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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 $A\subset B\in V_\alpha\implies ... 11 I will leave the explaining of large cardinals to someone more knowleadgeable and explain one place where they are useful in tidying up things: category theory. In category theory, you are constantly facede with proper classes (the category of all sets, of all groups, etc.). To make things worse, you want to form functor categories but due to the sizes of ... 10 No well-ordering of the reals is Lebesgue measurable. This is essentially due to Sierpiński, but one usually finds a weaker version in books (for example, in Rudin's "Real and Complex Analysis"), namely, that if CH holds, then a well-ordering of smallest order type is non-measurable. The proof is via Fubini's theorem. The argument can be made shorter, but ... 8 G. Rodrigues's specific answer gets at the general issue: large cardinals are used to examine how much more one can proof in ZFC set theory. The first time I discovered large cardinals (in Jech's 2000 book Set Theory), I was amazed. A large cardinal is just a "very big" set, after all, but I did not realize that the existence of such a set changed the ... 7 Well, how funny that I happen to be right here as you ask the question, although I don't have my article with me. But note that the elements of the powerset of$L_\kappa$in$L$appear at stages before (and unbounded in)$L_{\kappa^+}$, by the famous argument of Gödel showing the GCH in$L$. Thus, one application of powerset corresponds to the next ... 7 What you are looking for is the concept of Aronszajn tree. You can read about constructions of Aronszajn trees in any graduate level set theory text, and meanwhile, the Wikipedia page lists a summary of the basic facts: König's lemma states that$\aleph_0$-Aronszajn trees do not exist. The existence of Aronszajn trees ($=\aleph_1$-Aronszajn trees) was ... 7 I wrote the blurb that you quoted from Cantor's attic about the calamity of inconsistency, and what I had in mind was the following (although I may have overdone the purple prose there). One of the principal features of the large cardinal hierarchy is the fact that it is strictly increasing in consistency strength as one moves higher in the hiearchy. ... 7 The answer is negative, and those two models are not necessarily the same. The basic problem is that one might have an$\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}$, but of course it is in ... 7 For any set$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$are compatible; it is total, since it is dense to ... 6 Define a coloring$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$... 6 You don't need all of the Easton machinery to do this, as simple Cohen forcing will do the job. For example, one might start with an inaccessible cardinal$\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 ... 6 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 ... 6 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 ... 5 Although it is true under ZF+AD that there is no$\omega_1$-sequence of distinct reals, and also that$\omega_1$is a measurable cardinal, in fact you do not need any AD hypothesis to show that a measurable cardinal can never inject into$\mathbb{R}$. Theorem.(Assume ZF only) A measurable cardinal$\kappa$cannot inject into$\mathbb{R}$, or indeed, into ... 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 ... 5 What is meant here is a property$P$such that a cardinal$\kappa$satisfies$P$iff there is an$\alpha>\kappa$such that$V_\alpha\models\Psi(\kappa)$for some appropriate$\Psi$(that, of course, depends on$P$). Supercompactness and strongness are not like that, since both require the existence of arbitrarily large measures with certain properties. ... 5 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 ... 5 There is a standard approach to these results, that dates back to Lévy and Solovay. The point is that, for any$\lambda$, given an embedding witnessing$\lambda$-supercompactness of$\kappa$, there is a canonical way of lifting the embedding to a$\lambda$-supercompact embedding in$V^Q$(using simply that$Q$is "small"). Lévy and Solovay present this in ... 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 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 = ...

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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 $i\in ... 5 First note that since$\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$relation is something other than the real membership relation. Actually, this point is crucial in the development of Boolean-valued models. ... 5 I do not think Shelah cardinals are well understood yet, and they do not seem to have been studied much. They are beyond Woodin cardinals, so they are beyond the current threshold of "true understanding" provided by inner model theory. For this reason, currently there can be no known results for which Shelah cardinals are optimal consistencywise. But the ... 5 A cardinal$\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$is equivalent to a statement of the form ... 4 Let's let$B = \{ \alpha < \kappa : A_\kappa \cap \alpha = A_\alpha \}$. Suppose$C \in M$is a club subset of$\kappa$. We want to show that$C \cap B \neq \varnothing$. It follows that$j(C)$is a club subset of$j(\kappa)$, and also that$j(B) = \{ \alpha < j(\kappa) : j(A_\kappa) \cap \alpha = A_\alpha \}$. Now note two things:$\kappa \in ...

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I don't think you can get such an $X$, even if there is no $R$ to worry about. Suppose you had an $X$ as in the question. Since $X$ properly includes $V_\alpha$, it has elements of rank $\geq\alpha$. Since the rank function is absolute for transitive models, $X$ contains the rank of such a set; in particular it contains an ordinal $\geq\alpha$, and, being ...

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$\kappa$ is Mahlo is it has stationary many inaccessibles below it, or alternatively every normal (continuous in terms of limit ordinals) has a fixed point within $\kappa$. This is the same because one can define a CLUB set as an image of a normal function, so having stationary many inaccessibles below $\kappa$ means there's always a fixed point. In this ...

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No amount of hyperinaccessibility or hyperhyperinaccessibility and so on can be provably equivalent to Mahloness (unless those notions are inconsistent). The reason is that if $\kappa$ is Mahlo, then all its hyperinaccessibility and hyperhyperinacessibility properties and so on are expressible in the structure $\langle V_\kappa,\in\rangle$, once one knows ...

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The class of ordinals of a model $M$ will always be a proper class from the point of view of $M$. However, the model might miss some of the universe's ordinals. This is most clearly seen in rank-initial segments of the universe: if $\kappa$ is an inaccessible cardinal, say, then $V_\kappa$ is a model of ZFC, but its ordinals are (externally) just the ...

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Asaf, what you are saying is false. In $L[\mu]$, for example, there is a unique normal ultrafilter. (Unless I misunderstand. You are saying that for every $S$ stationary-costationary, you can find a normal $U$ with $S\in U$, right? Obviously, any normal $U$ satisfies that either $S$ or $\kappa\setminus S$ is in $U$, which is what you wrote.) Edit : This ...

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