All Questions
Tagged with axiom-of-choice constructive-mathematics
10
questions
5
votes
1
answer
266
views
Does weak countable choice imply that the Cauchy reals are Dedekind complete?
Assuming the axiom of weak countable choice, is the set of modulated Cauchy reals Dedekind complete?
The second theorem on this ncatlab page claims something equivalent, but it doesn't contain a proof ...
- 6,298
10
votes
0
answers
364
views
Why is the double negation of the axiom of choice rarely considered?
In constructive/intuitionistic mathematics, it is common to reject the axiom of choice, because it is highly nonconstructive and implies the law of the excluded middle by Diaconescu's theorem/Bishop's ...
- 1,843
11
votes
1
answer
438
views
A weak form of countable choice
Let $\Omega$ be the set/type of truth values. We're using constructive logic. Define
$AC_{0, 0} = \forall P : \mathbb{N}^2 \to \Omega, (\forall n \in \mathbb{N}, \exists m \in \mathbb{N}, P(n, m)) \to ...
- 263
25
votes
1
answer
2k
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In what ways is ZF (without Choice) "somewhat constructive"
Let me summarize what I think I understand about constructivism:
"Constructive mathematics" is generally understood to mean a variety of theories formulated in intuitionist logic (i.e., not assuming ...
- 30.8k
16
votes
2
answers
815
views
Cauchy real numbers with and without modulus
In constructive mathematics there are many possible inequivalent definitions of real numbers. The greatest variety seems to be in Dedekind-style approaches: in addition to "the" Dedekind real numbers ...
- 65.8k
14
votes
1
answer
2k
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Bishop quote stating that axiom of choice is constructively valid
This is about constructive mathematics, but it is not a research question. But since it may also be of interest for research mathematicians, I hope this question is appropriate for this forum. As ...
- 149
16
votes
1
answer
2k
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Axioms of Choice in constructive mathematics
There is a widely accepted opinion that the Axiom of Countable Choice (further, ACC)
$$ \forall n\in \mathbb{N} . \exists x \in X . \varphi [n, x] \implies \exists f: \mathbb{N} \longrightarrow X . \...
- 791
4
votes
1
answer
168
views
Does this axiom (a weak form of class valued choice) has a name?
At some point in my work (which has nothing to do with set theoretics foundation) I need to consider the following axiom:
For any set $X$, any class $V$ with a surjective map $f : V \...
- 40.8k
2
votes
2
answers
991
views
Axiom of choice and convergence
Hi fellows,
I was wondering. Is the axiom of choice used to show that $\mathbb{R}$ is complete? If yes, is there a way to construct monotonic bounded sequences that do not converge?
Thanks in ...
27
votes
4
answers
12k
views
Does constructing non-measurable sets require the axiom of choice?
The classic example of a non-measurable set is described by wikipedia. However, this particular construction is reliant on the axiom of choice; in order to choose representatives of $\mathbb{R} /\...
- 3,145