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Results tagged with lo.logic
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user 109573
first-order and higher-order logic, model theory, set theory, proof theory, computability theory, formal languages, definability, interplay of syntax and semantics, constructive logic, intuitionism, philosophical logic, modal logic, completeness, Gödel incompleteness, decidability, undecidability, theories of truth, truth revision, consistency.
9
votes
Accepted
Can one exhibit an explicit Kuratowski infinite set without invoking Replacement?
The answer is no. Let ZC' be ZFC without replacement and infinity and with the assertion there is a Kuratowski infinite set. We will construct a model $M$ of ZC' such that only hereditarily finite ele …
- 4,426
3
votes
Surreals and NSA: some foundational issues
Problem 1: There is a definable proper class saturated real-closed field $\mathbb{R}^*$, defined by a slight modification of your and Shelah's construction, such that there is an $\mathrm{OD}_p$ injec …
- 4,426
12
votes
Accepted
Is this set theory equivalent to ZFC?
Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with par …
- 4,426
9
votes
Accepted
Can a Vopenka cardinal be supercompact?
If $\kappa$ is almost huge with target $\lambda,$ then $V_{\lambda}$ thinks that $\kappa$ is a supercompact Vopenka cardinal.
I'll take for granted the standard facts about almost huge cardinals liste …
- 4,426
5
votes
Accepted
The difference between Baire 2 and 'effectively Baire 2'
It's provable in ZF that every Baire-2 function is effectively Baire-2. It suffices to prove the following:
(ZF) There is an explicit function which maps each Baire-1 function $f: \mathbb{R} \rightarr …
- 4,426
6
votes
Accepted
Stronger negation of AC given by rejecting "infinite hat" puzzles
Naturally, the more generalizations of the infinite hats puzzle we consider, the stronger it is to assert that none of them have a paradoxical solution. One of the variants you linked in your question …
- 4,426
10
votes
Accepted
Can we make ZF − infinity + “all ordinals are finite” as strong as ZFC?
ZFfin implies every set is finite, and in particular choice holds. Suppose there is an infinite set $X.$ By replacing every element of $\mathcal{P}_{\text{fin}}(X)$ with its cardinality, we see $\omeg …
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9
votes
Accepted
Must strange sequences wear Russellian socks?
For $n \ge 2$ and $\langle A_i \rangle$ such that $|A_i| = n$ for all $i,$ TFAE:
$|\prod A_i| \neq |\mathbb{R}|.$
$\bigsqcup A_i$ is uncountable.
$\bigsqcup A_i$ is un-orderable.
There are $k \in [2, …
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9
votes
Accepted
Building the real from Dedekind finite sets
Q1: There is no such partition. Let $\langle X_n \rangle$ be a countable partition of $\mathbb{R}.$ We will construct $n,$ an open interval $I,$ and an injection $g: \omega \rightarrow I \cap X_n$ wit …
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9
votes
Accepted
The Parity Principle and $\mathbf{C}_2$ (choice for $2$-sets)
Over ZFA, the Parity Principle is strictly weaker. We'll show it follows from Multiple Choice, the assertion that for any family of nonempty sets $\mathcal{F},$ there is $g: \mathcal{F} \rightarrow [\ …
- 4,426
9
votes
Global Choice bi-interpretable with Global Wellorder?
The answer to the first question is yes: Global Choice is bi-interpretable with Global Well-Ordering.
First, I will prove that a class linear order $(C, <)$ is a well-ordering iff every subset (rather …
- 4,426
2
votes
Is "There exists an unbounded non-measurable set but no bounded non-measurable set" consiste...
This is more of a long comment than an answer.
The "right" notion of an unbounded set being measurable in ZF is less than clear. Suppose $\mathbb{R}$ is a countable union of countable sets. Let $\lang …
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11
votes
Accepted
Is $\in$-induction provable in first order Zermelo set theory?
The answer is no. Take the standard model of Z and add in a $\mathbb{Z}$-sequence of objects, each of whose only element is the previous one. I.e., define
$M=\bigcup_{n<\omega} \bigcup_{m<\omega}\mat …
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12
votes
Accepted
Consistency of a strong Fubini type theorem for measure zero sets
ZFC refutes this principle. Let $\kappa=\text{non}(\mathcal{L}),$ i.e. the least cardinality of a set of reals of positive outer measure. Let $X \subset [0,1]$ be such that $|X|=\kappa$ and $\lambda^* …
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22
votes
What notable theorems cannot be automatically proven without choice using Shoenfield absolut...
$\text{ZF}+ \text{AC}_{\omega}$ is not $\Sigma^1_4$-conservative over ZF and ZF + DC is not $\Sigma^1_4$-conservative over $\text{ZF}+ \text{AC}_{\omega}.$
An example of the former: the sentence $\ome …
- 4,426