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23
votes
2answers
2k views

Similarities between Post's Problem and Cohen's Forcing

Remark: I have since learned that G.H. Moore addresses this question in the third reference listed at the end of this post, beginning on p. 157 in which he cites a letter from Kreisel to Gödel dated ...
54
votes
2answers
16k views

Is the analysis as taught in universities in fact the analysis of definable numbers?

Ten years ago when I studied in the university I had no idea about definable numbers, but I came to this concept myself. My thoughts were as follows: All numbers are divided into two classes: those ...
123
votes
15answers
24k views

Why worry about the axiom of choice?

As I understand it, it has been proven that the axiom of choice is independent of the other axioms of set theory. Yet I still see people fuss about whether or not theorem X depends on it, and I don't ...
35
votes
6answers
2k views

Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

If $V$ is given to be a vector space that is not finite-dimensional, it doesn't seem to be possible to exhibit an explicit non-zero linear functional on $V$ without further information about $V$. The ...
12
votes
1answer
804 views

Proof-Theoretic Ordinal of ZFC or Consistent ZFC Extensions?

Let the proof theoretic ordinal $\alpha$ of a theory $T$ be the least recursive ordinal such that $T$ does not prove that $\alpha$ is well-founded. This ordinal is intended to quantify in some sense ...
74
votes
9answers
21k views

solving $f(f(x))=g(x)$

This question is of course inspired by the question How to solve f(f(x))=cosx and Joel David Hamkins' answer, which somehow gives a formal trick for solving equations of the form $f(f(x))=g(x)$ on a ...
37
votes
7answers
7k views

Arguments against large cardinals

I started to learn about large cardinals a while ago, and I read that the existence, and even the consistency of the existence of an inaccessible cardinal, i.e. a limit cardinal which is additionally ...
22
votes
4answers
2k views

When $2^\alpha = 2^\beta$ implies $\alpha=\beta$ ($\alpha,\beta$ cardinals)

Sorry if this is a silly question. I was wondering, under what axioms of set theory is it true that if $\alpha$,$\beta$ are cardinals, and $2^\alpha=2^\beta$, then $\alpha=\beta$? Do people use these ...
14
votes
2answers
1k views

Does ZFC prove the universe is linearly orderable?

It is consistent with ZFC that the universe is well-ordered, e.g. in $V=L$ where global choice holds. I also know that it is consistent that global choice fails (although I have no immediate example ...
7
votes
2answers
1k views

How to construct a continuous finite additive measure on the natural numbers

I want to find some condition to construct a continuous finitely additive measure on the natural numbers, i.e. $f:P(\mathbb{N})\rightarrow [0,1]$ such that $f(\{n\})=0$, and $f$ is an additive ...
59
votes
16answers
15k views

Most 'unintuitive' application of the Axiom of Choice?

It is well-known that the axiom of choice is equivalent to many other assumptions, such as the well-ordering principle, Tychonoff's theorem, and the fact that every vector space has a basis. Even ...
82
votes
2answers
7k views

Does every non-empty set admit a group structure (in ZF)?

It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: for finite $S$ identify $S$ with a cyclic group, and for infinite $S$, the set of finite subsets of $S$ with the binary ...
42
votes
4answers
10k views

Non-Borel sets without axiom of choice

This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel sets would be a Hamel basis, which needs axiom of ...
29
votes
2answers
2k views

What interesting/nontrivial results in Algebraic geometry require the existence of universes?

Brian Conrad indicated a while ago that many of the results proven in AG using universes can be proven without them by being very careful (link). I'm wondering if there are any results in AG that ...
20
votes
3answers
2k views

Who needs Replacement anyway?

The set theory ETCS famously comes without the Replacement axiom schema (or an equivalent) that is part of ZFC. One (to me, not apparently useful) set that one cannot build in ETCS is $\coprod_{n\in ...
13
votes
8answers
3k views

Forcing as a tool to prove theorems

It is often mentioned the main use of forcing is to prove independence facts, but it also seems a way to prove theorems. For instance how would one try to prove Erdös-Rado, $\beth_n^{+} \to ...
29
votes
0answers
1k views

Concerning the various proofs from the axiom of choice that R^3 admits of surprising geometrical decompositions into circles, skew lines and so on: can we prove in any instance that there are no Borel such decompositions? Or that AC is required?

