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.

learn more… | top users | synonyms

2
votes
1answer
23 views

Regularity of $\omega_1$ and axiom of choice

Why is the regularity of the ordinal $\omega_1$ a consequence of the axiom of choice?
3
votes
0answers
44 views

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 ...
0
votes
1answer
91 views

Breaking the AC barrier using Kelley-Morse set theory

In her blog post Variants of Kelley-Morse set theory, Prof. Gitman proves that every model of the the common version of Kelley-Morse set theory ($\mathsf{KM}$) is a model of the Wikipedia version of ...
1
vote
1answer
45 views

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 ...
1
vote
1answer
27 views

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 ...
1
vote
0answers
44 views

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 ...
28
votes
2answers
1k views

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 ...
62
votes
7answers
11k views

Can you explain the “Axiom of choice” in simple terms?

As I'm sure many of you do, I read the XKCD webcomic regularly. The most recent one involves a joke about the Axiom of Choice, which I didn't get. I went to Wikipedia to see what the Axiom of ...
3
votes
3answers
90 views

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 ...
2
votes
2answers
18 views

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 ...
131
votes
0answers
4k views

Does the open mapping theorem imply the Baire category theorem?

A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice. On the other hand, the three ...
1
vote
2answers
73 views

Every II-finite set is III-finite

I need some help proving that if a set $X$ is II-finite then it is III-finite, i.e. if every non-empty family of subsets of $X$ which is linearly ordered by inclusion has a maximal element under ...
4
votes
1answer
34 views

Lebesgue measure without choice

From this question and this question (and their answers) I gather that it is consistent with ZF without The Axiom of Choice to assume that there exist countable sets $A_n$, $n\in \mathbb N$, such that ...
0
votes
1answer
34 views

Accumulation point is a limit of some sequences in first-countable space with Axiom of Choice.

I have a statement which says that Let $X$ be a first-countable space and $E$ be a subset of $X$. If $p$ is an accumulation point of $E$, then there exists a sequence in $E$ that converges to ...
1
vote
1answer
19 views

Finite complement topology over $\Bbb{R}$ is not second-countable under ZF?

Under $\mathsf{ZF+AC_\omega}$, the space $\Bbb{R}$ with finite complement topology is not second-countable, since a countable union of countable subsets of $\Bbb{R}$. However, such argument doesn't ...
3
votes
1answer
103 views

Without AC, it is consistent that there is a function with domain $\mathbb{R}$ whose range has cardinality strictly larger than that of $\mathbb{R}$?

I stumbled across this question earlier, and the top comment on the bottom answer asserts two claims: Without the Axiom of Choice, It is consistent that there exists a function with domain ...
0
votes
1answer
32 views

Every set with more than point admits a permutation with no fixed point and the Axiom of Choice [duplicate]

Assuming axiom of choice , for any set $S$ with more than one point , there exist a bijection $f:S \to S$ such that $f(s) \ne s , \forall s \in S$ . Is the converse true , i.e. Does the statement " ...
2
votes
2answers
153 views

Isomorphic Free Groups and the Axiom of Choice

When I read about free group, the proof which concerns about two free groups $F(X)$ and $F(Y)$ are isomorphic only if $\operatorname{card}(X) = \operatorname{card}(Y)$ has a sentence going as follows: ...
2
votes
1answer
53 views

Consequences of the negation of the Axiom of Dependent Choice

It seems to me that a proper reason to include The Axiom of Choice as a foundational axiom of set theory should be based on the observation that the negation of The Axiom of Choice has absurd ...
1
vote
1answer
64 views

In any commutative ring with unity, every prime ideal is finitely generated implies every ideal is finitely generated; can it be prove without A.C.?

Assuming Zorn's lemma, "In any commutative ring with unity, if every prime ideal is finitely generated, then every ideal is finitely generated". Is the converse true, i.e. if in any commutative ring ...
1
vote
1answer
46 views

In every ring with unity satisfying ACC, every ideal is finitely generated; can we prove it without assuming Axiom of choice?

