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This is true in $L$ by Exercise 5A.4 in Moschovakis $\textit{Descriptive Set Theory}$. In fact for all $n \geq 2$, $\Delta_n^1$ is a basis for $\Sigma_n^1$. The following is a sketch: Using the $\Delta_2^1$ well-ordering of $({}^\omega 2)^L$, you can show that $\Sigma_n^1$ has the uniformization property. Given $A$ which is $\Sigma_n^1$, consider $\{0\} ... 1 You can define the critical point in exactly the same way. First, note that even though the cardinalities of$V$may not be well-ordered, the cardinals and ordinals definitely will be. EDIT: BY "cardinality," I mean cardinality in the most general sense; by "cardinal," I mean cardinality of a well-orderable set, that is, "$\aleph$-number." Equivalently, by ... 2 The thing about "It is provable from$\sf ZF$that$L$is a model of$\sf ZF$" is that this is a bit misleading. For every axiom of$\sf ZF$, it is provable from$\sf ZF$that$L$satisfies that axiom. Therefore it is a meta-theorem that$L$satisfies$\sf ZF$. But that's fine for relative consistency argument, since those are meta-theory arguments anyway. ... 1 If$j: V \to L$were elementary, there would be a measurable cardinal in$V$hence$L$would think so too. But Scott showed that this is impossible. 3 You've asked a bunch of questions; let me give an overview of the answers. For more detailed information, you should look at Jech's book on set theory, which contains a ton of excellent information on large cardinals. So let's start with an elementary embedding$j: V\rightarrow M$. It's easy to see that, if$j$is nontrivial, then there is some least ... 3 There are two notions of "model" here. One is truly model-theoretic, using the relation of satisfaction which holds between structures and formulas. The other is merely proof-theoretic, using only the notion of provability of relativizations within a theory. In the first sense of "model", it is nontrivial that$ZF$proves that a theory has a model. For ... 0 Let$X$be any topological space and let$Y$have the same underlying set as$X$but the "sequential topology" (i.e., a subset of$Y$is closed iff it is sequentially closed in$X$). The identity map$f:X\to Y$is then sequentially continuous, but is only continuous at$x\in X$if$X$is "locally sequential" at$x$, meaning that$x\in \overline{A}$implies ... 2 Let$X=\beta\omega\setminus\omega$;$X$is compact Hausdorff. Moreover,$X$has no non-trivial convergent sequences, so every function on$X$is sequentially continuous. Finally,$w(X)=2^\omega$, so let$\mathscr{B}=\{B_\xi:\xi<2^\omega\}$be a base for$X$. Let$\{\langle\alpha_\xi,i_\xi\rangle:\xi<2^\omega\}$enumerate$2^\omega\times 2$. Given ... 5 Here's one sense in which the least measurable is much larger than the least inaccessible: Let$\kappa$be the smallest measurable cardinal. Then the set of inaccessible cardinals$<\kappa$has size$\kappa$. Jech's book has details on this (and much stronger) results. Let me sketch why the least measurable has to be so big. Let$\kappa\$ be measurable, ...