The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

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The least $\aleph$ that has no surjective map from $m$ to it.

Without $AC$. Let $\aleph^*(m)$ be the least aleph that $\not\leq^* m$. How to show that $\aleph^*(m)$ exists and $\aleph^*(m)= \{\alpha\in ON\mid\ \alpha\leq^*m\}$. $ON$ is the class of all ...
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dual Dedekind-infinity may not imply Dedekind-infinite without AC

It is written in wikipedia: https://en.wikipedia.org/wiki/Dedekind-infinite_set It is not provable (in ZF without the AC) that dual Dedekind-infinity implies that A is Dedekind-infinite. (For ...
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Does the principle of schematic dependent choice follow from ZFCU?

Let ZFCU be ZFC modified in the usual way to allow for urelements but without an axiom stating that there is a set of all urelements. Let the principle of Schematic Dependent Choice (SDC) be: $\...
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Are there weak versions of the axiom of choice equivalent to weak versions of Zorn's lemma and similar principles?

I recalled reading about other weaker forms of $AC$, for example countable choice, where we could make choices from a sequence $(S_{k})_{k \in \mathbb{N}}$ of non-empty sets. I also recalled mention ...
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Does there exist a model of $ZF¬C$ in which there is a function on $\mathbb R$ which is sequentially continuous at a point where it is not continuous? [duplicate]

Does there exist a model of $ZF¬C$ in which there is a function $f:\mathbb R \to \mathbb R$ such that $f$ is sequentially continuous at some $a \in \mathbb R$ but not $\epsilon-\delta$ continuous , i....
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“Sequential continuity is equivalent to $\epsilon$-$\delta $ continuity ” implies Axiom of countable choice for collection of subsets of $\mathbb R$?

"A function $f: \mathbb R \to \mathbb R$ is continuous at $x \in \mathbb R$ , if and only if it is sequentially continuous " , does this statement imply "the Axiom of Choice for countable collections ...
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Proof that every field $F$ has an algebraic closure $\bar F$

I am reading the book A First Course in Abstract Algebra written by Fraleigh and I do not really understand the proof of theorem 31.22, that every field $F$ has and algebraic closure $\bar F$. I ...
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126 views

First Uncountable Ordinal Cofinality: Needs AC?

Say $\omega_1$ is the first uncountable ordinal. The reason I care about $\omega_1$ is Any countable subset of $\omega_1$ is bounded (or if you prefer, there is no countable cofinal subset). This ...
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(Non-Hopfian) groups that only have quotients that are themselves or the trivial group.

A group is non-Hopfian provided it is isomorphic to a proper quotient. The classic, finitely presented, example of such a group is the Baumslag-Solitar group $$BS(2,3)= \langle x,t \mid t^{-1}x^2 t =x^...
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1answer
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ZF and the Existence of Finitely additive measure on $\mathcal{P}(\mathbb{R})$

My understanding is that Solovay (1970)'s relative consistency shows that if ZFC+I has a model then ZF+DC has a model in which every subset of the reals is Lebesgue measurable (and hence $\sigma$-...
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The non-existence of non-principal ultrafilters in ZF

In Hrbacek and Jech (1999, p.205), they point out that "it is known that the theorem [the extension of any filter to an ultrafilter] cannot be proved in Zermelo-Fraenkel set theory alone." And in Jech ...
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Do we know that we can't define a well-ordering of the reals?

Folklore has it that it is impossible to define a well-ordering of the reals explicitly. There exist pointwise definable models of ZFC where every set is definable without parameters: it is the ...
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undecidable statement of the form “$F$ is a choice function on $M$”

Are there unary predicates $\varphi(x), \psi(x)$ such that The formula that states "There is a set $M$ with: $\forall x[x\in M \leftrightarrow \varphi(x)]$" is provable in $ZFC$. The formula that ...
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1answer
75 views

Lindenbaum-Theorem only concerning sentential logic provable in ZF?

Is the Lindenbaum-Theorem of sentential logic (= propositional logic) provable in ZF (i. e. without the axiom of choice)? Lindenbaum's theorem of sentential logic states that every set $\Sigma$ of ...
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1answer
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If I assign a random number $r_x \in (0,1)$ to every $x \in (0,1)$ what are the odds that one of them will be a specific number?

I'll start by motivating by question with a simpler scenario to ensure I've at least understood that scenario properly. Scenario 1 : Imagine an infinite sequence of numbers where $i$ is the $i^{th}...
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Undetermined game of length $\omega_1+\omega$, without choice

On the following page, Taranovsky is talking about his "Determinacy Maximum" axiom: http://web.mit.edu/dmytro/www/DeterminacyMaximum.htm He also justifies the choice of the name, by pointing out that ...
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1answer
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Partition the set of positive real number into pairs without axiom of choice?

