Questions tagged [axiom-of-choice]
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. Use this tag in tandem with (set-theory).
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Countable Choice from Finite Sets
Consider the following 4 statements:
Axiom of countable choice
Axiom of countable choice from finite sets
Axiom of countable choice from Dedekind finite sets
Existence of a choice function for any ...
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A union of unions needn't be a union? (Sans AC) [duplicate]
For any collection $\mathscr C$ of sets, write $\Upsilon(\mathscr C)$ for the collection of arbitrary unions in $\mathscr C$. Now, I ask the innocent question of idempotency of $\Upsilon$:
Is $\...
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Does $A$-fold choice imply $|A| + |A| = |A|$ and $|A|\cdot |A| = |A|$?
Let $A$ be an infinite set. Then Zorn's lemma can be used to conclude that $A\times\{0, 1\}$, $A\times A$ and $A$ are all equinumerous (a proof is presented here). However, I am aware that for $A$ ...
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Is there a model of ZF not C where not every set of reals is Lebesgue measurable? [duplicate]
I know that there is a model of ZF set theory plus the negation of the axiom of choice where every set of reals is Lebesgue measurable. But is there also a model where not every set of reals is ...
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Principle in between BPI and AC
I have searched extensively in the literature, but all references I have consulted always place BPI (the Boolean Prime Ideal Theorem) as a sort of "cover" of the Axiom of Choice as far as ...
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Is every (infinite) permutation the composition of 2 involutions in ZF?
It is well known that any permutation on a finite set is the product of two involutions. I've wondered about what can happen in infinite sets.
Asuming the axiom of choice, every permutation is still a ...
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Discontinuous linear map and AC
The question arises when I am constructing an elementary proof for the following claim:
Given a normed vector space $V$, the following are equivalent:
$V$ is finite dimensional
Every linear map $T:V\...
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Axiom of Choice in characterizing openness in subspace
Below is the typical characterization of open sets in a subspace $Y$ of a metric space $X$.
$E$ is $Y$-open iff there exists an $X$-open $S$ such that $E = S \cap Y$.
The forwards direction usually ...
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Why is the Axiom of Choice Necessary in ZFC
Within the framework of Zermelo-Fraenkel set theory with the Axiom of Choice $(ZFC)$, when we considered the method of constructing the set of natural numbers, we regarded it as the smallest inductive ...
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Are the sets of the form $a=\{a\}$ different?
Suppose we adopt all the ZF axioms except the axiom of foundation. I suppose even adding the AC would not harm my question.
Somewhere on the internet, I read that in this case, the axiom of ...
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Truncated Tarski's theorem without axiom of choice
I read there that the fact about equivalence of $A$ and $A^2$ for any infinite set $A$ and the axiom of choice are equivalent. But what if we prove it only for sets that have cardinality of $\aleph_n,...
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uniform well-ordering and constructibility
When comparing between V=L and AC, one of the things that gets my attention is that, if we switch to an external perspective and don't care about first-order expressibility, in models of V=L we have a ...
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Is ACC implicitely involved in a construction?
Suppose that $X$ is a (edit: separable) complete metric space and $\{X_j:j\ge 1\}$ is a partition of $X$ (each $X_j\ne \emptyset$). Given a (continuous) function $f:X\to \mathbb{R}$, I construct
$$
h(...
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The relation between the cardinality of Bore $\sigma$-algebra and axiom of choice
Under axiom of choice, the cardinality of Borel $\sigma$-algebra $B$ is $\mathfrak{c}$.
In this proof axiom of choice is used three times: To prove
$\omega_1$-times recursion is sufficient,
each $|B_{...
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Construction of Proof: Zorn's lemma implies Axiom of choice
I have come across the prove that [Zorn's Lemma ==> AC] but am confused about the central statement, namely that we can take a set of all choice functions on subsets of X (lets just call it X, I ...
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Cantor-Bendixson theorem and AC
For context, the Cantor-Bendixson theorem states that a closed subset $A$ of a Polish space can be written as the union of a perfect subset and a countable set $A=P\cup C$.
