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Results tagged with axiom-of-choice
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user 7206
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.
10
votes
Accepted
Axiom of choice and vector spaces over a given field
Allow me to steal my answer to a question from math.SE on the same topic. With minor changes
In their paper, Howard and Tachtsis discuss these sort of questions. The paper was published rather rece …
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12
votes
Accepted
Subsets of $M$ such that $2M \cong M$
No.
Simply because if $A$ is a set of any cardinality, there is $B$ such that:
$|A|\leq|B$, and
$|B|+|B|=|B|$, and in fact we can require $|B\times B|=|B|$.
Just take $B=A^\omega$, all the functi …
- 39.8k
8
votes
When does Skolemization require the axiom of choice?
This implies the axiom of choice.
Let $\{A_i\mid i\in I\}$ be a family of non-empty sets, consider the language with predicates $R_i$, for $i\in I$, with the model whose universe is $\bigcup A$ and $ …
- 39.8k
5
votes
Accepted
A question about Cantor's Power Set theorem without the Axiom of Choice
For the first equality, the answer is true.
It is quite easy to construct examples where the set of finite subsets is strictly larger. For example if $X$ is an infinite Dedekind-finite set which is t …
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1
vote
Accepted
Axiom of dependent choice (up to $\omega_1$) and group rank
Without sitting to verify the details in full, here is a sketch of a proof:
Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of …
- 39.8k
2
votes
Accepted
Some definitions without full choice
Without any appeal to the axiom of choice, if you have a vector space which is well-ordered then it has a basis. Moreover, if you have a generating set which is well-ordered then the vector space has …
- 39.8k
3
votes
Group morphism and axiom of choice
Well, this depends on what do you mean by "a form of choice". If you mean a statement of the form "Every family with property ... admits a choice function", then the answer, to my best of knowledge, i …
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4
votes
Why doesn't choice imply global choice (in NBG)?
As Noah S said, you have to use global choice to choose a well-order for each $V_\alpha$.
But we can have a concrete example for this sort of failure. Consider an inaccessible cardinal $\kappa$, and …
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4
votes
Accepted
Well-Ordering theorem of cardinal$\kappa$
The counterpart you may intend to use is that every cardinal is comparable with $\kappa$. This means that every infinite set which is not smaller than $\kappa$ has a subset of cardinality $\kappa$.
T …
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7
votes
1
answer
440
views
For models of ZF, if for some $A$ we have $L[A] = L$, what can we deduce about $A$?
Suppose $V$ is a model of ZF. Within $V$ we have $L$ which is a model of ZFC, furthermore $L[A]$ is a model of choice for every $A\in V$.
Suppose $A=\emptyset$ then clearly $L[A]=L$, furthermore if $ …
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8
votes
Accepted
Does ZF bound countable unions of countable sets?
In Jech The Axiom of Choice, problem 14 on chapter 5 states:
Let $M$ be a transitive model of ZFC, there exists $M\subseteq N$ with the same cardinals as $M$ [read: initial ordinals] and the followin …
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4
votes
Axiom of Choice and Vitali's theorem
The existence of a non-measurable set is completely too weak to prove the axiom of choice.
It is consistent with ZF (without large cardinals at all) that the real numbers are a countable union of co …
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13
votes
Consistency of a non-measurable set of reals when the continuum cannot be well-ordered
Shelah proved that assuming $\sf ZF+DC$, if every set of reals is Lebesgue measurable, then $\omega_1$ is inaccessible to reals.
This alone should hint you that it is easy to arrange models where $2^ …
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3
votes
Accepted
Can the axiom of choice or its weaker versions be (dis)proved using reflection principles?
The question seems to me asking if sufficiently large cardinals defined by indescribability properties will prove the axiom of choice or its weak variants hold below such cardinals. (Disproving is moo …
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8
votes
Accepted
Countably generated $\sigma$-algebras of ${\mathcal P}({\mathbb R})$ and choice
We can cheat in the following way:
Suppose $\lbrace A_n\rbrace$ is a countable collection of countable sets, let $\lbrace B_n\rbrace$ be an enumeration of the countable collection of open intervals w …
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