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forcing, large cardinals, descriptive set theory, infinite combinatorics, cardinal characteristics, forcing axioms, ultrapowers, measures, reflection, pcf theory, models of set theory, axioms of set theory, independence, axiom of choice, continuum hypothesis, determinacy, Borel equivalence relations, Boolean-valued models, embeddings, orders, relations, transfinite recursion, set theory as a foundation of mathematics, the philosophy of set theory.

19 votes
1 answer
665 views

Models of ZFC and the Borel hierarchy

The collection of binary relations $R$ on the natural numbers such that $(\mathbb{N},R) \models ZFC$ forms a Borel set, neither closed nor open -- assuming Con(ZFC). Can you show it's not $F_\sigm …
Bjørn Kjos-Hanssen's user avatar
18 votes
4 answers
2k views

Complete Boolean algebra not isomorphic to a $\sigma$-algebra

Does there exist a complete Boolean algebra that is not isomorphic to any $\sigma$-algebra? If so, what is an easy or canonical example or construction?
Bjørn Kjos-Hanssen's user avatar
11 votes
Accepted

Usual technical term for replacing a set by the set of singletons of its members?

$A'$ is the discrete partition of $A$. That is, we think of it as a partition of $A$ induced by the finest equivalence relation, the identity relation.
Bjørn Kjos-Hanssen's user avatar
10 votes
2 answers
1k views

$\aleph$ looks like $\mathbb N$?

We all know the notation $\aleph_\lambda$ for the $\lambda$th (or, I guess, $\lambda+1$st) infinite cardinal number; in particular $\aleph_0$ is the cardinality of the the set of natural numbers $\mat …
Bjørn Kjos-Hanssen's user avatar
10 votes
Accepted

What non-standard model of arithmetic does Hofstadter reference in GEB?

My first guess is that the triples come from the fact that nonstandard countable models of PA look like $$\mathbb N + \mathbb Z\times\mathbb Q$$ and elements of $\mathbb Q$ can be represented by pairs …
Bjørn Kjos-Hanssen's user avatar
10 votes
4 answers
2k views

Axiom of Infinity needed in Cantor-Bernstein?

Can one prove the Cantor-Bernstein (or Schröder-Bernstein) theorem without using the Axiom of Infinity?
Bjørn Kjos-Hanssen's user avatar
9 votes
Accepted

Choice sets from above and below

Let $\cal S=\{\{1,2\},\{2,3\},\{3,1\}\}$. Then $\cal S$ has no choice set, whatsoever. So there is no asymmetry -- not every shy set is contained in a choice set, and not every gregarious set contai …
Bjørn Kjos-Hanssen's user avatar
9 votes

Natural examples of $\bf\Sigma^0_3$ equivalence relations

Turing equivalence of real numbers.
Bjørn Kjos-Hanssen's user avatar
9 votes
1 answer
396 views

VC dimension of Borel sets [duplicate]

Can there be an uncountable set $S\subseteq\mathbb R$ such that for each subset $D\subseteq S$, there is a Borel set $U$ with $D=S\cap U$? I'm asking merely out of curiosity, but I'll mention that thi …
Bjørn Kjos-Hanssen's user avatar
8 votes

Is it consistent with ZFC (or ZF) that every definable family of sets has at least one defin...

Yes, fix a definable relation $\le_L$ that well-orders all of $L$. If $V=L$ then every definable nonempty set $A$ has a definable member $a$, namely: $a :=$ the $\le_L$-least element of $A$.
Bjørn Kjos-Hanssen's user avatar
8 votes
1 answer
252 views

Automorphisms of power set lattice mod finite

Let $N$ be a countably infinite set and let $\mathcal P$ denote power set. I get that the automorphisms of $(\mathcal P(N),\subseteq)$ are all induced by permutations of $N$. But what can be said abo …
Bjørn Kjos-Hanssen's user avatar
8 votes
1 answer
479 views

VC dimension of standard topology on the reals

Can there be an uncountable set $S\subseteq\mathbb R$ such that for each subset $D\subseteq S$, there is an open set $U$ with $D=S\cap U$? I'm asking merely out of curiosity, but I'll mention that thi …
Bjørn Kjos-Hanssen's user avatar
7 votes
Accepted

Is there a name for this equivalence relation?

$\mathscr F$-indistinguishability. In analogy with Topological indistinguishability.
Bjørn Kjos-Hanssen's user avatar
7 votes
Accepted

The consistency of ZFC + CH gives the ability to travel to a universe which models ZFC + \ne...

The empty set $\emptyset$ here is the forcing condition that has the least amount of information about the generic object being constructed. Since it has the least information, you might think it sh …
Bjørn Kjos-Hanssen's user avatar
6 votes
Accepted

Translates of meager sets

No, there is no such set. The situation for meager sets is dual to that described by Pietro Majer in a comment on Translates of null sets, "I was vaguely thinking to Hausdorff measures w.r.to gaug …
Bjørn Kjos-Hanssen's user avatar

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