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Results tagged with lo.logic
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user 11115
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.
28
votes
Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)?
The answer to your question is (almost) yes (almost is because of the addition of DC to the statement).
Recently Gabriel Goldberg has proved
''Con(NBG+DC+Reinhardt)$ \implies$ Con(ZFC+I0)''.
…
- 32.2k
23
votes
Can infinity shorten proofs a lot?
Consider the following question of Erdos and Hajnal:
Question (Erdos-Hajnal) Is there a finite $K_4$-free graph which, when the edges colored by $2$ colors, always contains a monocolored triangle.
T …
- 32.2k
22
votes
Accepted
What is known about the consistency of $2^{\aleph_\alpha} = \aleph_{\alpha+\gamma}$ for all ...
By a result of Patai, $\gamma$ should be finite (this is exercise 5.15 in Jech's book).
For any finite $n>0, H_n$ is consistent, see Merimovich's paper A power function with a fixed finite gap every …
- 32.2k
18
votes
What is the large cardinal strength of the assertion that every $\kappa$-complete filter on ...
Let us call a cardinal $\kappa, \kappa-$compact, if every $\kappa-$complete filter on $\kappa$ extends to a $\kappa-$complete ultrafilter.
The following is proved in Gitik's paper On measurable cardi …
- 32.2k
17
votes
Does every set of reals contain a measure-zero set of the same cardinality? Does it contain ...
In the paper "Uncountable sets of real numbers with no uncountable subsets of measure zero"
it is proved that under the continuum hypothesis, there is an uncountable set of reals which has no uncounta …
- 32.2k
16
votes
Accepted
Fubini without CH
See Cardinal Conditions for Strong Fubini Theorems,
Joseph Shipman
Transactions of the American Mathematical Society
Vol. 321, No. 2 (Oct., 1990), pp. 465-481.
In general: Let $(X,A,μ)$ and $(Y,B,ν …
- 32.2k
16
votes
Accepted
Can $\kappa^\lambda$ be large if $2^\lambda$ is small and $\lambda<\mathrm{cof}(\kappa)$?
It is consistent that such a pair exists, see my paper Singular cofinality conjecture and a question of Gorelic.
To show that some large cardinals are needed, suppose for example $\lambda=\aleph_0 < \ …
- 32.2k
15
votes
Accepted
A proper class of ordinals without an infinite constructible subset
Stanley, M. C., A cardinal preserving immune partition of the ordinals, Fundam. Math. 148, No. 3, 199-221 (1995). ZBL0843.03028.
An infinite set (or class) of ordinals is said to be immune if it nei …
- 32.2k
15
votes
Latest status of core model theory?
The following may not be an answer to your question, but I think it is related.
I have taken it from the introduction of a joint work I am doing with James Cummings and Sy Friedman (which has now appe …
- 32.2k
14
votes
Non-set-theoretic consequences of forcing axioms
Indeed there is a vast of applications, for example:
Using Martin's axiom, Shelah showed that there is a non-free Whitehead group. The book ``
Consequences of Martin's Axiom'' contains many other exam …
- 32.2k
14
votes
Accepted
Is it consistent that $2^{(\cdot)}$ is "surjective" on the class of uncountable ordinals?
By Konig's lemma $cf(2^\kappa) > \kappa,$ so for example $\aleph_\omega$ can never be of the form $2^\kappa$ for any $\kappa.$
- 32.2k
13
votes
Accepted
Stationarity and Fodor's lemma for a (nice) poset?
You may look at the paper Regressive functions and stationary sets by Karsten Steffens (In: Müller G.H., Scott D.S. (eds) Higher Set Theory (Proc. Conf., Math. Forschungsinst., Oberwolfach, 1977), pp. …
- 32.2k
12
votes
Accepted
Can there be a tree of height $\omega_2$ having all levels countable, with no cofinal branch?
The following theorem of Kurepa answers the question.
Theorem (Kurepa) Suppose that $\kappa$ is regular, $\lambda< \kappa$ , and $T$
is a $\kappa$-tree each of whose levels has cardinality less …
- 32.2k
12
votes
If $\kappa$ is weakly inaccessible and $A\subset\kappa$, can $L[A]$ violate $\kappa^{\lt\kap...
By Lemma 2.2. of my paper Shelah's strong covering property and CH in V[r], we can show that:
Claim. If $A\subseteq \kappa,$ and if $Y\in L[A]$ is a bounded subset of $\kappa,$ then there exists a …
- 32.2k
12
votes
Accepted
Is injectivity of $2^{(\ldots)}$ weaker than $\mathsf{GCH}$?
No, take Merimovich's model in which $2^\kappa=\kappa^{+3}$ for all cardinals $\kappa.$
Merimovich, Carmi A power function with a fixed finite gap everywhere. J. Symbolic Logic 72 (2007), no. 2, 361 …
- 32.2k