axiom-of-choice's questions - English 1answer

326 axiom-of-choice questions.

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 ...

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 ...

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 $\...

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-...

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} /\...

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 ...

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\...

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 ...

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 $A$...

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 ...

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?, ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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, ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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?...

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 (...

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 ...

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 ...

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 ...

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| ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 (...

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 ...

It is well known that from a free (non-principal) ultrafilter on $\omega$ one can define a non-measurable set of reals. The older example of a non-measurable set is the Vitali set, a set of ...

There are a lot of ways to build a model where DC fails. However, all of them that I'm aware of involve adding at least a messy set of reals (or rather, taking a forcing extension and then passing to ...

It is interesting that to prove the transfer principle for the definable hyperreal field, one requires no more choice than for proving, for instance, the countable additivity of the Lebesgue measure. ...

(This question is already posted on Math SE but it isn't answered, so I ask same question on this site.) The following form of a continuum hypothesis occurs in Rathjen's paper "Indefiniteness in semi-...

There is a widely accepted opinion that the Axiom of Countable Choice (further, ACC) $$ \forall n\in \mathbb{N} . \exists x \in X . \varphi [n, x] \implies \exists f: \mathbb{N} \longrightarrow X . \...

Let $\ell^\infty(\mathbb{N})$ denote the set of bounded real sequences $(a_n)_{n\in\mathbb{N}}$. The $\lim$ operator is a partial linear operator from $\ell^\infty(\mathbb{N})$ to $\mathbb{R}$. With ...

Let $V$ denote the vector space of sequences of real numbers that are eventually 0, and let $W$ denote the vector space of sequences of real numbers. Given $w \in W$ and $v \in V$, we can take their "...

I'm looking for references for the following facts concerning the ultrafilter lemma (~ "there exist non-principal ultrafilters"): The ultrafilter lemma is independent of ZF. ZF + the ultrafilter ...

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|^{\...

The following is known: Theorem. Suppose $V[G]$ is a generic extension of $V$ by a set forcing, and let $N$ be a model of $ZFC$ with $V\subseteq N\subseteq V[G].$ Then $N$ is a generic extension of $...

The question is in the title, amenability being understood as the existence of a left-invariant finitely additive probability measure on the group of interest. The case of countable groups is treated ...

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