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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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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