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Results tagged with lo.logic
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user 5903
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.
10
votes
Does "compact iff projections are closed" require some form of choice?
This can be done choiceless and quite elementarily as follows:
- if $F$ is closed in $X\times Y$ and $y\notin\pi[F]$ then, as $F$ is closed its complement is the union of a family of basic open sets, …
- 11.4k
1
vote
Is there anything against this function j being injective?
Once you have such a $j$ you can make another one that is not injective: take a bijection $b:A\to A\times A$ and the projection $\pi:A\times A\to A$ onto the first coordinate.
Then the composition $j\ …
- 11.4k
2
votes
Accepted
Hahn-Banach theorem and ultrafilter lemma
I'm using the notation of the original paper.
In hindsight one could have used the principal ultrafilter generated by $\{n\}$, where $n$ is the index of the extension $F$ found in the proof. But that …
- 11.4k
5
votes
Difference between Laver's and Mathias's forcing
In this paper Alan Dow compares Laver and Mathias forcing regarding their effects on the algebra $\mathcal{P}(\omega)/\mathit{fin}$. He considers four iterations of length $\omega_2$: iterate $L$, ite …
- 11.4k
6
votes
A unique ultrafilter extending a union of filters?
The following is due to Alan Dow:
In any model obtained by adding $\aleph_2$ many Cohen reals to a model
of $\mathsf{CH}$ the statement is false.
We force with $\mathbb{P}=\operatorname{Fn}(\omega_2, …
- 11.4k
9
votes
Accepted
Is the set of $\kappa$-complete ultrafilters closed in $\beta X$?
If $\kappa=\aleph_0$ then yes: every ultrafilter is $\aleph_0$-complete.
If $\kappa>\aleph_0$ then no, if $\lambda X$ is nonempty. Split $X$ into countably many sets $\{X_n:n\in\mathbb{N}\}$, of the s …
- 11.4k
12
votes
Does Urysohn's Lemma imply Dependent Choice?
In Versions of normality and some weak forms of the axiom of choice Paul Howard et al exhibit a model of MC (Multiple Choice) and not-DC, see page 381.
In that model Urysohn's Lemma (NU) holds, so it …
- 11.4k
11
votes
Which topological spaces admit a nonstandard metric?
These are $\omega_\mu$-metrizable spaces, where $\omega_\mu$ is the cofinality of ${}^*\mathbb{R}$. Take a decreasing sequence $\langle x_\alpha:\alpha<\omega_\mu\rangle$ of positive elements that con …
- 11.4k
8
votes
Accepted
Can totally inhomogeneous sets of reals coexist with determinacy?
In Rigid Borel sets and better quasiorder theory (Logic and combinatorics, Proc. AMS-IMS-SIAM Conf., Arcata/Calif. 1985, Contemp. Math. 65, 199-222 (1987), zbMath review here) Fons van Engelen, Arnold …
- 11.4k
7
votes
Iterated forcing and CH
Here is a sketch.
We may assume that each $\dot Q_\alpha$ has $\omega_1$ as its universe; in which case the underlying set of $P_{\alpha+1}$ can be taken to be $P_\alpha\times\omega_1$ and the orderin …
- 11.4k
11
votes
"Transitivity" of the Stone-Cech compactification
The answer to Q1 is even more `no': Kunen showed that there are x and y such that x cannot be mapped to y and y cannot be mapped to x, see this review. This has been strengthened by Rudin and Shelah. …
- 11.4k
4
votes
Accepted
Simplified method of building an Aronszajn tree
Your argument is basically Kurepa's proof from his thesis Ensembles ordonnées et ramifiés, see page 96 (a footnote has Aronszajn's construction).
As noted in the comments you need to show that what yo …
- 11.4k
5
votes
Accepted
Stone-Čech compactification of a Boolean subalgebra of $\{0,1\}^S$
A partial answer.
for your specific example, the algebra $B$ generated by the rational intervals, the answer is negative. This algebra is countable and dense in $2^\mathbb{R}$. It is countable and ze …
- 11.4k
7
votes
Accepted
Embedding ordinals with the order topology into connected $T_2$-spaces
The answer is no: if $\lambda$ is larger than $\omega^2$ and if $X$ contains $\lambda+\omega$ then it also contains $\lambda+\omega+\omega$.
To see this observe that $\lambda+1$ is homeomorphic with $ …
- 11.4k
2
votes
Accepted
Large chromatic number in hypergraphs with large edges
For $\kappa=\aleph_0$ yes: there are (many) models with ultrafilters of character less than $\mathfrak{c}$. Let $E\subseteq[\omega]^\omega$ be a base for an ultrafilter, say $|E|=\aleph_1<\mathfrak{c} …
- 11.4k