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An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty ...

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2answers
432 views

Can we have an infinite sequence of decreasing cardinality all terms of which have equal sized power sets?

Is the following consistent with $\text{ZF}$? There exists a set $S=\{x_1,x_2,x_3,...\}$ such that: $|x_{i+1}| < |x_i|$ $\forall m,n \in S (|P(m)|=|P(n)|)$ Where cardinality $``||"$ is ...
6
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2answers
433 views

Are there known examples of sets whose power set is equal in size to power set of larger sets only in absence of choice?

The question of existence of sets $x,y$ such that $$|x|<|y| \wedge |P(x)|=|P(y)|$$ is known to be independent of $\text{ZFC}$! But are there known examples of sets fulfilling the above condition ...
42
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1answer
2k views

When does $A^A=2^A$ without the axiom of choice?

Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$ However without the axiom of choice this doesn't ...
13
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1answer
305 views

Do choice principles in all generic extensions imply AC in $V$?

It's well-known that not all choice principles are preserved under forcing, e.g. in this answer https://mathoverflow.net/a/77002/109573 Asaf shows the ordering principle can hold in $V$ and fail in a ...
11
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1answer
822 views

The Tall Tale of Terminating Transfinite Towers

The transfinite tower of iterative automorphisms of a group $G$ is simply definied to be the following chain of the groups where $G_{\alpha+1}=Aut(G_{\alpha})$ for each ordinal $\alpha$ and the direct ...
159
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15answers
34k 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 ...
10
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1answer
232 views

The partial preorder on $\mathbb N$ generated by the finite axioms of choice

Let $\mathsf C_n$ denotes the statement: for any family $\mathcal F$ of $n$-element sets there exists a choice function (i.e., a function $f:\mathcal F\to\bigcup\mathcal F$ such that $f(F)\in F$ for ...
16
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1answer
1k views

How to construct a basis for the dual space of an infinite dimensional vector space?

Let $V$ be an infinite-dimensional vector space over a field $K$. Then it is known that $\dim V < \dim V^*$. More precisely, by a result attributed to Kaplansky and Erdos, we have $\dim V^* = |K|^{\...
11
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1answer
730 views

Is $\mathbb{R}$ a $\mathbb{C}$-module without AC?

Assuming ZFC. We can make $(\mathbb{R},+)$ into a nontrivial (scalar multiplication is not identically zero) $\mathbb{C}$-module. Now my questions are? 0.Is it consistent with $ZF$ that $\mathbb{R}$ ...
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0answers
922 views

The Rise and Fall of Dictators & How it Depends on Our Choice

This question is loosely inspired by the following paper of Shelah on Arrow property in which he answered a question of Gil Kalai affirmatively. Shelah, Saharon, On the Arrow property. Adv. in Appl. ...
6
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1answer
136 views

Countable (?) dependent choice

In some circumstances I've been using a form of choice over the first uncountable ordinal knowing a priori that only a countable number of choices were going to be made (without any a priori upper ...
10
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0answers
828 views

Kaplansky's theorem and Axiom of choice

Kaplansky in his paper titled by Projective Modules gave an important and essential theorem as follow: Theorem: Let $R$ be a ring, $M$ an $R$-module which is a direct sum of (any number of) countably ...
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1answer
353 views

Are there paradoxes in ZF + (the Axiom of Choice for finite sets)? [closed]

The good intuitive reasons that the Axiom of Choice (AC) for arbitrary sets-- that a function f with f(i) in S(i) for each i in {i}, for any non-empty sets {i} and all S(i), exists-- doesn't follow ...
14
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1answer
486 views

Does every model of ZF-foundation have an extension, with no new well-founded sets, where every set is bijective with a well-founded set?

This question follows up on an issue arising in Peter LeFanu Lumsdaine's nice question: Does foundation/regularity have any categorical/structural consequences, in ZF? Let me mention first that my ...
5
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0answers
184 views

On the Choice Content of Carathéodory's Conformal Mapping Theorem

The Schoenflies theorem, as a variant of the well-known Jordan curve theorem, states that the interior and the exterior planar regions determined by a simple closed curve (aka Jordan curve) in $\...
4
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0answers
101 views

Graphs without maximal vertex-transivite subgraphs

The axiom of choice is of no use when trying to prove that every vertex-transitive subgraph is contained in a maximal vertex-transitive subgraph, because a union of an ascending chain of vertex-...
13
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4answers
7k views

Does constructing non-measurable sets require the axiom of choice?

