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6

The definition of forcing is the same with and without the axiom of choice. And the truth lemma holds with and without the axiom of choice. Namely, $$p\Vdash_\Bbb P\varphi\iff\text{For every }V\text{-generic } G\subseteq\Bbb P\text{ with }p\in G: V[G]\models\varphi$$ You can also consider iterations without choice, at least with a two-step iteration this ...

6

I think that your claim is false. I will try to construct a counterexample. Suppose that $\kappa$ is measurable and $U$ is a normal measure on $\kappa$. Let $j:V\to M$ be the ultrapower by $U$. First, we build $M\prec H_{\kappa^+}$ such that $U'=U\cap M$ is $M$-amenable. Let $M_0$ be any transitive elementary substructure of $H_{\kappa^+}$ of size $\kappa$. ...

4

This can be proved in several ways. The easiest I know is to use Cohen's second model (you can find it in Jech The Axiom of Choice in Chapter 5) in which there is a sequence $\langle A_n\mid n<\omega\rangle$ such that: $A_n\subseteq\mathcal P(\Bbb R)$. $A_n$ has two elements. If $n\neq m$, then $A_n\cap A_m=\varnothing$. $\prod A_n=\varnothing$. ...

3

Following up on my sketch in the comments, here's a proof that axioms (1) and (2) (where $\phi$ has no parameters) are conservative over $MK$ plus the choice schema: (CC) $\forall x\exists X\phi(x, X) \to \exists Y\forall x\phi(x, Y_x)$ where $Y_x = \{y:\langle x, y\rangle\in Y\}$, and the $\omega$-dependent choice schema: (DC) \$\forall X\exists Y\phi(X, ...

1

Start with the (excellent but outdated) handbook of set theoretic topology and then move onto (recently revised) open problems in set theoretic topology. This has evolved into a vast subject so it is hard to recommend specific papers too strongly. My advice is: Learn the basic techniques (elementary submodels, forcing, combinatorial principles), fall in love ...

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