# Tag Info

## Hot answers tagged axiom-of-choice

17

Impossible, by the result from Hickman (1978). He proves that the existence of a Dedekind-finite infinite field (that is, an infinite field does not have any countable subset) is consistent with ZF. More precisely, he proves the following theorem: Theorem. For a prime $p$ and a sequence of natural numbers $m_0\mid m_1\mid\cdots$, it is consistent with ZF ...

16

The argument does not really use the axiom of choice. Fix a countable basis $\{B_n\mid n\in\Bbb N\}$, now note the following is true. If $U,V$ are open subsets of $\Bbb R^n$, then $U=V$ if and only if $\{n\mid B_n\subseteq U\}=\{m\mid B_m\subseteq V\}$. This gives us an injection from the open sets to subsets of $\Bbb N$. Now using Cantor-Bernstein, ...

11

You don't need choice here at all - every open set $U$ has a canonical description in terms of basic open sets, $code(U)=\{B_\eta: B_\eta\subseteq U\}$ (where $\{B_\eta: \eta<\omega\}$ is the set of all balls with rational radius and center). We clearly have $code(U)=code(V)\iff U=V$, and the set of codes has size continuum.

10

The main problem here is that the name "Axiom of Choice" leads people to think that the axiom says something about our ability to choose things. Then, whenever we choose something, like one instance of an axiom schema, they think the axiom of choice is involved. The axiom of choice is not about our abilities at all. It is about the existence of certain ...

8

Your colleague is not using the term "choice principle" with the technical sense it has is set theory; probably he is not aware that those words have a specialized technical meaning, but has heard (or read) arguments that use it and tried to reconstruct their meaning from the everyday English meaning of "choice". That doesn't go well. In set theory "choice ...

8

Here's a sketch for a concrete construction of a counterexample, probably close to the result by Hickman that Hanul Jeon found: If I remember field theory right, ZF proves that there exists: A nested sequence of finite fields $$\mathbb F_p = K_0 \subseteq K_1 \subseteq K_2 \subseteq \cdots$$ such that $K_n$ has $p^{2^n}$ elements, and the only ...

5

The part about atlases is, as Asaf noted, a duplicate, but let me answer the general question whether uniqueness lets you avoid the axiom of choice. The first counterexample that comes to mind is the smallest ordinal number whose cardinality (i.e., the number of smaller ordinals) equals the cardinality of the real line. The existence of this ordinal is ...

4

The definition of forcing is the same with and without the axiom of choice. And the truth lemma holds with and without the axiom of choice. Namely, $$p\Vdash_\Bbb P\varphi\iff\text{For every }V\text{-generic } G\subseteq\Bbb P\text{ with }p\in G: V[G]\models\varphi$$ You can also consider iterations without choice, at least with a two-step iteration this ...

4

No, this reflects a total misunderstanding of what the axiom of choice is. The axiom of choice is no more needed to invoke an instance of an axiom scheme than it is needed to say that $43$ is a natural number (after all, that is also "choosing" an element of an infinite set). In almost all situations where you use an instance of an axiom scheme, you ...

4

No, choice is irrelevant here. First of all, as long as your language is countable, there are only countably many formulas, so in particular the set of formulas is well-ordered and no choice is needed to select one. More importantly, though, proofs are finite: even if we allow uncountable language, we only ever need to make finitely many choices, and this ...

4

This can be proved in several ways. The easiest I know is to use Cohen's second model (you can find it in Jech The Axiom of Choice in Chapter 5) in which there is a sequence $\langle A_n\mid n<\omega\rangle$ such that: $A_n\subseteq\mathcal P(\Bbb R)$. $A_n$ has two elements. If $n\neq m$, then $A_n\cap A_m=\varnothing$. $\prod A_n=\varnothing$. ...

2

The Axiom of Choice is not being invoked when choosing an instance of an axiom scheme. You are selecting one element from an infinite set. The Axiom of Choice concerns choosing an infinite set. If $\{A_\alpha\}$ is an infinite collection of distinct sets, the Axiom of Choice allows us to an element of each set.

2

As Asaf points out in the comments, $\cal L$ is already a $\sigma$-algebra in $\sf ZF$, so the problem is trivial and satisfied by $\Sigma=\cal L$. (Edit: This statement is followed by a long proof, which reminds me of a certain joke.) We start with the definition of $\mu^*(A)$, the Lebesgue outer measure, defined as ...

2

Not necessarily. You can prove in $\sf ZFC$ that the algebraic closure of $\Bbb Q$ exists and it is unique up to isomorphism. And while you can always prove that $\Bbb Q$ has an algebraic closure, you cannot prove it is unique without appealing to the axiom of choice. Another example in the same vein is the well-defined notion of a dimension of a vector ...

2

I believe that's right (without specifying the exercise, I can't know for sure; but this is a feature of the simplest permutation model, so it's probably true). Note, however, that other models can be constructed where $U$ can be linearly ordered, but not well-ordered. I don't understand what you are asking when you say you "would like to know how to obtain ...

1

I believe the answer is yes, as follows: Start with a model of ZF+atoms, $M$, with a set of atoms $A$ which forms a group isomorphic to $\mathbb{D}/\mathbb{Z}$, where $\mathbb{D}$ is the set of dyadic fractions: $\mathbb{D}=\{{p\over 2^k}: p, k\in\mathbb{Z}\}$. Let $G$ be the group of automorphisms of $A$, and consider the symmetric submodel $N$ of $M$ ...

1

I'm curious what is wrong with the following argument then: Agree in addition upon the choice of representatives of functions $\mathbb N\to\mathbb N$ where two functions are equivalent if they differ in finitely many places. Do as before and open all "other mathematicians" boxes. You get the sequence of numbers $x(m'), m'\ne m$ as before. Take its ...

1

There's no definite way to create a set of sock pairs that don't admit a choice function, using the usual constructions allowed in set theory. After all, it is known that the Axiom of Choice is consistent with the usual axioms of set theory, meaning that it cannot be proved that such at sock set exists at all -- and an explicit construction of one would ...

1

Because a choice function, working on $\emptyset$, would be a function for which $f(\emptyset)\in\emptyset$, which is obviously untrue.

1

If I recall correctly, the intent of the book's phrasing was not to imply anything about whether it is actually strictly stronger. I certainly haven't ever seen a proof that 3.8.3 implies the stronger statement where $Y$ isn't a set; but I don't think I've ever seen a proof that it doesn't either. Which means that I guess it is an open problem. I suspect ...

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