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Results tagged with set-theory
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user 491676
forcing, large cardinals, descriptive set theory, infinite combinatorics, cardinal characteristics, forcing axioms, ultrapowers, measures, reflection, pcf theory, models of set theory, axioms of set theory, independence, axiom of choice, continuum hypothesis, determinacy, Borel equivalence relations, Boolean-valued models, embeddings, orders, relations, transfinite recursion, set theory as a foundation of mathematics, the philosophy of set theory.
5
votes
0
answers
122
views
Large set of almost disjoint functions on a product space
Given an increasing sequence of cardinals $\langle\kappa_\alpha\mid \alpha\in\kappa\rangle$, let $K=\prod_{\alpha\in\kappa} \kappa_\alpha$, then we call $f,g\in K$ eventually different if there exists …
- 393
4
votes
2
answers
239
views
Extending normal filters
If $F$ is a $\kappa$-complete filter on some set $S$, and $F$ is generated by a basis of size $\lambda$, then $F$ extends to a $\kappa$-complete ultrafilter on $S$ when we assume that $\kappa$ is $\la …
- 393
3
votes
1
answer
116
views
Forcing equivalence and equal generic extensions
Two forcing notions $\Bbb P$ and $\Bbb Q$ could be defined to be forcing equivalent if the associated complete Boolean algebras are isomorphic (so, the CBA's formed by considering the regular opens of …
- 393
7
votes
0
answers
128
views
What is the forcing $\bf U$ from Bartoszyński-Judah?
In Set Theory - on the structure of the Real Line by Bartoszyński & Judah, a forcing notion $\bf U$ is mentioned on page 339, allegedly corresponding to $\rm{cof}(\cal N)$ as it has several properties …
- 393
6
votes
0
answers
272
views
Referring to the countability of $\Bbb Q$ as "Cantor's first diagonal argument"
I had a discussion with one of my students, who was convinced that they could prove something was countable using Cantor's diagonal argument. They were referring to (what I know as) Cantor's pairing f …
- 393
4
votes
1
answer
128
views
Coherent sequence of ultrafilters in iterated forcing extensions
Remember that if $\kappa$ is strongly compact, then any ${<}\kappa$-complete filter extends to a ${<}\kappa$-complete ultrafilter.
Let $\Bbb P_\delta=\langle\Bbb P_\alpha,\dot{\Bbb Q}_\alpha\mid \alph …
- 393