# Tag Info

22

I disagree that most branches of mathematics are just an application of set theory and logic. The fact that most areas of mathematics use set related notions and employ logic does not mean they are applications of these areas. For instance, would you say that English Literature = English Words + English Grammar? After all, every piece of English literature ...

7

No. For example, if we spend the time to formalize group theory in a set theory (take ZFC as an example), then by the end of this long and arduous exercise, we probably haven't learned a whole lot of new concepts in group theory. We've just worked out how to implement the old concepts in ZFC. Furthermore, we're probably no better at discovering new ...

6

The statement is equivalent to Zorn's lemma. It implies Zorn's lemma quite easily, because they have the same requirements from the partial order, and if there is a maximal element above each point, then there's certainly a maximal element. On the other hand, assuming Zorn's lemma, and given a partial order $(P,\leq)$ satisfying these requirement, consider ...

4

Here is how to construct an explicit bijection. First we construct a bijection $\alpha$ from $\mathbb{N}$ to $\mathbb{Q}$. (For definiteness I will take $\mathbb{N}$ to include $0$, this makes no real difference.) For brevity, call $\alpha(i)$ by the name $r_i$. Now we proceed semi-formally. Map all numbers which are not rational or of the form ...

4

We proceed by transfinite induction to show that $\left|V_\alpha\right|<\kappa$ for every $\alpha < \kappa$. For successor ordinals we use (and need) the strong inaccessibility: $$\left|V_{\alpha+1}\right|=2^{\left|V_\alpha\right|}<\kappa$$ since $\left|V_\alpha\right|<\kappa$. For limit ordinals,  \left|V_\alpha\right|=\left| ...

3

The process works, but in general you have to repeat it many times through a transfinite process. Call $\mathcal C_0=\mathcal C$, and $\mathcal C_\alpha$, for $\alpha>0$ an ordinal, the result of taking complements of sets in $\bigcup_{\beta<\alpha}\mathcal C_\beta$, and all possible finite and countable unions of sets in ...

3

Fix an enumeration of $\Bbb Q$, $q_n$, and find a countably infinite subset of $\Bbb I$, $r_n$. Now find a map which fixes all the points which are not $r_n$'s, and maps the union $\{q_n,r_n\mid n\in\Bbb N\}$ into $\{r_n\mid n\in\Bbb N\}$. The above can be translated quite neatly to cardinal arithmetic. Write $\Bbb I$ as $A\cup B$ where $A$ is countably ...

2

In what follows, I write $\mathfrak c$ for $|\mathbb R|$, as customary. That $\mathfrak c=2^{\aleph_0}$ can be proved in several ways. For example, the Cantor set $C$, by construction, has size $2^{\aleph_0}$. This shows that $2^{\aleph_0}\le\mathfrak c$. On the other hand, it is easy to see that $\mathbb R$ has the same size as any open interval, and ...

2

Hint: When he says that $\langle\lambda_\gamma\mid\gamma<\delta\rangle$ is a rearrangement of $\langle\kappa_\alpha\mid\alpha<\beta\rangle$, he means that there is a bijective function $\sigma:\beta\to\delta$ such that for all $\alpha<\beta$ we have $\kappa_\alpha=\lambda_{\sigma(\alpha)}.$ Do you see how to use $\sigma$ construct the desired ...

1

Generally, "the full Solovay model" means that there was $\kappa$ which was inaccessible in $M$ and $G$ is a generic filter for $\operatorname{Coll}(\omega,<\kappa)$. Since this is a homogenous forcing, it doesn't change $\rm HOD$, so $\rm HOD$ of $M[G]$ is the same as in $M$. If $\kappa$ is inaccessible in $M$ then it is regular and strong limit. In ...

Only top voted, non community-wiki answers of a minimum length are eligible