customary in mathematics ... In 1921/22, independently and almost simultaneously, two different methods were offered [by Fraenkel and Skolem] for replacing in the axiom of ...


3

Since the $X_i$ are distinct, there is an $m\in\omega$ such that the sets $X_i\cap m$ are distinct. It follows that the $k$ sets $X_i\cap n$ are distinct for each $n\ge m$. Thus, for each $n\ge m$ there is an $f_n\in{}^{\wp(n)}\wp(n)$ such that $f_n[X_i\cap n]=Y_i\cap n$ for each $i<k$, and $\{\langle n,f_n\rangle:n\ge m\}$ is an infinite subset of ...


2

I don't have an answer to your question, but I did search through quite a few set theory books this morning and I made notes of what I found in case you or others are interested. The topic seems less covered in books than I expected, and I suspect you'll have to consult journal articles to find much of significance (unless you can read Hessenberg's and ...


2

The trivial examples are a one-point subspace and an infinite discrete subspace of $\Bbb R$. The simplest non-trivial examples for $A$ are the rationals, the Cantor set, the irrationals, and the Cantor set minus any one of its points: each is homeomorphic to its square. In none of these cases is this entirely trivial to prove. The Cantor set is ...


2

The set of cardinals below $\kappa$ is always closed in $\kappa$ (it might not be closed if $\kappa$ is a limit cardinal). The reason is that if $A$ is a set of cardinals, then $\sup A$ is a cardinal as well. So if $\sup A<\kappa$, it is a cardinal smaller than $\kappa$. Therefore $\{\lambda<\kappa\mid\lambda\text{ is a cardinal}\}$ is a closed set. ...


1

Consider Hamkins' and Yang's paper "Satisfaction is not Absolute". There they show the following result: $M_1$,$M_2$$\vDash$ZFC $\mathbb N^{M_1}$=$\mathbb N^{M_2}$ $M_1$ believes $\mathbb N$$\vDash$$\theta$ $M_2$ believes $\mathbb N$$\vDash$$\lnot$$\theta$. Without the assumption of the "higher-order ontological commitment, going strictly beyond the ...


1

Hope this reference helps: "Broadening the Iterative Conception of Set" by Mark F. Sharlow, Notre Dame Journal of Formal Logic, Volume 42, Number 3, 2001, pp.149-170. According to the abstract, "the modified conception maintains most of the features of the iterative conception of set, but allows for some non-wellfounded sets. It is suggested that this ...


1

You might look at the Springer Encyclopedia of Mathematics. I think it is not so easy to give a definite "birth date" to something such as ZFC, because many slightly different forms of the axioms appear over a period of several years.


1

Let $\alpha$ be some ordinal such that $f : \alpha \to {}^*\mathbb{R}$ is a bijection (i.e. a well-ordering of the hyperreal numbers). Let $I$ be the subset of $\alpha$ defined by $$ I = \{i \in \alpha : f(i) \ge f(j) \text{ for all } j \le i\} \quad (f(i) \ge f(j) \text{ as hyperreals.}) $$ It suffices to show that $I$ is uncountable, and we will have ...


1

I’ve seen only one construction. Let $T={}^{<\omega}2$ be the binary tree of height $\omega$. For each $f\in{}^\omega2$ let $B_f=\{f\upharpoonright n:n\in\omega\}$ be the branch determined by $f$, and let $\mathscr{B}=\{B_f:f\in{}^\omega2\}$; $\mathscr{B}$ is an AD family in $\wp(T)$, so it can be extended to a MAD family $\mathscr{A}$. Clearly ...


1

Even when free the sets have a natural well order to them, the countable union of countable sets is not necessarily countable. For example, in some models of $\sf ZF$ the first uncountable ordinal, $\omega_1$ is the countable union of countable ordinals. And no, the countable product of finite sets doesn't have to be non-empty without choice, let alone ...



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