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Let's solve the general problem surrounding the question, with a few observations, each of them easy to see. Every $V_\alpha$ for any ordinal $\alpha$ satisfies Extensionality and Foundation, since all transitive sets satisfy Extensionality and Foundation. Every $V_\alpha$ satisfies Separation, for the simple reason that $A\subset B\in V_\alpha\implies ... 12 I will leave the explaining of large cardinals to someone more knowleadgeable and explain one place where they are useful in tidying up things: category theory. In category theory, you are constantly facede with proper classes (the category of all sets, of all groups, etc.). To make things worse, you want to form functor categories but due to the sizes of ... 11 No well-ordering of the reals is Lebesgue measurable. This is essentially due to Sierpiński, but one usually finds a weaker version in books (for example, in Rudin's "Real and Complex Analysis"), namely, that if CH holds, then a well-ordering of smallest order type is non-measurable. The proof is via Fubini's theorem. The argument can be made shorter, but ... 9 One can use the class of indiscernibles to generate elementary embeddings of$L[A]$to itself with critical point above the supremum of$A$, in much the same way as the indiscernibles of$L$give us embeddings of$L$into itself. By results of Kunen (for$L$, but they generalize straightforwardly), the existence of an embedding$j:L[A]\to L[A]$that has ... 9 I wrote the blurb that you quoted from Cantor's attic about the calamity of inconsistency, and what I had in mind was the following (although I may have overdone the purple prose there). One of the principal features of the large cardinal hierarchy is the fact that it is strictly increasing in consistency strength as one moves higher in the hiearchy. ... 9 I do not think Shelah cardinals are well understood yet, and they do not seem to have been studied much. They are beyond Woodin cardinals, so they are beyond the current threshold of "true understanding" provided by inner model theory. For this reason, currently there can be no known results for which Shelah cardinals are optimal consistencywise. But the ... 8 Well, how funny that I happen to be right here as you ask the question, although I don't have my article with me. But note that the elements of the powerset of$L_\kappa$in$L$appear at stages before (and unbounded in)$L_{\kappa^+}$, by the famous argument of Gödel showing the GCH in$L$. Thus, one application of powerset corresponds to the next ... 8 For any set$X$, we may consider the partial order$\mathbb{P}$consisting of all finite partial functions from$\omega$to$X$, ordered by extension. If$G\subset\mathbb{P}$is$V$-generic for this forcing notion, then$f=\cup G$is a function from$\omega$onto$X$. It is a function, since conditions in$G$are compatible; it is total, since it is dense to ... 8 Crossposted from MO. Allow me to make some comments as someone who converted to the universeful approach recently; but take it with a pinch of salt, as I have only been studying category theory for 2½ years. I should briefly mention the trigger that led me to the pro-universe camp: about 6 months ago, I started learning about quasicategories and became ... 8 You don't need all of the Easton machinery to do this, as simple Cohen forcing will do the job. For example, one might start with an inaccessible cardinal$\kappa$in$L$, and forcing with the finite partial functions from$\kappa$to$\{0,1\}$(i.e., "adding$\kappa$Cohen reals") will result in a model where$2^{\aleph_0}=\kappa$. This forcing is mild, in ... 8 G. Rodrigues's specific answer gets at the general issue: large cardinals are used to examine how much more one can proof in ZFC set theory. The first time I discovered large cardinals (in Jech's 2000 book Set Theory), I was amazed. A large cardinal is just a "very big" set, after all, but I did not realize that the existence of such a set changed the ... 8 No, this is not inconsistent, at least not relative to the existence of a Mahlo cardinal. Suppose that$\kappa$is a Mahlo cardinal, then$V_\kappa$has a club of ordinals$\alpha$such that$V_\alpha\prec V_\kappa$. Therefore there is a stationary set of inaccessible cardinals satisfying this. Simply enumerate these inaccessible cardinals and let$\alpha$... 7 Your theory is equiconsistent with what is known as the Levy scheme, or Ord is Mahlo, which is strictly weaker in consistency strength than the existence of a Mahlo cardinal. On the one hand, your theory implies Ord is Mahlo, since if$C$is any proper class definable club (definable with parameters), then it will follow that$C$is unbounded in the ... 7 Let me address the question as stated first. Why does$\sf CH$has no determinate provability from any of the axioms we throw at it? This is false. As remarked in the comments. Plenty of axioms prove$\sf CH$or disprove it. Things like$V=L$or$\lozenge$imply$\sf CH$whereas things like$\sf PFA$and similar forcing axioms imply its negation (these in ... 7 The person you want to ask about consistency strength is Ralf Schindler. Years ago ("Successive weakly compact or singular cardinals", Symbolic Logic 64 (1999), no. 1, 139–146) Ralf showed that the existence of two consecutive singular cardinals implies the existence of an inner model with a Woodin cardinal. The paper assumes an additional background ... 7 The first statement is equiconsistent with$\sf ZF$, without large cardinals. The proof is due to Spector, Mitchell Spector, The$\kappa $-closed unbounded filter and supercompact cardinals, J. Symbolic Logic 46 (1981), no. 1, 31--40. The outline of the model is not hard. Consider$\Bbb P$to be the forcing which is the lottery sum of all forcings ... 7 What you are looking for is the concept of Aronszajn tree. You can read about constructions of Aronszajn trees in any graduate level set theory text, and meanwhile, the Wikipedia page lists a summary of the basic facts: König's lemma states that$\aleph_0$-Aronszajn trees do not exist. The existence of Aronszajn trees ($=\aleph_1$-Aronszajn trees) was ... 7 The answer is negative, and those two models are not necessarily the same. The basic problem is that one might have an$\omega$-sequence of ordinals very high up, above$\kappa$, and if this sequence isn't sufficiently definable in$V$, then it will not be in$\text{HOD}^{V[G]}_{{}^\omega\omega}$, but of course it is in ... 6 the problem with your argument is when you say that$\kappa\in j_{\mathcal U}(A)$iff the class of the identity is a "member" of the class of the constantly-equal-to-$A$function. This assumes that the identity represents$\kappa$, which is equivalent to the normality of${\mathcal U}$. 6 The$U_i$sequence in Mitchell's definition is playing the role of your function$g$, and his formulation relies on normality in that it assumes that$U_i$concentrates on$i$, which would be true when$W$is normal. Without any normality assumption on$W$, you can say$U$is Mitchell below$W$if and only if there is$A\in W$and measures$U_i$for$i\in ...

