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15

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]$that has ... 9 I wrote the blurb that you quoted from Cantor's attic about the calamity of inconsistency, and what I had in mind was the following (although I may have overdone the purple prose there). One of the principal features of the large cardinal hierarchy is the fact that it is strictly increasing in consistency strength as one moves higher in the hiearchy. ... 8 For any set$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$are compatible; it is total, since it is dense to ... 8 Well, how funny that I happen to be right here as you ask the question, although I don't have my article with me. But note that the elements of the powerset of$L_\kappa$in$L$appear at stages before (and unbounded in)$L_{\kappa^+}$, by the famous argument of Gödel showing the GCH in$L$. Thus, one application of powerset corresponds to the next ... 8 You don't need all of the Easton machinery to do this, as simple Cohen forcing will do the job. For example, one might start with an inaccessible cardinal$\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 ... 7 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 ... 7 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 ... 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 ... 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 ... 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 = ...

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

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 ...

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

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 ...

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

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 $i\in ... 5 Very much so. Measurable cardinals were considered by Banach and Kuratowski in 1929 and 1930. In 1943 Tarski proved that measurable cardinals are weakly compact, although the definition by tree properties and by coloring were given only in 1961 by Erdos and Tarski. Hausdorff worked with weakly inaccessible cardinals, and Mahlo defined what is now known as ... 5 An extender is a system of compatible ultrafilters. The compatibility manifests itself directly: We have projections from ultrafilters on "larger" spaces to those on "smaller" spaces, or natural ways of extending "small" ones into "larger" ones. But, more relevantly, the compatibility means that the elementary embeddings you can form by taking the ... 5 Although it is true under ZF+AD that there is no$\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 ...

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