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Sep
30
awarded  Explainer
Sep
13
comment A question about $\dot{S^Q}$-semiproperness and revised countable support iterated forcing of length a limit ordinal
Your revised version does not make sense. There is no demand on $\dot \xi$, other than being a name decided by $p$.
Sep
13
comment A question about $\dot{S^Q}$-semiproperness and revised countable support iterated forcing of length a limit ordinal
Namba forcing may be semiproper in the traditional sense (i.e., $\{\aleph_1\}$-semiproper). If you let $\kappa=\aleph_3$, $\lambda=\aleph_4$, and you compose Namba with a forcing that collapses $\kappa$ to $\aleph_1$ with countable functions, then the whole forcing is $\lambda$-cc and hence satisfies your version of semiproperness, if I am not mistaken. Alternatively, let $\alpha$ be a measurable cardinal and use Prikry forcing.
Sep
11
comment A question about $\dot{S^Q}$-semiproperness and revised countable support iterated forcing of length a limit ordinal
Perhaps I am misunderstanding the question. Can't you just let $Q_0$ be an antichain of size $\alpha:=\aleph_2$, and then continue with an iteration which at the $Q_0$-generic coordinate makes $cf(\alpha)=\omega$ (and is otherwise trivial)?
Aug
30
comment reference on aperiodicity and cluster
To give some context: It seems that the source of this picture is this website: grahamshawcross.com
Aug
21
comment Is there a “hereditary” construction for $L$?
What do you mean by "similar"? I think that $L$ is "similar" to OD, because both are defined using quite explicitly definability relations. (And clearly "hereditarily in L" is the same as "in $L$".)
Aug
19
comment Are there fragments of set theory which are axiomatized with only bounded (restricted) quantifiers used in axioms?
This was my 100th answer. Where are the fireworks?
Aug
19
comment Are there fragments of set theory which are axiomatized with only bounded (restricted) quantifiers used in axioms?
@IoachimDrugus: if you are interested in working with infinite sets, adjunction won't get you very far. There is a good reason why we need a union axiom, an infinity axiom (in your theory that could be done by a constant) and (usually) a power set axiom (that could be a unary operation). (Btw, you meant $(x;y)=x$ iff $y\in x$, not if $x\in y$.)
Aug
19
comment Are there fragments of set theory which are axiomatized with only bounded (restricted) quantifiers used in axioms?
@AndreasBlass and @ Francois: An obvious drawback of ZFU is that it says nothing about the objects that are not in $U$. A minimalistic strengthening would be to demand $\forall x: x\in U \vee x=U$.
Aug
18
answered Are there fragments of set theory which are axiomatized with only bounded (restricted) quantifiers used in axioms?
Aug
17
comment Are there fragments of set theory which are axiomatized with only bounded (restricted) quantifiers used in axioms?
The tag "set theory" indicates to me that you might be interested in a theory that includes a version of the axiom of infinity. But the references in the wikipedia article seem to deal with computations with hereditarily finite sets only.
Aug
17
comment Are there fragments of set theory which are axiomatized with only bounded (restricted) quantifiers used in axioms?
What language are you thinking of? ZF is often written in a purely relational language (with no constant symbols), so there won't be any closed formulas in which all quantifiers are bounded. It seems to me you need at least the constant $\omega$. Which function symbols do you want to allow? Power set? Smallest your favorite cardinal above $\alpha$?
Aug
8
comment Can we unify addition and multiplication into one binary operation? To what extent can we find universal binary operations?
The preprint On the last question of Stefan Banach by Pasha Zusmanovich points out that the result in my answer was already proved by Jerzy Łoś in 1950 (Fund.Math 37, p. 84-86). Łoś uses this theorem of Sierpiński: Every countable set of unary functions on a given set X is included in the set generated by two unary functions.
Jul
17
awarded  Custodian
Jul
17
reviewed Approve suggested edit on Any rigorous way to claim that sums with repeat summands are few?
Jul
2
awarded  Curious
Jun
10
comment How can an ultrapower of a model of ZFC be “ill-founded” yet still satisfy ZFC?
That is what we often (implicitly) do. Rather than talking about arbitrary models of set theory (ZFC), we talk about transitive $\in$-models, which up to isomorphism are the same as well-founded models. --- However, this concerns only set theory itself; in mathematics outside set theory, already full first order ZFC is usually more than needed, and there is (usually, or at least often) no need to consider second order extensions.
May
13
answered Unconventional types of induction
May
6
awarded  Yearling
May
5
comment Examples of mathematics motivated by technological considerations
Some of us do speak German, and we see nothing wrong with that. All of us speak English, and there is nothing wrong with that, either. To avoid misunderstandings (in particular: giving the impression of being rude; I am not claiming that you are), you could change the first paragraph to something like "One of the reasons the Allies won WW II..."