This question follows up on a comment I made on Joseph O'Rourke's recent question, one of several questions here on mathoverflow concerning surprising geometric partitions of space using the axiom of ...
9
votes
1answer
621 views

Resembling the Levy Collapse

Suppose $\kappa$ is a weakly compact cardinal. Is there a $\kappa$-c.c. forcing $\mathbb{P}$ such that $\mathbb{P} \subseteq V_\kappa$ and $\Vdash_{\mathbb{P}} \kappa = \aleph_1$, where $\mathbb{P}$ ...
89
votes
9answers
14k views

Solutions to the Continuum Hypothesis

Related MO questions: What is the general opinion on the Generalized Continuum Hypothesis? ; Completion of ZFC ; Complete resolutions of GCH; How far wrong could the Continuum Hypothesis be?; When was ...
56
votes
2answers
5k views

Is every sigma-algebra the Borel algebra of a topology?

This question arises from the excellent question posed on math.SE by Salvo Tringali, namely, Correspondence between Borel algebras and topology. Since the question was not answered there after some ...
40
votes
7answers
2k views

How would one even begin to try to prove that a simple number-theoretic statement is undecidable?

This question is closely related to this one: Knuth's intuition that Goldbach might be unprovable. It stems from my ignorance about non-standard models of arithmetic. In a comment on the other ...
27
votes
5answers
5k views

Completion of ZFC

I attended a talk given by W. Hugh Woodin regarding the Ultimate L axiom and I wanted to verify my current understanding of what the search for this axiom means. I find it to be a fascinating topic ...
19
votes
2answers
3k views

Large cardinal axioms and Grothendieck universes

A cardinal $\lambda$ is weakly inaccessible, iff a. it is regular (i.e. a set of cardinality $\lambda$ can't be represented as a union of sets of cardinality $<\lambda$ indexed by a set of ...
34
votes
3answers
1k views

Is the fixed point property for posets preserved by products?

Recall that a partially ordered set (poset) $P$ has the fixed point property (FPP) if any order preserving function $f:P\longrightarrow P$ has a fixed point. Theorem. Suppose $P$ and $Q$ are posets ...
29
votes
3answers
2k views

Even XOR Odd Infinities?

Modular Arithmetic (MA) has the same axioms as first order Peano Arithmetic (PA) except $\forall x (Sx \ne 0)$ is replaced with $\exists x(Sx = 0)$. ...
19
votes
7answers
1k views

Does every set admit a rigid binary relation? (and how is this related to the Axiom of Choice?)

Let us say that a set B admits a rigid binary relation, if there is a binary relation R such that the structure (B,R) has no nontrivial automorphisms. Under the Axiom of Choice, every set is ...
6
votes
3answers
905 views

Tractability of forcing-invariant statements under large cardinals

It is usual to mention theorems of the kind: Th. Assume there is a proper class of Woodin cardinals, $\mathbb{P} $ is a partial order and $G \subseteq \mathbb{P}$ is V-generic, then $V \models \phi ...
15
votes
3answers
668 views

Singularizing forcing of “small” cardinality?

Can there be a large cardinal $\kappa$ and a forcing of size $\kappa$ that makes $\kappa$ a singular cardinal? The motivation is that the standard Prikry forcing does not have a dense set of size ...
15
votes
3answers
887 views

Questions about $\aleph_1-$closed forcing notions

"Foreman`s maximality principle" states that every non-trivial forcing notion either adds a real or collapses some cardinals. This principle has many consequences including: 1) $GCH$ fails ...
11
votes
4answers
3k views

Pi1-sentence independent of ZF, ZF+Con(ZF), ZF+Con(ZF)+Con(ZF+Con(ZF)), etc.?

Let ZF1 = ZF, ZFk+1 = ZF + the assumption that ZF1,...,ZFk are consistent, ZFω = ZF + the assumption that ZFk is consistent for every positive integer k, ... and similarly define ZFα ...
20
votes
2answers
895 views

If ZFC has a transitive model, does it have one of arbitrary size?

It is known that the consistency strength of $\sf ZFC+\rm Con(\sf ZFC)$ is greater than that of $\sf ZFC$ itself, but still weaker than asserting that $\sf ZFC$ has a transitive model. Let us denote ...
16
votes
1answer
1k views

Gödel's Constructible Universe in Infinitary Logics (A Possible Approach to HOD Problem)

Gödel's constructible universe ($L$) is defined using definable power set operator in first order logic ($\mathcal{L}_{\omega ,\omega}$). One can produce such a universe in infinitary logics in ...
16
votes
4answers
2k views

Why is there no Borel function mapping every countable set of reals outside itself?