The classic example of a non-measurable set is described by wikipedia. However, this particular construction is reliant on the axiom of choice; in order to choose representatives of $\mathbb{R} /\...
3
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1answer
86 views

Ascending chain of vertex-transitive graphs

For any set $X$ we set $[X]^2 = \big\{\{x,y\}: x, y\in X\text{ and } x\neq y\big\}$. Let $G=(V,E)$ be a simple, undirected graph, and suppose ${\cal V}$ is a collection of subsets of $V$ such that ...
9
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2answers
523 views

Choosing subsets of $\mathbb R$ of cardinality $\frak c$, who wins?

Consider the following infinite game: two players, I and II, are alternating and choosing a descending sequence of subsets of $\mathbb R$ of cardinality $\frak c$, so I chooses a set $A_1\subseteq\...
7
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1answer
480 views

Relations of axioms of choice

We start with $ZF$. The axiom of countable choice, $AC_\omega$, says that any set product of nonempty sets with a countable index set is nonempty. For any $ZF$-definable set $A$, we should be able ...
20
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1answer
1k views

Are classes still “larger” than sets without the axiom of choice?

Classes are often informally thought of as being "larger" than sets. Usually, the notion of "larger" is formalized via an injection: $B$ is "at least as large" as $A$ iff there is an injection from $A$...
17
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2answers
711 views

Can There be a 1 dimensional Banach-Tarski paradox in the absence of choice

Let $\mathbb{R}$ act on itself by translation. Then there is no finite decomposition of a unit interval into pieces which, when translated, yields two distinct unit intervals. More formally does ...
2
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1answer
282 views

Measurable functions with non measurable image

I am just curious about examples of measurable functions $f:[0,1]\to[0,1]$ such that $f[0,1]$ is not measurable. This is motivated by the question Is measure preserving function almost surjective?, ...
4
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1answer
135 views

Does the existence of a unique chromatic (possibly transfinite) number for every (possibly non-finite) simple graph imply the axiom of choice?

Assuming the axiom of choice I can write for any cardinal number $\kappa$ and any simple graph $G$ that a function $f$ is a $\kappa\text{-coloring}$ of $G$ if and only if the cardinality of the image ...
14
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1answer
433 views

In what ways is ZF (without Choice) “somewhat constructive”

Let me summarize what I think I understand about constructivism: "Constructive mathematics" is generally understood to mean a variety of theories formulated in intuitionist logic (i.e., not assuming ...
8
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1answer
1k views

Does the axiom of choice follow from the statement “Every simple undirected graph is either connected, or its complement is connected”?

Using the Well-Ordering Principle, which is equivalent to the Axiom of Choice, it can be proved that (S): for every simple, undirected graph $G$, finite or infinite, either $G$ or its ...
3
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0answers
231 views

Can we prove the epsilon theorems without the axiom of choice?

Hilbert's epsilon $\epsilon$ is a quantifier. It follows the rule that if $\exists x. p(x)$, for some predicate $p$, we can infer $p(\epsilon x. p(x))$. Semantically, it represents picking some ...
15
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5answers
1k views

Forcing over models without the axiom of choice

In the vast majority of papers forcing is always developed over ZFC. Not surprisingly too, since infintary combinatorial principles are often used to prove results based on properties such as chain ...
11
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1answer
336 views

A model of ZF without a well-ordering of the reals in which any two sets of reals are comparable

Let us say that two sets $A$ and $B$ are comparable if there is an injection from $A$ to $B$ or there is an injection from $B$ to $A$. Obviously, in a model of ${\rm ZFC}$ any two sets are comparable ...
34
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1answer
1k views

Chromatic number of a topological space

Here is a question I asked myself years ago. Since it is not really in my field, I hope to find some (partial) answers here... Since it was unclear, I precise that I am looking for an answer in ZFC, ...
44
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6answers
3k 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 ...
17
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1answer
4k views

Are there any non-linear solutions of Cauchy's equation ($f(x+y)=f(x)+f(y)$) without assuming the Axiom of Choice?

