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Results tagged with lo.logic
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user 1946
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.
3
votes
Accepted
Can one say that there are equal numbers of sets satisfying formulas in Second Order Arithme...
Perhaps I've misunderstood, but isn't the answer easily yes? You can express that $\phi$ has exactly $n$ realizers by saying: there is a class $Y$ coding a list of $n$ classes (for example, $Y$ consis …
- 12.9k
13
votes
Essential reads in the philosophy of mathematics and set theory
Last fall I taught a course in the Philosophy of Set Theory at NYU and you can find the reading list available on my web page. This course was more narrowly focused on the question of realism and plur …
- 236k
17
votes
2
answers
2k
views
Do the surreal numbers enjoy the transfer principle in ZFC?
The surreal field $\newcommand\No{№}\No$ is definable in ZFC, and it is easy to see that the surreal order is $\kappa$-saturated for every cardinal $\kappa$, precisely because we fill any specified ga …
- 13.1k
14
votes
Accepted
Does synonymy seep down to the fragments of theories?
It's a very nice question!
The answer is negative. For a counterexample, consider:
Let $T$ be the theory of a partial order $\leq$, that is, a reflexive, transitive, antisymmetric relation.
Let $H$ b …
- 236k
7
votes
How many colors do we need?
You are asking for the smallest cardinal $\lambda$ for which $2^{|\newcommand\R{\mathbb{R}}\R|}<\lambda^{|\R^3|}$.
The answer is $\lambda=(2^{|\R|})^+$.
First of all, this cardinal works, since $\lamb …
- 236k
2
votes
Accepted
To which arithmetic\set theory this theory is bi-interpretable?
Your theory is true in the one-element universe $\{a\}$ in which $a<a$ is true and $a\in a$ is false. The order transitivity holds trivially in this case; the finiteness axiom holds vacuously; and the …
- 236k
24
votes
Accepted
Which recursively-defined predicates can be expressed in Presburger Arithmetic?
Presburger arithmetic admits elimination of quantifiers, if one expands the language to include truncated minus and the unary relations for divisibility-by-2, divisibility-by-3 and so on, which are de …
- 4,713
7
votes
Are there models of ZF in which all uncountable sets are super/hyper/ultra-singular?
The ordinal $\omega_1$ is uncountable, even in ZF, but it can never be hyper-singular, since otherwise we would have $\omega_1=\cup x$ where $\omega_1>|x|$ and so $x$ is countable, and the elements $y …
- 236k
10
votes
What is the relationship between non-existence of those kinds of singular sets and AC?
Yes, it is equivalent to axiom of choice to say that there is no supersingular set.
It is easy to see with AC that there is no supersingular set.
Conversely, suppose that there is no supersingular set …
- 236k
20
votes
Accepted
Can there exist a definable "ultrafilter" on the ordinals?
As Gabe pointed out in the comments, your requirements are too strong, since your properties will imply that the definability hierarchy collapses, as you require that every definable class of ordinals …
- 236k
11
votes
Accepted
If existence of a pre-isomorphism implies existence of an isomorphism, would AC follow?
You say
Now using axiom of choice one can easily prove that for any distinct sets if there exists a pre-isomorphism between them then there exists an isomorphism between them.
But this is not true. …
- 236k
2
votes
Can this semi-constructible structure satisfy existence of a measurable cardinal?
Update. This answer is not correct, because it is a subtler matter to ensure that the levels are closed under relative constructibility while also maintaining $L_{\alpha+1}\cap\mathfrak{L}_\alpha=L_\a …
- 236k
12
votes
Accepted
Does this ZFC+V=L like theory, have a limit on large cardinal properties?
The answer is no, you cannot have measurable cardinals consistently with your theory.
Your theory includes the axiom "V=L or V=L[c] for an $L$-generic Cohen real $c$". This statement is provable from …
- 236k
6
votes
Accepted
Can there exists a model of ZFC with permutation that sends successor infinite stages to the...
The answer is no.
If $j$ maps each $V_{\omega+n+1}^M$ onto $V_{\omega+n}^M$ for $n\in\omega^M$, then $j[V_{\omega+\omega}^M]=V_{\omega+\omega}^M$ and so it cannot be that
$j[V_{\omega+\omega+1}^M]=V_{ …
- 236k
16
votes
Accepted
Example of a forcing notion with finite-predecessor condition that does not add reals
The answer is yes, because every forcing notion is equivalent to a forcing notion with finite predecessors.
Theorem. Every forcing notion is forcing equivalent to a forcing notion with finite conditio …
- 236k