A choice function maps every set (in its domain) to an element of itself. This question concerns existence of an anti-choice function defined on the family of countable sets of reals. In an answer to ...
12
votes
1answer
2k views

Aleph 0 as a large cardinal

The first infinite cardinal, $\aleph_0$, has many large cardinal properties (or would have many large cardinal properties if not deliberately excluded). For example, if you do not impose ...
11
votes
2answers
297 views

Sets that are not $\infty$-Borel

I have seen a few techinques for proving that certain sets of real numbers are $\infty$-Borel (definition) but it just occurred to me that I don't know of any way to prove that a set of real numbers ...
10
votes
3answers
888 views

How elementary can we go?

It is a theorem of A. Levy, if $\kappa$ is an inaccessible cardinal, then $V_\kappa\prec_{\Sigma_1} V$ namely $V_\kappa$ is an elementary submodel when considering only $\Sigma_1$ sentences. One ...
8
votes
5answers
842 views

Measurable cardinals under Axiom of Determinacy

I seem to remember reading somewhere that ZF+AD proves that omega-1 and omega-2 are measurable cardinals. Is that right? If so, can someone [point me to or give here] a [sketch or proof] of these ...
6
votes
2answers
691 views

Is there one binary operation foundational for set theory?

The membership relationship "$\epsilon$" is foundational for set theory, in the sense that the axioms of any set theory are formulated in the language of "$\epsilon$". Naturally, the question arises ...
6
votes
1answer
392 views

Infinite graphs isomorphic to their line graph

The only finite connected graphs $G$ that are isomorphic to their line graph $L(G)$ are the cycle graphs $C_n$ (see this link for example). There are connected countable graphs that are isomorphic to ...
14
votes
3answers
465 views

Does every set $X$ have a topology for which the only continuous self-surjection is the identity map?

This question is a special case of Dominic van der Zypen's question Reconstructing relations with the image relation of a topology, as discussed in the comments, particularly the comment of Eric ...
2
votes
1answer
161 views

Does $|(X\times\{0\}) \cup (X\times\{1\})| \leq |X|$ for $X$ infinite imply ${\sf AC}$?

Consider the statement For any infinite set $X$ there is an injection $\varphi$ from $(X\times\{0\}) \cup (X\times\{1\})$ into $X$. Does this imply the ${\sf AC}$?
2
votes
1answer
501 views

Compactness and completeness in Gödel logic

The standard proof of the completeness theorem in first-order Gödel logic is based on a first-order countable language. I want to know that is there any proof of the completeness theorem in ...
155
votes
27answers
19k views

What are some reasonable-sounding statements that are independent of ZFC?

Every now and then, somebody will tell me about a question. When I start thinking about it, they say, "actually, it's undecidable in ZFC." For example, suppose A is an abelian group such that every ...
51
votes
5answers
10k views

Inaccessible cardinals and Andrew Wiles's proof

In a recent issue of New Scientist (16 Aug 2010), I was surprised to read that a part of Wiles' proof of Taniyama-Shimura conjecture relies on inaccessible cardinals. Here's the link: ...
55
votes
18answers
7k views

Proofs of the uncountability of the reals.

Recently, I learnt in my analysis class the proof of the uncountability of the reals via the Nested Interval Theorem. At first, I was excited to see a variant proof (as it did not use the diagonal ...
59
votes
5answers
3k views

How do the compact Hausdorff topologies sit in the lattice of all topologies on a set?

This question is about the space of all topologies on a fixed set X. We may order the topologies by refinement, so that τ ≤ σ just in case every τ open set is open in σ. ...
41
votes
2answers
5k views

A question about ordinal definable real numbers

If ZFC (Zermelo-Fraenkel set theory with the Axiom of Choice) is consistent, does it remain consistent when the following statement is added to it as a new axiom? "There exists a denumerably ...
30
votes
13answers
3k views

Cardinalities larger than the continuum in areas besides set theory

It seems that in most theorems outside of set theory where the size of some set is used in the proof, there are three possibilities: either the set is finite, countably infinite, or uncountably ...
25
votes
10answers
50k views

What practical applications does set theory have?

I am a non-mathematician. I'm reading up on set theory. It's fascinating, but I wonder if its found any 'real-world' applications yet. For instance, in high school and we were learning the properties ...
18
votes
3answers
2k views

Is there a computable model of ZFC?

Background Assuming ZFC is consistent, then by downward Löwenheim–Skolem, there is a countable model (M,$\in$) of ZFC. Since the universe M is countable, we may as well think of it as actually being ...