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15h
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 $L$".)
2d
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?
2d
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 $x\in y$.)
2d
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..."
May
5
comment Which linearly ordered sets have the property that their completion is equipotent with their powerset?
Really $({\le 2})$-to-1.
May
4
comment Which linearly ordered sets have the property that their completion is equipotent with their powerset?
It is possible to use an order-theoretic "derivative" to prove this: identify two points of L if the interval between them is finite. Do the same on the quotient L', etc. After $\alpha<\omega_1$ steps you either you have found a homomorphic image $\mathbb Q$ (representatives will again have order type $\mathbb Q$), or the order was scattered and you end up with a finite order. In going from L to L' you lose only countably many unfilled cuts: some unfilled cuts $C$ in L (i.e., without supremum) may have an image which is filled by an element $x_C$ of L', but the map $C\mapsto x_C$ is 1-1.
May
4
comment Is the decomposition of an algebra into irreducible components essentially unique?
Bjarni Jónsson's example is very nice. But it seems to me that the group operation is not necessary; the unary function suffices.
May
4
comment Consistency Results Separating Three Cardinal Characteristics Simultaneously
I am not so familiar with historic iteration (front.math.ucdavis.edu/math.LO/9607227), but it seems to me that this is quite different from our paper. Our forcings are not ccc, and our forcing looks more like a product than an iteration, in the sense that there is no obvious linear order on the coordinates; each "iterand" produces a real which is somewhat generic over all the others. Also (bug or feature?) we do not increase $\mathfrak d$. At the moment.
May
4
answered For which cardinal numbers $\kappa$ is it consistent with ZFC that $\kappa^{\mathrm{cf}(\kappa)} < \kappa^\kappa$?