Assuming Zorn's lemma, we can prove that in every ring with unity satisfying Ascending-Chain-Condition, every ideal is finitely generated. Is this statement equivalent to Zorn's lemma? Can we prove it ...
2
votes
2answers
52 views

$\forall \alpha \exists \beta: \beta > \alpha$ where $\alpha$ and $\beta$ cardinals

I have to prove ZF $\vdash$ $\forall \alpha \exists \beta:\beta > \alpha$, where $\alpha, \beta-$ cardinal numbers. I can prove it only in ZFC. Let's fix some cardinal number $\alpha$. By ...
1
vote
1answer
40 views

Assuming the axiom of choice ,how to prove that every uncountable abelian group must have an uncountable proper subgroup?

Assuming the axiom of choice , how to prove that every uncountable abelian group must have an uncountable proper subgroup ? Related to Does there exist any uncountable group , every proper subgroup ...
2
votes
1answer
54 views

Can you verify this proof of the Schroeder-Bernstein theorem?

I'm a freshman in college and my professor challenged us to find a proof of this theorem. Please don't give me the answer but please verify if this proof works or, if not, if it is the start of a ...
1
vote
1answer
62 views

Proving that “Every non-trivial ring (i.e. with more than one element ) with unity has a maximal ideal” implies axiom of choice is true

I know that assuming axiom of choice or equivalently Zorn's lemma , it can be proved that every non-trivial ring with unity has a maximal ideal (two sided ) . The wiki article on axiom of choice says ...
2
votes
1answer
30 views

Formal proof that $N$ is the smallest infinite set [duplicate]

I wish to write a formal proof of the following statement: For any infinite set $X$, there exists an injection $f:\mathbb{N}\to X$. I'd like the proof to explicitly use the full axiom of choice (for ...
6
votes
2answers
2k views

Zorn's Lemma And Axiom of Choice

How can I prove Zorn's lemma is equivalent to Axiom of choice?
2
votes
1answer
62 views

Acyclicity of Flasque sheaves without A.C.

I say that a sheaf on a space X, is flasque if the restriction maps are surjective, that is any local section extend to a global section. Now it is a fact that if $F_1$ is flasque and if $0 \to F_1 ...
3
votes
4answers
829 views

Axiom of Choice and Right Inverse

I read an Theorem that states: Let $A$ and $B$ be non-empty sets, and let $f:A \to B$ be a function, then the function $f$ has a right inverse if and only if $f$ is surjective. The Theorem ...
4
votes
1answer
44 views

Cardinality of sets of reals without choice

Assuming just ZF (no axiom of choice): Does $\aleph_n\leq|\mathbb{R}|$ for all $n<\omega$ imply $\aleph_\omega\leq|\mathbb{R}|$? (with $\kappa\leq|\mathbb{R}|$ meaning that there is a set of reals ...
1
vote
1answer
26 views

Conceptual problem in Noetherian rings and Zorn Lemma

I've got a problem, and it's that I can't see the difference of one of the definitions of Noetherian ring, and supposing the zorn lemma for that ring. I mean, if I use the definition of the maximal ...
2
votes
1answer
131 views

Closure of a set in a “Topology of finite complement”

Well, I was reading this article by Kelley and when reached the point where he say that $X_a$ is closed in $Y_a$ I had to stop, probably mine is just a stupid misunderstand but can't figure out how to ...
5
votes
2answers
689 views

Second-countable implies separable/Axiom countable choice

Let $(X,\mathscr T)$ be a topological space, and $(B_n)_{n\ge1}$ a countable basis for X. Under this assumptions, X is separable. The proof of this assertion is as follows: We can assume ...
2
votes
1answer
74 views

Did I use axiom of choice in my proof?

I have two different affine open covers for a scheme $X$, say $X = \cup_{i \in I} U_i$ and $X = \cup_{j \in J} V_j$. For each $p \in X$, we know there exist some $i(p)$ and $j(p)$ such that $p \in ...
-2
votes
1answer
68 views

On axiom of choice [closed]

Suppose we have a countable set of intervals (i.e. I= $\{[a,b]|a,b \in \mathbb{R}\}$, does axiom of choice guaranteed us to find one with the maximal length?
0
votes
1answer
52 views

Every epimorphism in Sets is split: why is it equivalent to axiom of choice?

