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Results tagged with axiom-of-choice
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user 113405
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.
5
votes
Accepted
Does n-well ordered choice schema imply the axiom of choice?
$2$-well ordered choice is enough to imply AC.
Let $α$ be any ordinal, and look at $\mathcal P^2(α)\setminus\{\emptyset\}$.
We have $x\in^2 \mathcal P^2(α)\setminus\{\emptyset\}⇒∃z⊆\mathcal P(α)\;(x\i …
- 1,676
6
votes
0
answers
164
views
Weak trichotomy principle in the absence of choice
It is well known that the trichotomy property of cardinals ($∀κ,λ\in\operatorname{Card}(κ<λ∨κ=λ∨κ>λ)$) is equivalent to the axiom of choice.
D. Feldman and M. Orhon had defined in [1] a generalization …
- 1,676
12
votes
1
answer
533
views
Building the real from Dedekind finite sets
It is well known that the real numbers can be countable union of countable sets by starting with GCH and taking a finite support permutations while collapsing all of $\aleph_n$ for natural $n$.
The id …
- 1,676
11
votes
1
answer
768
views
Scott's trick without regularity
In ZF(C), one can easily get a class partition of $V$, we can even get an $\mathrm{Ord}$-partition using the Cumulative hierarchy: $P=\{V_{α+1}\setminus V_α\mid α∈\mathrm{Ord}\}$, such a partition let …
- 1,676
11
votes
Simpler proofs using the axiom of choice
There is an interesting class of problems that are provable in $ZF$ by "coincidence" in the sense that either (a fragment of) $AC$ holds and then it is provable from the proof in $ZFC$, or there speci …
14
votes
1
answer
489
views
Injection into a proper class and choice without regularity
In $\sf ZF$, we have that the axiom of choice is equivalent to:
For all sets $X$, and for all proper classes $Y$, $X$ inject into $Y$
and
For all sets $X$, and for all proper classes $Y$, $Y …
- 1,676
4
votes
Accepted
Does cardinal definable choice imply AC?
Over ZF yes, it does.
Let $X$ be any family of nonempty sets, and let $κ$ be cardinal such that $ρ(κ)>ρ(X)$, then $V_{ρ(κ)}\setminus\{∅\}$ is cardinal definable, hence has a choice function that induc …
- 1,676
4
votes
1
answer
206
views
Exactly how much (and how little) can partial ordered sets (classes) embed to the cardinalities
In the paper "Convex Sets of Cardinals", Truss mentioned a result of Jech:
If $M$ is a countable transitive model of ZFC, and $(P,<)∈M$ is a poset, then there exists a Cohen extension of $M$ such tha …
- 1,676
4
votes
Accepted
Automorphisms of vector spaces and the complex numbers without choice
This is not a full answer, but it is too long to be a comment.
Let $B(F)$ for field $F$ be the statement "every vector space over $F$ has a basis" and let $AL19(F)$ be the statement "for every vector …
- 1,676