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Results tagged with axiom-of-choice
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user 1946
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.
173
votes
Most 'unintuitive' application of the Axiom of Choice?
I have enjoyed the other answers very much. But perhaps it
would be desirable to balance the discussion somewhat with
a counterpoint, by mentioning a few of the
counter-intuitive situations that can o …
116
votes
Why worry about the axiom of choice?
Yes, many people continue to fuss about the Axiom of Choice.
At least part of the explanation for why people continue to fuss as they do over the Axiom of Choice is surely the historical fact that the …
62
votes
Accepted
Zorn's lemma: old friend or historical relic?
I agree with almost everything in your post. But still, I believe I know why people use Zorn's lemma.
My answer. Zorn's lemma encapsulates succinctly many of the consequences of AC via transfinite rec …
- 236k
49
votes
0
answers
3k
views
Concerning proofs from the axiom of choice that ℝ³ admits surprising geometrical decompositi...
This question follows up on a comment I made on Joseph O'Rourke's
recent question, one of several questions here on mathoverflow
concerning surprising geometric partitions of space using the axiom
of …
- 236k
43
votes
Accepted
Cardinality: Why is there no "ℵ½"?
The point is that without the Axiom of Choice, cardinalities are not linearly ordered, and it is possible under $\neg AC$ that there are additional cardinalities to the side of the $\aleph$'s. Thus, t …
- 236k
41
votes
Accepted
Are all sets totally ordered ?
In the paper Dense orderings,
partitions and weak forms of choice, by Carlos G. Gonzalez FUNDAMENTA MATHEMATICAE 147 (1995), the author states the following theorem, where AC is the Axiom of Choice, …
- 236k
41
votes
2
answers
2k
views
On the difference between two concepts of even cardinalities: Is there a model of ZF set the...
An interesting question has arisen over at this
math.stackexchange
question
about two concepts of even in the context of infinite
cardinalities, which are equivalent under the axiom of
choice, but whi …
- 236k
39
votes
3
answers
3k
views
Can one show that the real field is not interpretable in the complex field without the axiom...
We all know that the complex field structure $\langle\mathbb{C},+,\cdot,0,1\rangle$ is interpretable in the real field $\langle\mathbb{R},+,\cdot,0,1\rangle$, by encoding $a+bi$ with the real-number p …
- 236k
32
votes
Accepted
How much of the axiom of choice do you need in mathematics?
Your hypothesis is in a sense stronger than just assuming ZFC outright.
Namely, if we have $\lambda$-DC for some inaccessible cardinal $\lambda$, and ZF in the background, then in particular, we will …
- 236k
32
votes
Is the statement that every field has an algebraic closure known to be equivalent to the ult...
As I mentioned in a comment to Eivind Dahl's answer, it
seems that there is also an easy argument directly from the
Compactness theorem of first order logic. Since you said you are looking for alterna …
- 236k
30
votes
Applications of Zorn’s lemma that aren’t chain-complete/directed-complete?
On the one hand, one might expect that there can be no fully satisfactory example of the phenomenon, in light of the observations mentioned in the comments, namely, that every partial order fulfilling …
- 236k
28
votes
Non-Borel sets without axiom of choice
If you assume the countable axiom of choice, then most sets
of reals are not Borel. Under AC, what you get is that
there are continuum many Borel sets, that is,
$2^{\aleph_0}$ many, but $2^{2^{\aleph_ …
- 236k
26
votes
Accepted
Are classes still "larger" than sets without the axiom of choice?
Yes, your remarks about incomparability of sets and classes without the axiom of choice are correct.
Yes, in ZF (or in GB), the axiom of choice is equivalent to the assertion that every set injects …
- 236k
25
votes
Axiom of choice and non-measurable set
No, the existence of a non-Lebesgue measurable set does not imply the axiom of choice. If ZF is consistent, then set-theorists can construct models of ZF having a non-Lebesgue measurable set, but stil …
- 236k
24
votes
Does $2^X=2^Y\Rightarrow |X|=|Y|$ imply the axiom of choice?
Here is a little progress towards AC.
Theorem.
ICF implies the dual Cantor-Schröder-Bernstein
theorem, that is $X$ surjects onto $Y$ and $Y$ surjects onto $X$,
then they are bijective.
Proof. You e …
- 236k