All Questions
Tagged with axiom-of-choice lo.logic
273
questions
14
votes
2
answers
1k
views
Proof/Reference to a claim about AC and definable real numbers
I’ve read somewhere on this site (I believe from a JDH comment) that an argument in favor of AC (I believe from Asaf Karagila) is that without AC, there exists a real number which is not definable ...
- 243
10
votes
1
answer
613
views
Infinitary logics and the axiom of choice
Suppose we want to enhance ZF by allowing for infinitary formulas instead of just first-order ones in our axiom schema of separation and/or replacement. It seems that we don't need much power in our ...
- 4,857
-2
votes
1
answer
162
views
Is discriminative choice provable in ZFC?
Let $\phi$ be a formula defining an equivalence relation.
Definitions:
The $\phi$-cardinality of a set $X$ be the cardinality of $X/\phi$. That is, the cardinality of the set of all equivalence ...
- 10.2k
10
votes
1
answer
488
views
Must strange sequences wear Russellian socks?
This is an attempt to make more precise a vague guess at the end of this answer of mine. We work in $\mathsf{ZF}$ throughout.
Say that a sequence $\mathcal{A}=(A_i)_{i\in\omega}$ of disjoint sets is ...
- 20.7k
6
votes
2
answers
458
views
Does Well-Ordered Interval Power Set "WOIPS" principle , prove $\sf AC$ in $\sf ZFA$?
Does $\sf ZFA + WOIPS$ prove $\sf AC$?
Where $\sf WOIPS$ is phrased as: for every infinite set $X$ the set of intervening cardinals between $|X|$ and $|\mathcal P(X)|$ is well-ordered.
In $\sf ZF$, I ...
- 10.2k
5
votes
1
answer
149
views
The equivalence of Dedekind-infinite and dually Dedekind-infinite as a weak form of (AC)
A set $X$ is Dedekind-infinite if there is an injective map $f: X\to X$ that is not surjective.
A set $X$ is dually Dedekind-infinite if there is a surjective map $f: X\to X$ that is not injective.
In ...
26
votes
1
answer
897
views
A cardinal inequality for finiteness
Nearly ten years ago, I explained in a blog post that, assuming only ZF, a cardinal number $\mathfrak{n}$ is finite if and only if it satisfies this monstrous inequality:
$$2^{2^{2^{2^{\mathfrak{n}}}}}...
- 43.8k
1
vote
1
answer
298
views
Proof of the axiom of choice for finite sets in ZF [closed]
Let the set $A$ be finite and $\emptyset \notin A$. How can I, without using the axiom of choice, prove by mathematical induction that there exists a function $f : A \rightarrow \bigcup A$ satisfying $...
- 21
3
votes
0
answers
161
views
Periodicity in the cumulative hierarchy
Under Reinhardt cardinals in ZF, the cumulative hierarchy exhibits a periodicity in that for large enough $λ$, certain properties of $V_λ$ depend on whether $λ$ is even vs odd. See Periodicity in the ...
- 6,365
7
votes
0
answers
180
views
Large cardinals beyond choice and HOD(Ord^ω)
Are Reinhardt and Berkeley cardinals (and even a stationary class of club Berkeley cardinals) consistent with $V=\mathrm{HOD}(\mathrm{Ord}^ω)$ ?
It seems natural to expect no, but I do not see a proof....
- 6,365
4
votes
0
answers
144
views
Consistency of definability beyond P(Ord) in ZF
Is it consistent with ZF that the satisfaction relation of $L(P(Ord))$ is $Δ^V_2$ definable? More generally, is it consistent with ZF that there is a $Δ^V_2$ formula (taking $α$ as a parameter) that ...
- 6,365
5
votes
1
answer
270
views
Would strengthening Foundation and Choice in NBG, make it equi-consistent with MK?
This is a follow up to an earlier question about strengthening of foundation in relation to proving the consistency of ZFC. It was shown that it would achieve that, but it may fail short of MK. Here, ...
- 10.2k
5
votes
1
answer
230
views
Long chains of Dedekind finite sets
This is a variation on this question with amorphous cardinals replaced with dedekind finite sets.
Dedekind finite sets are sets that have no countable subset, and it is well known that this is a ...
- 384
13
votes
1
answer
549
views
Long chains of amorphous cardinalities
An amorphous set is an infinite set that cannot be partitioned into 2 infinite subsets. An amorphous cardinality is the cardinality of an amorphous set. Working in $\sf ZF$, it is consistent that ...
- 384
12
votes
1
answer
506
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 ...
- 1,654