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11

A standard argument to show that $\mathcal{L}(\mathbb R)$ has the same size as $\mathcal{P}(\mathbb R)$ is to note that the Cantor subset of $[0,1]$ has the same size as $\mathbb R$ and measure 0, so any of its subsets also has measure 0. You can use the same idea to find the size of $\mathcal P(\mathbb R)\setminus\mathcal L(\mathbb R)$: Fix a nonmeasurable ...


5

The obvious answer is that you get the consistency of $\sf ZFC$ with an inaccessible. But you get more. You get the existence of a transitive model of this theory. But you actually get more. Since an inaccessible cardinal is the limit of worldly cardinals,1 you get a transitive model of $\sf ZFC$+There exists an inaccessible cardinals+There is a proper class ...


4

Let me address a point which is implicit in the question but hasn't been squarely addressed by the other answers. Your description of the "axiom of consistency" is not precise: it is not at all obvious how to encode this axiom in the language of set theory. In particular, there is no direct way to encode the self-reference involved in saying "this axiom". ...


3

It has not turned out to be very helpful to prove that specific theorems are provable indirectly, when no actual proof of the theorem was previously known. There are two settings in mathematical logic, however, where we do show indirectly that particular theorems are provable in particular systems or in particular ways. The difference is that, when we work ...


3

Suppose $\bigcup Y$ is a successor ordinal, say the successor of some ordinal $\alpha$. Then $\alpha\in \bigcup Y$, so $\alpha\in y$ for some $y\in Y$. But $\alpha+1$ is the least ordinal containing $\alpha$, so $y\geq \alpha+1$. Since $y\in Y$, $y\subseteq \bigcup Y=\alpha+1$, so $y\leq \alpha+1$. Thus $y=\alpha+1=\bigcup Y$. This means that $\bigcup Y$...


2

Let $\Bbb P={^{<\omega}\{0,1\}}$, ordered by $\subseteq$; certainly $\langle\Bbb P,\subseteq\rangle$ is ccc. For $\varphi\in{^\omega\{0,1\}}$ let $$A_\varphi=\left\{(\varphi\upharpoonright n)^\frown\big(1-\varphi(n)\big):n\in\omega\right\}\;;$$ $A_\varphi$ is clearly an antichain in $\Bbb P$. To see that $A_\varphi$ is maximal, let $s\in{^n\{0,1\}}$. ...


2

If $Y$ is a set of ordinals, then $\bigcup Y$ is an ordinal: it is transitive (since $Y$ is a set of transitive set, therefore $\bigcup Y$ is a transitive set), and it is a set well-ordered by $\in$ (well, every element of an element of $Y$ is an ordinal, since $Y$ is a set of ordinals, and ordinals are themselves sets of ordinals; so $\bigcup Y$ is again a ...


2

For every $\alpha$ there is a proper class of subsets of $X$ which can be mapped onto $V_\alpha$, and for each of those there are many surjections. You need to choose a set for each $\alpha$ simultaneously and choose a surjection, and do it in a coherent way. Not to mention, that in order to prove that there is a bijection between $X$ and $V$ you need to ...


2

Since $1_P \Vdash |\tau| \leq \check{\lambda}$, there is some name $f$ such that $1_P \Vdash f:\tau \to \check{\lambda}\text{ is an injection}$. In particular, if $p\Vdash\sigma\in\tau$, then we can pick a maximal antichain $A$ of elements of $P$ below $p$ which determine the value of $f(\sigma)$. The function which takes $a\in A$ to the $\alpha<\lambda$...


2

The point you're missing is that in a general topos, the internal logic is not necessary Boolean, so $\neg$ does not necessarily give complements. Topological spaces are a good source of examples for intuitionistic logic; in a lattice of open sets, $\neg$ gives the exterior of an open set. Usually, the complement of the set is not open, and thus does not ...



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