# 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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### Is the axiom of choice used to order the orbits of the truncation function through periodic 2-adic numbers using their Lyndon words?

Let $X$ be the periodic two adic integers. Let $f$ be the truncation function, i.e. $f(x)=(x-1)/2$ for odd numbers and $x/2$ for even. Let $X/f$ be the set of transitive orbits of $f$ in $X$ These ...
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### What does it mean to explicity exhibit something (modulo a proof of its existence) which cannot be explicitly exhibited?

I was reading this answer that no free ultrafilter can be exhibited on the natural numbers. I have as a theorem that if the Collatz conjecture is true then the following is a free ultrafilter on the ...
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### There is no surjective function from a set in the Hartogs number of its power set

I'm trying to prove an equivalent state of the Axiom of Choice : Given two non-empty sets, there is a surjective function from one of them into the other one. If we prove that for any non-empty set $X$...
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### Partition of non-well-orderable sets

Call a non-well-orderable set "simple'' if whenever it is partitioned into two pieces, one of them is well-orderable. Does ZF prove that every non-well-orderable set has a simple non-well-...
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### Does the Axiom of Choice imply the existence of all the choice functions of a set?

We know that, given a set $X$, there exists at least one choice function $f:X\rightarrow\cup X$ thanks to the Axiom of Choice (AC). Can we conclude that all choice functions for a generic set $X$ ...
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### Questions on Jech's proof of the independence of AC from the ordering principle

In the the book The Axiom of Choice section 5.5, Jech presents a proof of the independence of the axiom of choice from the ordering principle (every set can be linearly ordered). There are some ...
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### $\bigcap\limits_{\varphi\in E^*}\ker(\varphi)$ and the Axiom of Choise

Context. Give a nonzero $K$-vector space $E$, it is known that $\displaystyle \bigcap_{\varphi\in E^*}\ker(\varphi)=0$ under AC. It is also known that, without AC, there are models of ZF in which some ...
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### Is choice provable if we add parameter free definability to ZF?

If we add the following $\omega$-inference rule to $\sf ZF$, would that entail the axiom of Choice? $\textbf{Definability: }$ if $\phi_1,\phi_2, \phi_3,...$ are all formulas in which only symbol "...
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### Iff propositions where both directions require choice?

Recently, I have been revising a basic course on noncommutative rings and modules over them. One proposition proven early on is all left modules over $R$ are free iff $R$ is a division ring and an ...
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### What is the relationship between the Boolean Prime Ideal Theorem and the Countable Axiom of Choice?

The base theory is $\mathsf{ZF}$. The Boolean Prime Ideal Theorem ($\mathsf{BPI}$) states Every Boolean algebra contains a prime ideal. The Countable Axiom of Choice ($\mathsf{AC}^{\omega}$) states ...
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### Relation between Axiom of Foundation and $\in$-induction

At the end of an intro to set theory course, we were introduced the Axiom of Foundation and the Principle of $\in$-induction as one of its consequences. I found it easy to prove that, assuming the ...
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The theorem of transfinite recursion states the following (A quick introduction to basic set Theory). Theorem 4.4 (Transfinite recursion). For every ordinal $\kappa$, set $A$, and map $^3 F: \... 0 votes 1 answer 47 views ### Infinite finitely splitting tree and AC The problem is: Using the Axiom of Choice, prove that if$(X,\leq)$is an infinite finitely splitting tree, then$(X,\leq)$has an infinite path. Be explicit where you use the Axiom of Choice. I have ... 11 votes 2 answers 331 views ### Is every complete consistent theory satisfiable in ZF? It is well-known that the compactness theorem for FOL is equivalent to the Boolean Prime Ideal Theorem over ZF, so it is a weak choice principle. The whole strength of BPIT in the proof of compactness ... 2 votes 1 answer 125 views ### Is AC invoked in the claim "the product/sum of a sequence of separable metrizable spaces is separable"? The Claim:The product/sum of a sequence of separable metrizable spaces is separable. Here the sum of a family$((X_i, d_i))_{i\in I}$of metric spaces is defined (up to isometry) as follows: By ... 0 votes 0 answers 52 views ### Does a total or well ordering has a countable cofinal I read a post sometime earlier asking about the cardinality of total ordering on a set, which I forgot about the detail but led me to this question. For a total/well-ordered set$A$does there exist ... 0 votes 0 answers 35 views ### Do we need the axiom of choice when extracting a linearly independent spanning set from any spanning set of a separable Hilbert space? [duplicate] Let$H$be a separable Hilbert space and$\{v_n\}$be some countable subset of$H$such that its linear span is dense in$H$.$\{v_n\}$need NOT be linearly independent. (Here, linear independence of ... 0 votes 0 answers 52 views ### Does probability theory need necessarilly Axiom of choice? [duplicate] I had seen the posting "https://math.stackexchange.com/questions/29381/picking-from-an-uncountable-set-axiom-of-choice?rq=1" but I'm still curious about it especially on the matter of the ... 2 votes 2 answers 107 views ### Constructing A Choice Function (Munkres Lemma 9.2) In Munkres, there's a proof for the existence of a choice function. I understood what was going on until the line underlined in red at the bottom. It seems that all the$B'$partition$\mathcal{B} \...
I have a set $B$ that can be written as $B = \cup_{\nu < \lambda} B_{\nu}$ where $\kappa$ is the cardinality of $B$, that is uncountable, and $\vert B_{\nu} \vert < \kappa$ with \$ \...