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be s.t. $f(x+y) = f(x) + f(y), \ \forall x, y$ It is quite obvious that this implies $f(cx)=cx$ for all $c \in \mathbb{Z}$ and even further: $\forall c \in ...
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5answers
4k views

Subset of the plane that intersects every line exactly twice

In a comment to this question, Tim Gowers remarked that using the axiom of choice, one can show that there exists a subset of the plane that intersects every line exactly twice (although it has yet to ...
7
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1answer
390 views

Totally bounded spaces and axiom of choice

Wikipedia article on totally bounded spaces states "... the completion of a totally bounded space might not be compact in the absence of choice." Where is the axiom of choice used, and do you need it ...
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0answers
104 views

Selection in a small category

I came across the following problem. I do not know if these notions are known (I would actually be interested to know), so the names might not be the canonical ones. Given a small category $C$, a ...
15
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2answers
443 views

Cauchy real numbers with and without modulus

In constructive mathematics there are many possible inequivalent definitions of real numbers. The greatest variety seems to be in Dedekind-style approaches: in addition to "the" Dedekind real numbers ...
9
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1answer
519 views

Forcing, cuts, and Dedekind-finite cardinalities

Tl;dr version: there are two natural classes of cuts in the nonstandard model of arithmetic consisting of the Dedekind-finite sets (if, in fact, they constitute such a model); both these classes are ...
6
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1answer
416 views

Consistency of a non-measurable set of reals when the continuum cannot be well-ordered

Can it be shown, on the assumption that $ZF$ is consistent, that there is a model of $ZF$ in which the reals cannot be well-ordered but there does exist a set of reals which is not Lebesgue measurable?...
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15answers
2k views

Objects which can't be defined without making choices but which end up independent of the choice

It happens a lot of times that when one defines a new object (ring, module, space, group, algebra, morphism, whatever) out of given data, one first chooses some additional structure. And sometimes (...
7
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1answer
448 views

Are these large cardinals properties equivalent?

Consider the three following large cardinal axioms: there exists a nontrivial elementary embedding $j:V\to V$. there exists a n.e.e. $j:V\to M$ such that $M^{j^\omega(crit(j))}\subseteq M$. there ...
15
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0answers
417 views

Do all countable $\omega$-standard models of ZF with an amorphous set have the same inclusion relation up to isomorphism?

In my recent paper with Makoto Kikuchi, J. D. Hamkins and M. Kikuchi, The inclusion relations of the countable models of set theory are all isomorphic. manuscript under review. (arχiv) we proved ...
2
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3answers
485 views

Why doesn't choice imply global choice (in NBG)?

I thought ZFC proved the existence of an inductive well-ordering that is itself a set for any stage of V. NBG with only the regular AC should then prove/assert the existence of a class R of ordered ...
3
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1answer
342 views

Choice sets and the axiom of choice

Let $X\neq \emptyset$ be a set. We say ${\cal C} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ is a cover if $\bigcup {\cal C} = X$. A subset $D\subseteq X$ is a choice set for ${\cal C}$ if $|D\cap c| ...
8
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1answer
278 views

Undetermined games of “overdetermined” type

This is motivated by a previous question of mine, but I think it is ultimately more interesting (and hopefully easier to answer in the positive). In that question, a class of games (on $\omega$, of ...
15
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1answer
1k views

The axiom of choice as a consequence of a stronger semantics?

I've never had a problem with the axiom of choice, but it has often confused me how many authors find full choice so much different from finite choice. In my head they seem quite similar. We are ...
9
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0answers
359 views

Is it known whether every $\omega$-tree with an infinite antichain has an infinite chain in $\mathsf{ZF}$?

In this paper by Good and Tree, the following result is mentioned without proof as part of Proposition 6.5: Each of the following statements imply those beneath it. The countable union of ...
20
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7answers
2k views

Unique Existence and the Axiom of Choice

The axiom of choice states that arbitrary products of nonempty sets are nonempty. Clearly, we only need the axiom of choice to show the non-emptiness of the product if there are infinitely many ...
17
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0answers
258 views

Hahn-Banach and the “Axiom of Probabilistic Choice”

Stipulate that the Axiom of Probabilistic Choice (APC) says that for every collection $\{ A_i : i \in I \}$ of non-empty sets, there is a function on $I$ that assigns to $i$ a finitely-additive ...
8
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2answers
296 views

Relation between well-orderings of $\mathbb{R}$, and bases over $\mathbb{Q}$

The following question arose from a discussion about the definability of bases of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. (ZF without AC) something we can note is that the existence of a (...
9
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1answer
434 views

Axiom(s) of choice and bases of vector spaces

I'm not sure this question is more suitable for MO or for MSE, so feel free to move it to MSE if necessary. I work here in ZF theory. Consider the following statements: $(C)$ Axiom of choice: for ...