Backround. Yesterday I made a comment on Achuille hui's answer and then I discuss on the chat with Paul Plummer about the way to partition the set of positive real number into pairs $\Bbb{R}^+=\...
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1answer
101 views

Is there a constructive discontinuous exponential function? [duplicate]

It is well-known that the only continuous functions $f\colon\mathbb R\to\mathbb R^+$ satisfying $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb R$ are the familiar exponential functions. (Prove $f(x)=f(1)^x$...
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Box topology and axiom of choice

Below is the definition of box topology: Given an indexed family of topological spaces $X_\alpha $, the collection of all sets of the form $$\prod_{\alpha\in J} U_\alpha,$$ where $U_\alpha$ is open ...
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1answer
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Could the Hamel basis of $\mathbb{R^Z}$ be the set $\mathbb{R^Z}-{\mathbf{\{0\}}}$?

This is the follow up question to this question (*) According to page 2 of this link 1 and this link 2, $\mathbb{R^Z}$ (which is referred as $\mathbb{R^\infty}$ in link 1) has elements of the ...
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1answer
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For an infinite cardinal $\kappa$, $\aleph_0 \leq 2^{2^\kappa}$

I'm trying to do a past paper question which states: $$ \text{For all infinite cardinals $\kappa$, we have } \aleph_0 \leq 2^{2^\kappa}. $$ I'm supposed to be able to do this without the axiom of ...
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1answer
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Zorn's lemma converse? (Context: Maximal proper subgroups)

So, in my qual prep class a pretty simple question popped up: "Prove that for any nontrivial finite group there exists a maximal proper subgroup." So of course, my natural inclination was to ...
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$|\alpha\times\alpha|=|\alpha|$ for infinite ordinal $\alpha$, definably

I am trying to prove Proposition 1.7 of http://math.bu.edu/people/aki/7.pdf: Proposition 1.7 (Halbeisen–Shelah). If $\aleph_0\le|X|$, then $|{\cal P}(X)|\not\le|{\rm Seq}(X)|$. In words, ...
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What's wrong with this proof of the inconsistency of the axiom of choice?

Let $\mathscr{T}$ be the (countable) collection of all theorems provable in ZFC. Define an equivalence relation on $\mathscr{T}$ by $\phi\sim\psi$ iff $(\phi \iff \psi)$. In other words, two theorems ...
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2answers
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Choice function for collection of arbitrary finite sets. AC required?

I understand how we can show the existence of a choice function for any (finite or infinite) collection of (finite or infinite) subsets of, say, $\mathbb{N}$ or $\mathbb{Z}$ without using the axiom of ...
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1answer
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Looking for explanation of Banach-Tarski Proof, preferably by visual methods “Video, Pictures, Diagrams…”

could someone please explain the four steps of Banach-Tarski? 1- Find a paradoxical decomposition of the free group in two generators. 2- Find a group of rotations in 3-d space isomorphic to the free ...
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1answer
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Is AC necessary to show that in metric spaces $x\in\operatorname{closure}(A)$ implies $\exists\{a_n\}_{n=1}^\infty\subseteq A$ s.t. $\lim a_n=x$?

Let $(X,d)$ be a metric space. Let $x\in\operatorname{closure}(A)$, where $A\subseteq X$. Then for each $n\in\mathbb{N},\exists x_n\in B_{\frac{1}{n}}(x)\cap A$, where $B_\varepsilon(x)$ is the open ...
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1answer
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If $|A|<|B|$ does $B$ surject onto $\aleph(A)$?

After reading Proving existence of a surjection $2^{\aleph_0} \to \aleph_1$ without AC I became curious if there is a generalization to arbitrary cardinals. That is, if $\frak m<n$, does it follow ...
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1answer
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How to show countability of $\omega^\omega$ or $\epsilon_0$ in ZF?

I know that with choice, the countable union of countable sets is countable, making $\omega^\omega$ and $\epsilon_0$ both countable. Can we show this without choice? E.g. in the case that $\omega_1$ ...
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1answer
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How to prove that $\sf CH$ implies $2^{\aleph_0}=\aleph_1$

Of course, most of you will, upon reading the title, exclaim "But isn't that the definition of the continuum hypothesis?" So I need to be a little more careful about the exact definitions. Let ${\...
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Does “$(\exists f:A\twoheadrightarrow B)\implies(\exists f:B\hookrightarrow A)$” implies the axiom of choice? [duplicate]

Let $P$ denotes the property that if there exists a surjection from set $A$ to set $B$, then there exists an injection from $B$ to $A$. It's apparent that $P$ can be proved in ZFC. My question is that ...
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1answer
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Can we prove, without axiom of choice, that the set of all zero divisors (including $0$) of a commutative ring with unity contains a prime ideal?

