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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.
53
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1
answer
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When does $A^A=2^A$ without the axiom of choice?
Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$
However without the axiom of choice this doesn't …
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53
votes
1
answer
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Does $2^X=2^Y\Rightarrow |X|=|Y|$ imply the axiom of choice?
The Generalized Continuum Hypothesis can be stated as $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. We know that GCH implies AC (Jech, The Axiom of Choice, Theorem 9.1 p.133).
In fact, a relatively weak form …
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0
answers
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How many algebraic closures can a field have?
Assuming the axiom of choice given a field $F$, there is an algebraic extension $\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is …
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38
votes
Accepted
Linear Algebra without Choice
Some things about vector spaces which are consistent with the failure of choice:
Vector spaces may have bases of different cardinality. In particular, this means that the notion of "dimension" is no …
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30
votes
Accepted
How to construct a basis for the dual space of an infinite dimensional vector space?
It is consistent with the axioms of $\sf ZF$ that this is impossible. Specifically, if you consider $\Bbb R[x]$, then its dual space is just $\Bbb{R^N}$. And it is consistent with $\sf ZF$ that $\Bbb{ …
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27
votes
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answer
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If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?
Suppose $V$ is a vector space, we say that $\mathcal B$ is a basis for $V$ if:
Every $v\in V$ can be written as a linear combination of elements of $\mathcal B$;
If $\sum\alpha_i b_i = 0$, where $\a …
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26
votes
Accepted
How much of GCH do we need to guarantee well-ordering of continuum?
Yes, you are right. This is a theorem of Specker. If there are no intermediate cardinals between $A,\mathcal P(A)$ and $\mathcal{P(P(}A))$, then $A$ can be well-ordered.
You can find nice details in:
…
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26
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2
answers
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Sizes of bases of vector spaces without the axiom of choice
Assuming the axiom of choice does not hold we have that there is a vector space without a basis. The situation can be, in some sense, worse. It is consistent that there are vector spaces that have two …
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22
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Hahn's Embedding Theorem and the oldest open question in set theory
I don't know the answer to (1), and would be glad to give it some thought later this week. Regardless to (1) the answer to (2) is semi-negative.
There are two conjectures which seem to be slightly ol …
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22
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3
answers
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Half Cantor-Bernstein without choice
I had a discussion with one of my teachers the other day, which boiled to the following question:
Assume ZF. Let $A,B$ be sets such that there exist $f\colon A\to B$ which is injective and $g\colon A …
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22
votes
1
answer
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The Continuum Hypothesis and Countable Unions
I recently edited an answer of mine on math.SE which discussed the implication of the two assertions:
$AH(0)$ which is $2^{\aleph_0}=\aleph_1$, and
$CH$ which says that if $A\subseteq 2^{\omega}$ and …
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22
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4
answers
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What sort of large cardinal can $\aleph_1$ be without the axiom of choice?
Assuming the axiom of choice it is very easy to see that $\aleph_1$ is a regular Joe of a successor cardinal. It is not very large in any way except the fact that it is the first uncountable cardinal. …
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21
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Compactness of the Hilbert cube without the Axiom of Choice
If by the Hilbert cube you mean only $[0,1]^\mathbb N$ then the answer is yes. There is such proof, you can find it in Herrlich's The Axiom of Choice as Theorem 3.13.
If you mean the general case of …
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20
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Accepted
Relation between the Axiom of Choice and a the existence of a hyperplane not containing a ve...
It is not hard to see that this statement is equivalent to "In every vector space, for every vector $v$ there is a functional $f$ such that $f(v)=1$".
If $\cal P$ holds, then the projection onto $k\ …
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20
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Accepted
Proof/Reference to a claim about AC and definable real numbers
The argument, given to me by Hugh Woodin over a drink in Barcelona in September 2016, is a nice retort to "AC is obviously false since there cannot be a well-ordering of the real numbers". It is argua …
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