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Results tagged with lo.logic
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user 7206
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.
27
votes
Accepted
Is there a theorem whose only known proof uses "$A$ or not $A$" for undecidable $A$
Here is an example:
It is provable from $\sf ZF$ that there exists four infinite cardinals, $\frak p,q,r,s$ with $\frak p<q, r<s$ such that $\frak p^r=q^s$. (Here cardinals do not mean just finite …
- 39.8k
3
votes
1
answer
156
views
Rigid structure which is generically homogeneous
Is it possible to have a structure $T$ in some language which is rigid in $V$, but in a cardinal-preserving extension $T$ is homogeneous (in a suitable sense of the word)?
If this is not possible, is …
- 39.8k
8
votes
When does Skolemization require the axiom of choice?
This implies the axiom of choice.
Let $\{A_i\mid i\in I\}$ be a family of non-empty sets, consider the language with predicates $R_i$, for $i\in I$, with the model whose universe is $\bigcup A$ and $ …
- 39.8k
4
votes
Reverse Skolem's paradox
While the original question has been answered, remember that Skolem's paradox says that if $M$ is a countable model of set theory, then it knows only about countably many real numbers.
The reverse Sk …
- 39.8k
6
votes
1
answer
238
views
Definability of defining classes
Suppose that $M$ is a transitive class, denote by $\mathrm{HOD}(M)$ the class of all those sets which are hereditarily definable from ordinals and parameters in $M$.
Some trivial examples include $\m …
- 39.8k
1
vote
Accepted
Axiom of dependent choice (up to $\omega_1$) and group rank
Without sitting to verify the details in full, here is a sketch of a proof:
Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of …
- 39.8k
2
votes
Accepted
Some definitions without full choice
Without any appeal to the axiom of choice, if you have a vector space which is well-ordered then it has a basis. Moreover, if you have a generating set which is well-ordered then the vector space has …
- 39.8k
4
votes
Accepted
Is definability of a basis for $\mathbb{R^N}$ independent of ZFC?
The answer, if I understand the question correctly, is negative. That is, if we understand "definable" as "can be defined from ordinals [and a real number]", or in simpler words, sets which are in $\s …
- 39.8k
4
votes
First-order vs second-order provability
Replacing the induction with the Replacement Schema, it is a nice theorem that if $V_\kappa$ is a model of ZFC with second-order replacement axiom then $\kappa$ is inaccessible.
This is not true for …
- 39.8k
1
vote
Set Theory and Definability
We want to show that $L_\alpha$ is a set for every $\alpha\in Ord$.
$L_0 = \emptyset$ which is definitely a set.
If $L_\alpha$ is a set, and $L_{\alpha+1} \subseteq \mathcal{P}(L_\alpha)$ then it i …
- 39.8k
12
votes
Accepted
Fixed points of injective self-maps
As Yair suggests, a strongly amorphous set has this property.
Recall that an amorphous set is a set which cannot be split into two infinite sets. A strongly amorphous set is a set such that in additi …
- 39.8k
5
votes
Accepted
Is existence of this set consistent with Zermelo set theory minus choice?
First note that if $s\in S$, then there is a unique $r$ such that your condition holds. Simply because $\mathcal P(X)=\mathcal P(X')$ if and only if $X=X'$. So we can write $s^-$ for that $r$ and call …
- 39.8k
7
votes
Accepted
Does $|\kappa^{<\kappa}|=|\lambda^{<\lambda}|$ imply $\kappa=\lambda$?
No.
Consider when $2^{\aleph_0}=2^{\aleph_1}=2^{\aleph_2}=\aleph_3$, with $\kappa=\aleph_1$ and $\lambda=\aleph_2$.
If you allow for one of these to be singular, then consider $\kappa=\beth_\omega$ a …
- 39.8k
4
votes
Accepted
Can a stage of the cumulative hierarchy violate the partition principle?
If you are asking whether or not a $V_\alpha$ could violate the partition principle, the answer is easily yes.
As we all know, it is always the case that $\Bbb R$ can be partitioned into $\aleph_1$ pa …
- 39.8k
4
votes
Is acyclic ZF consistent?
It is impossible to have a rank that is both:
Ill-founded, and
successive steps are obtained by the power set operation.
The reason is simple, it would define a "Specker-tree" which is not a tree. R …
- 39.8k