6

A cardinal $\kappa$ is $\Sigma_2$ reflecting if whenever a sentence $\varphi$ is true in some $V_\alpha$, then there is an $\alpha\lt\kappa$ such that $V_\alpha\models\varphi$. (This can be seen to be equivalent to the "reflecting" version of $\Sigma_2$-reflecting, since every $\Sigma_2$ statement $\psi$ is equivalent to a statement of the form ...

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First note that since $\mathsf{ZFC}$ is just a formal first-order theory, it doesn't pick out exactly what form a model's interpretation of the $\in$ symbol takes. So there are models of $\mathsf{ZFC}$ where the $\in$ relation is something other than the real membership relation. Actually, this point is crucial in the development of Boolean-valued models. ...

6

First of all, by elementarity $\operatorname{rank}(j(x))=j(\operatorname{rank}(x))$. So it suffices to show that there is some $x$ whose rank is moved. Now we can prove the following by induction: Suppose that $\operatorname{rank}(j(x))=\operatorname{rank}(x)$ and for all $y$ such that $\operatorname{rank}(y)<\operatorname{rank}(x)$, $j(y)=y$, then ...

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Define a coloring $c:[\kappa]^2 \rightarrow \{0, 1\}$ as follows $c(\{\alpha, \beta\}) = 0$ iff $f(\alpha) = f(\beta)$. Let $d:\kappa \rightarrow \{0, 1\}$ be defined by $d(\alpha) = 0$ if $A_{\alpha} = \{\beta < \kappa: c(\{\alpha, \beta\}) = 0\} \in U$ and $d(\alpha) = 1$ if $B_{\alpha} = \{\beta < \kappa : c(\{\alpha, \beta\}) = 1\} \in U$. Then $d$ ...

6

Just a supplement to Joel's answer: The exact boundary for existence of $0^\sharp$ was pinned down by Klaus Gloede: $0^\sharp$ exists if and only if there is a cardinal $\kappa$ such that every constructible partition of $[\kappa]^{<\omega}$ has an uncountable homogeneous set. Gloede, Klaus, Ordinals with partition properties and the constructible ...

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The borderline seems to be very near the $\omega_1$-Erdős cardinals. The Wikipedia page on $0^\sharp$ explains it thus: The existence of $\omega_1$-Erdős cardinals implies the existence of $0^\sharp$. This is close to being best possible, because the existence of $0^\sharp$ implies that in the constructible universe there is an $\alpha$-Erdős cardinal ...

6

Two-valued measures behave very differently from real-valued measures. For example, suppose $\mathcal{U}$ is a countably complete ultrafilter on a set $X$ and suppose that $f:X\to2^\omega$ is an injection. There is a $b \in 2^\omega$ such that $$B_n = \{ a \in X : f(a)(n) = b(n) \} \in \mathcal{U}$$ for every $n < \omega$. By countable completeness, $B = ... 5 It is a good question. The answer is that real-valued measurable cardinal need not be strongly inaccessible, whilst every measurable cardinal is strongly inaccessible. Indeed, it is consistent that the continuum itself is a real-valued measurable cardinal, but the continuum can never be a measurable cardinal, since every measurable cardinal is strongly ... 5 No infinite set of Silver indiscernibles is in$L$. (Of course, every finite set of Silver indiscernibles is in$L$since$L$contains all finite sets of ordinals.) To see this, assume$0^\#$exists in$V$and suppose$(\eta_i)_{i<\omega}$is an infinite increasing sequence of Silver indiscernibles. I will show that$0^\#$is definable from ... 5 Okay let me try and answer all your questions in order: First the terminology: Let$\kappa$be a cardinal,$\kappa$is a successor cardinal if there exists a cardinal$\lambda$such that, if$\alpha<\kappa$then$|\alpha|\leq\lambda$.$\kappa$is a weak limit cardinal if it is$\aleph_\alpha$for a limit ordinal$\alpha\$. Namely, it is the limit of a ...

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