Now, I know two proofs of ...
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Why can't $\mathbb{R}/\mathbb{Q}$ be linearly-ordered without Axiom of Choice?
This Question has an answer which is the only source that I can find about how $\mathbb{R}/\mathbb{Q}$ cannot be linearly ordered. I couldn't manage to open either of the source links provided in the ...
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Is the Axiom of Dependent Choice necessary in this proof?
While typically, the Axiom of Choice and its peripheral arguments are not emphasized in one's first exposure to Real Analysis, I am trying to be as rigorous as possible in my learning as an axiomatic ...
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Use of (weak forms of) AC for elementary embeddings proof
I encounter this issue when going through equivalent characterizations of measurable cardinals. For completeness, let me reproduce the statement:
For ordinal $\kappa$, the following are equivalent.
...
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Is the existence of a discontinuous linear map from a Hilbert space equivalent to the Axiom of Choice?
All constructions of a discontinuous linear map from an Hilbert Space, that i have seen, rely on using the Axiom of Choice. A lot of theorems that heavily rely on AC are equivalent to AC itself so i ...
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Could the Axiom of Choice be viewed as a restricted version of the Axiom of Union?
Given the ZFC Axiom of Union in the form :
$$\forall x \exists y \forall z (z \in y \iff \exists w (z \in w \land w \in x)) $$
or consider :
$$\forall x \exists y \text{ ("separate" every element ... ...
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Point-separating fields of clopen sets on compact spaces without Choice
In Matthew Dirk's Paper on Stone's representation theorem there is a proof of
Lemma 3.8. If X is a Stone space and F is a separating field of clopen subsets of
X, then F is the dual algebra of X; that ...
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How does one prove without the axiom of choice that the product of a collection of nonempty well-ordered sets is nonempty?
Suppose $\{X_{\alpha}\}_{\alpha\in\mathcal A}$ is an indexed family of nonempty well-ordered sets, where $X_{\alpha}=(E_{\alpha},\le_{\alpha})$ for each $\alpha$. It seems intuitively obvious that we ...
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Linear independence of tensor products $\{v_i \otimes w_j\}$ without choice
Let $V$, $W$ be vector spaces. Suppose $\{v_i\} \subseteq V$ and $\{w_j\} \subseteq W$ are linearly independent. Then $\{v_i \otimes w_j\} \subseteq V \otimes W$ is linearly independent. The usual ...
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Exterior and symmetric powers without choice
Let $R$ be a commutative ring, $F$ a free $R$-Module and $n\in \mathbb{N}$. Can it be proven in ZF that the canonical projections $F^{\otimes n}\twoheadrightarrow \bigwedge^n(F)$ and $F^{\otimes n}\...
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Is every set an image of a totally ordered set?
It is known that the statement "Every set admits a total order" is independent of ZF. See here, for example. However, can it be proven in ZF that for every set $Y$, there exists a totally ...
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How to construct a countable series of sets in $\mathbb{R}$ with no rational differences and complete coverage
I had this question in the test and unfortunately I didn't know how to prove it, I would appreciate some help:
Assuming Axiom of Choice, show that there exists a countable series of sets $A_0,A_1,A_2,...
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Proof Without Axiom of Choice: Infiniteness of Union
Without using any form of the axiom of choice, prove that if $A$ is infinite, then the set $\bigcup A$ is also infinite.
I have encountered this proposition in my studies and find it intriguing, yet I'...
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Dependence of the equation $a+1=a$ for infinite cardinals $a$ on the axiom of choice
let $A$ be a set such that for all $n \in $ N $ A ≉ N_n$
where $N_n = \{ 0 ,1 ,2 ...... n-1\} $
and $a$ be the Cardinality of $A$ meaning ($|A| = a$)
is it possible to prove that $a+1=a$ without ...
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Intuition for why the Power Set Axiom can not be used to derive the Axiom of Choice
Using the Axiom of Replacement, every set E, with elements e, has a mirror set E' with the property :
$$ E' := \{\langle E,e\rangle \mid e \in E \} $$
Again using the Axiom of Replacement, for any set ...
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