Let $R$ be a commutative ring with unity , I know that assuming axiom of choice , if $A$ is the set of all zero divisors (including $0$ ) then it is a union of prime ideals so it contains a prime ...
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1answer
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The existence of the sequence corresponding to some asymptotic sequence

The following proof of the axiom of choice by induction is obviously false: Let $(\Lambda)_{i=1, 2, \ldots}$ be an infinite sequence of nonempty sets. When $i=1$, self-evident. We will assume this ...
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Does a direct sum decomposition of an infinite-dimensional vector space require Zorn's lemma?

Let $V$ be an infinite-dimensional vector space and $V'\subset V$ a subspace. Does it require Zorn's lemma to write $V=V'\oplus V''$ for some other subspace $V''\subset V$?
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Axiom of choice : continuous function and uniformly continuous

How I proof that every continuous function f in [0,1] is uniformly continuous, without axiom o choice? I took this from the book Axiom of Choice from Horst Herrilich He had a observation that ...
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Does the proof of Bolzano-Weierstrass theorem require axiom of choice?

When selecting the terms of subsequence from each bisections, I thought axiom of choice might be required. But I'm not so sure whether or not, so please tell me. [edited] I'm sorry for the lack of ...
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1answer
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Can we prove AC from the statement “There is no $\aleph$ cardinal strictly between $\operatorname{CARD}(X)$ and $\operatorname{CARD}(2^X)$”?

If $X$ is a set, let $\operatorname{CARD}(X)$ denote the Cardinal number of $X$. Let GCH(1) be the statement "If $K$ is an infinite initial ordinal number, then there exists no initial ordinal number $...
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1answer
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Is it consistent without the axiom of choice that every permutation of some infinite set have fixed points?

A "permutation" of a non-empty set means an injective mapping of the set onto itself. Let $S(1)$ be the statement "There exists a permutation of every set containing at least two elements, which has ...
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A question about infinite sets and Cantor's Power Set theorem

Let $\operatorname{Card}(X)$ denote the cardinal number of the set $X$. The standard proof of Cantor's Power Set theorem stating that "$\operatorname{Card}(X) < \operatorname{Card}(2^X)$" is simple,...
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Choosing a Cauchy sequence for a real

It is easy to form in ZF, for each real $a$, a "canonical" Cauchy sequence that converges to $a$. For example, one can take the sequence of finite initial segments of the decimal expansion of $a$, ...
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1answer
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Axiom of choice for singletons

Let $\mathscr C \subseteq \mathscr P (\mathbb R )$ be a family of singletons, i.e.: each element of $\mathscr C$ contains exactly one real number. Let $f: \mathscr C \to \mathbb R $ be the function ...
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1answer
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Zorn's Lemma's chain condition

Zorn's Lemma requires that every chain in a partially ordered set $X$ has an upper bound. In this article Gowers uses Zorn's Lemma to find a maximal linearly independent (over $\mathbb{Q}$) subset of $...
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Regularity of $\omega_1$ and axiom of choice

Why is the regularity of the ordinal $\omega_1$ a consequence of the axiom of choice?
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Basis of $\mathbb{F}[[x]]$ over $\mathbb{F}$ without AC

Does the ring of formal power series $\mathbb{F}[[x]]$ as a vector space over $\mathbb{F}$ admit a basis without assuming the Axiom of choice, at least in some special cases of $\mathbb{F}$? I'm ...
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1answer
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Difference between Zorn's Lemma and the ascending chain condition

Let $S$ be a non-empty partially ordered set with respect to a relation $\leq$. Then: Zorn's Lemma: If $S$ has the property that any totally ordered subset $U\subset S$ has an upper bound, then $S$...
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Choice of a skeleton

Suppose we are in presence of a strong enough axiom of choice (e.g., choice for conglomerates). I know that any category has a skeleton, but I would like to know if I can choose a skeleton which ...
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In P. Cohen's models (or others) may we have $\neg\mathsf{AC}+\mathsf{CH}$? May we have $\neg \mathsf{AC} + \neg \mathsf{CH}$?

I know we have the consistency of $\mathsf{ZF} + \mathsf{AC} + \mathsf{GCH}$, $\mathsf{ZF} + \neg \mathsf{AC}$, and $\mathsf{ZF} + \mathsf{AC} + \neg \mathsf{GCH}$. What about $\mathsf{ZF} + \neg \...
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Is the axiom of choice really all that important?

According to this book: The Axiom of Choice is the most controversial axiom in the entire history of mathematics. Yet it remains a crucial assumption not only in set theory but equally in modern ...
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Equivalence of countable choice for subsets of the reals and “second countable $\implies$ Lindelöf”

Looking for a proof that in a second countable space every open cover has a countable subcover -- i.e. every s.c. space is a Lindelöf space -- I bumped into this question. That answered my question. I ...
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Prove there exists a strictly increasing function from the natural numbers to a partially strictly ordered set

Let $P$ be a non-empty partially strictly ordered set and assume no element of $P$ is maximal (i.e. for every $x \in P$ there exists $y\in P$ with $x < y$). Show there exists a function $f\colon\...