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5

The regularity is necessary, but you can circumvent it as follows: If $\kappa$ has an uncountable cofinality, then every stationary subset of $\kappa$ can be split into $\operatorname{cf}(\kappa)$ disjoint stationary sets. First of all, for singular cardinals of countable cofinality the usual notion for clubs is meaningless, since there are ...


4

HINT: It suffices to find a surjection from $2^\kappa$ onto the natural numbers, now look at cardinalities of finite sets. You can also show that equality is never possible. But that is besides the point. You are asked to find an injection, not to prove there are no bijections.


3

It's not clear how to phrase "compute the Hanf number" as a problem for which decidability makes sense. The Hanf number problem takes in a logic, and outputs a cardinal; decidability, or computability, makes sense in the context of natural numbers. In certain cases, the Hanf number is completely understood. For example, for a first-order language $L$, the ...


3

Since you didn't specify a context, I'll assume the question is about ultrafilters on arbitrary semigroups. Then the answer is yes. Just take a semigroup that has an identity element $e$ (i.e., a monoid, for example a group) and let $\mathcal V$ be the principal ultrafilter that contains $\{e\}$.


3

Unfortunately, it is consistent that if $\kappa>\aleph_0$, then $\kappa<\kappa^{<\kappa}$. It is consistent from large cardinals (one and two) that for every infinite cardinal, $2^\kappa=\kappa^{++}$. And then we have the following: If $\kappa=\lambda^+$, then $\kappa^{<\kappa}=\kappa^\lambda=\lambda^{++}=\kappa^+>\kappa$. If $\kappa$ is ...


2

Let $\langle A_\alpha:\alpha<\omega_1\rangle$ witness $\Diamond$. Let $\langle\eta_\xi:\xi<\omega_1\rangle$ be an increasing enumeration of the limit ordinals in $\omega_1$. Let $S=\{\xi<\omega_1:\sup A_{\eta_\xi}=\eta_\xi\}$. For $\xi<\omega_1$ let $$A_{\eta_\xi}'=\begin{cases} A_{\eta_\xi},&\text{if }\xi\in S\\ ...


1

You can get any finite set of sets from the pairing axiom and unions. In the language of set theory, it is difficult to write the general axiom formally, but we can prove individual cases as needed just from pairing and unions. Basically, there is no need for it, and it adds a complicated "axiom scheme" to our theory. ...


1

It depends on what you mean by number of sets. If it is a finite natural number, then you can get it by the axioms of pairing and induction. If you want possibly infinitely many sets, then you either already have some set containing all of those sets as elements, or you have some sequence of sets, namely a function on an index set, in which case what you ...


1

This is a standard Fubini type argument for category. Letting B to be a Cohen name for an F-sigma meager set, consider the set W of all pairs (x, y) of reals such that some initial segment of x forces y to be in B. Check that W is a meager borel set (use Kuratowski-Ulam theorem which is Fubini theorem for category). Hence if A is any non meager set in V, ...



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