Suppose that $f: A \rightarrow B$ is epic in Sets. One can construct a section $s: B \rightarrow A$ of $f$ as follow: Let us define an equivalence relation $R$ on $A$ as follow: $aRa'$ iff $a, a' \in ...
2
votes
1answer
53 views

Forcing reference

Who first proved that, over ZF, the statement (1) The reals are well-orderable is strictly stronger than the statement (2) Every real-indexed family of nonempty sets of reals has a choice ...
1
vote
1answer
29 views

Every Partially ordered set has a maximal independent subset

I am working on this problem:- Every Partially ordered set has a maximal independent subset. Definition:Let $\langle E,\prec\rangle$ be a partially ordered set. A subset $A\subset E$ is called ...
3
votes
1answer
74 views

Why in Teichmüller-Tukey lemma finitness is essential?

First we will state a Teichmüller-Tukey Lemma: Let $A$ be a set and $\phi$ be a property defined on all finite subset of $A$. Assume that $B$ is a subset of $A$ such that each finite subset of $B$ ...
3
votes
1answer
80 views

Finitely-additive measure over $\Bbb{N}$

On the set of natural number, we can consider the finitely-additive measure defined as: $$\mu(A) = \lim_{n\to\infty}\frac{\#(A\cap [1,n])}{n}.$$ However, there is a definable (by PA, or some ...
0
votes
2answers
52 views

Is the axiom of choice used here?

I was reading the following How do i prove that every open set in $\mathbb{R}^2$ is a union of at most countable open rectangles? The answer by "user4594" basically considers a k-cell with rational ...
6
votes
0answers
72 views

One more AC equivalence question

Is "Every vector space admits a norm" weaker than AC? I know that the statement follows from "Every vector space has a basis", which is equivalent to AC.
7
votes
1answer
127 views

Outer automorphisms of the infinite symmetric group

Denote by S$_\infty$ the group of permutations of $\mathbb N$. Question: Does there exist an outer automorphism of S$_\infty$, and if so, can one be exhibited? Does this depend on the continuum ...
2
votes
0answers
28 views

Example of a maximal idea

Let $A$ be the set of bounded continuous functions from the set of real numbers to itself. Then $A$ is a ring under pointwise addition and multiplication. The set $I$ of all functions $f \in A$ ...
2
votes
1answer
40 views

Question related to ordinal number without using Axiom of Choice.

Can we proof this result without using Axiom of Choice :- $$A\cap \alpha=\emptyset \,\,\,\, \mbox{and}\, \, \, A\times \alpha \sim A\cup \alpha$$ then there is an $A^{'} \subset A$ such that $\alpha ...
0
votes
2answers
36 views

Is this an accurate way to write the axiom of choice using transversals?

Is this an accurate way to write the axiom of choice using transversals? $\forall \mathcal{C} \exists T\, (T\subseteq \cup\,\mathcal{C} \land \forall X \exists s \forall t\, ((X\in\mathcal{C} \land ...
2
votes
0answers
47 views

$Gal(\bar{\mathbb Q}/\mathbb Q)$ without choice, and constructive Galois theory

By this question: Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice? We have that over ZF, algebraic closures of $\mathbb Q$ aren't unique. Are their Galois groups as extensions over ...
4
votes
0answers
42 views

Construction of injective hulls without axiom of choice

Motivation: It is known that without the axiom of choice (AC), it is not provable that all categories of modules have enough injectives, let alone injective hulls. Still, there are examples of rings ...
1
vote
3answers
38 views

Recursive use of the Axiom of Choice

In a standard proof that any sequence-compact metric space $(X,d)$ has a (finite) $\varepsilon$-net, the approach is the following: Make a sequence $(x_n)$ such that $$ x_{n+1}\notin\bigcup_{i=1}^n ...
2
votes
0answers
92 views

What is the name of proofs with (without) Axiom of Choice

In many contexts we distinguish between proofs using AC and proofs which do not use AC. (To phrase this somewhat differently: If there is a proof without AC, this proof is usually preferred.) I would ...