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Results tagged with axiom-of-choice
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An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty sets. The axiom of choice is related to the first of Hilbert's problems.
21
votes
Accepted
Does completeness of the theory of a bijection without finite orbits depend on choice?
There are several ways how to prove the completeness of this theory (let me call it $S$) without choice:
Prove in a purely syntactic way (by induction on the complexity of a formula) that $S$ has qua …
- 47.1k
7
votes
Accepted
When does Skolemization require the axiom of choice?
Upon rereading the question, I sense there may be a terminological or conceptual confusion in play.
The terms “second-order logic” and “higher-order logic” are unfortunately used to denote two vastly …
- 47.1k
6
votes
Accepted
Does ZF prove that proximity spaces are completely regular?
The answer is no. In fact, it is consistent with ZF that $(*)$ there exists an infinite compact Hausdorff space $X$ such that every continuous function $f\colon X\to\mathbb R$ is constant, so that $X$ …
- 47.1k
9
votes
Involutions in the absolute Galois group (and the Axiom of Choice)
Let me spell out a completely explicit elementary proof that visibly makes no use of choice.
Lemma 1. Let $\sigma$ be an automorphism of order $2$ of a field $K$ of characteristic $\ne2$, and let $F$ …
- 47.1k
8
votes
Is there a model of ZF set theory with a set that does not inject into the cardinals?
This is a summary of my comments above showing that there is such a model of set theory with atoms (ZFA).
If $X$ and $B\colon a\mapsto B_a$ are as in the question, let me call $B$ an assignment funct …
- 47.1k
27
votes
Does the "three-set-lemma" imply the Axiom of Choice?
To complement godelian’s answer, the three-set lemma is not provable in ZF alone, as it implies the axiom of choice for families of pairs. This holds even if we allow any finite (or even just well ord …
- 47.1k
6
votes
Does k(X) have a k-basis for every set X, without AC?
While the question is still unanswered, let me make a couple of observations, showing that the problem is essentially equivalent to construction of symmetric bases (this is already alluded to in Is th …
- 47.1k
4
votes
Accepted
Do saturated models require choice?
On request, I summarize here some of the results mentioned in the comments (now unfortunately deleted), even though they do not really answer the question.
For definiteness, I assume that $M$ is satur …
- 47.1k
38
votes
Accepted
Can one show that the real field is not interpretable in the complex field without the axiom...
An interpretation of $(\mathbb R,+,\cdot)$ in $(\mathbb C,+,\cdot)$ in particular provides an interpretation of $\DeclareMathOperator\Th{Th}\Th(\mathbb R,+,\cdot)$ in $\Th(\mathbb C,+,\cdot)$. To see …
- 47.1k
12
votes
Accepted
Weakest choice principle required for Robertson-Seymour Graph Minor Theorem?
As far as I can see, the Robertson–Seymour theorem is provable in plain ZF.
First, the restricted version of the theorem for graphs whose vertices are natural numbers can be written as a $\Pi^1_1$ se …
- 47.1k