Took me way too long than I'd like to admit to figure out how to invite you to a room @Akiva

Anyway, do you know what $H(\kappa)$ is for a cardinal $\kappa$?

No

Ok so the proper definition is that $H(\kappa)=\{x\mid |\mathrm{trcl}(x)|<\kappa\}$

(It's not even obvious that $H(\kappa)$ is a set from this definition but turns out that $H(\kappa)\subseteq V_\kappa$)

What's trcl

For $\kappa$ regular, which is the case one is usually interested in, $H(\kappa)$ is the set of sets hereditarily of cardinality less than $\kappa$

I'm guessing from the letters it's "transitive closure" but I dunno what that means

@AkivaWeinberger transitive closure. smallest transitive set containing $x$

@AlessandroCodenotti So like, the number of brackets is less than kappa?

where {{}} has two brackets (but one element) for example

(the transitive closure exist because the $V_\alpha$ are all transitive and every set is somewhere in the Von Neumann hierarchy)

Er, pairs of brackets

@AkivaWeinberger Hmmm yes I suppose, even though $\omega_3$ brackets isn't really something you can write down :P

Yeah true

I also forget what a transitive set is

In particular $H(\omega)=V_\omega$ is the set of hereditarily finite sets, while $H(\omega_1)$ is the set of hereditarily countable sets (and is way smaller than $V_{\omega_1}$)

Does transitive mean every element is a subset?

Yes

Ah OK

Equivalently $x$ is transitive if $z\in y\in x$ implies $z\in x$ which makes the meaning of the name "transitive" clearer

Elements of elements are elements

I see

So the smallest transitive set contain $x$

would be the set of its elements (aka x) as well as the elements of its elements as well as the elements of those, etc?

Like $x\cup(\bigcup x)\cup(\bigcup\bigcup x)\cup\dotsb$

Yeah, if you define $\bigcup^n x$ to be $\bigcup\cdots\bigcup x$ with $n$ unions then the transitive closure if $\bigcup_{n\in\omega}\bigcup^nx$

So, yes

and then instead of counting brackets, we're counting elements plus elements-of-elements plus el.s-of-el.s-of-el.s, etc

which I guess is the same as counting brackets

(pairs of)

Right OK so I think I'm up to speed on the definitions now

yep

Erm

well not really 'cause there can be repeats

like {x,{x}}

x would be counted twice from what I just said

but whatever, I don't think it changes much

The $H(\kappa)$ are interesting because if $\kappa$ is regular and uncountable then $H(\kappa)$ is a model of $\mathsf{ZFC}^-$ ($\mathsf{ZFC}$ without powerset)

(uncountable is needed for infinity, regular for replacement)

If $\kappa$ is strongly inaccessible then $H(\kappa)$ models $\mathsf{ZFC}$, but this shouldn't be surprising, because under this assumption $H(\kappa)=V_\kappa$

OK I'm back

@AlessandroCodenotti Right, 'cause you need powerset to prove uncountable sets exist

By the way looking at $H(\omega_1)$ is the usual way to prove that $\mathsf{ZFC}-$Powerset$+$"every set is countable" is consistent relative to $\mathsf{ZFC}$

I just realized that for the $\Bbb N\to\mathcal P(\Bbb N)$ argument I also need a couple of facts on posets and generic filters on a poset so I'm afraid I don't have time to explain the whole thing ...

I think I know how those work

more-or-less

Upwards and meet closed

I don't know what "generic" means actually

So if $(P,\leq)$ is a poset we say that $D\subseteq P$ is dense if for all $p\in P$ there is $q\in D$ with $q\leq p$

(this is the same as dense in the topology where the open sets are the downward closed sets)

So the upwards-closure of $D$ is $P$?

yes

If $\mathcal D$ is a family of dense subsets of $P$ we say that a filter $G$ on $P$ is $\mathcal D$-generic if $G\cap D\neq \varnothing$ for every $D\in\mathcal D$

OK

If $M$ is a transitive model of $\mathsf{ZF}^-$, $P$ is a poset in $M$, $G$ a filter on $P$ ($G\in V$, not necessarily $G\in M$), we say that $G$ is $P$-generic over $M$ if $G$ meets every dense subset of $P$ which is also an element of $M$

Existence of such generic filters is independent of ZFC in general I believe, but we can show that if $\mathcal D$ is a countable family of dense subsets of a poset $P$ then there always is a $\mathcal D$-generic filter over $P$. So the trick is to pick $M$ to be a transitive countable model of $\mathsf{ZF}^-$ (which exist by Löwenheim-Skolem, assuming there is a model at all)

OK

I'm not sure why this is a useful concept but OK

They are the kind of filters you need for forcing

The idea is that (apart from trivial cases such as $P$ being atomic), $G$ is not in $M$ and you can build a generic extension $M[G]$ containing it

Actually let me tell you about the first example which is quite interesting and can be done skipping all technical details

Let $P$ be the poset of finite partial functions $\omega\to 2$, ordered by reverse inclusion, so $f\leq g$ iff $f\supseteq g$

So two-colorings of subsets of the integers

Suppose that $M$ is a transitive model of $\mathsf{ZF}^-$ and that $G$ is a $P$-generic filter over $M$ (so $G$ meets every dense subset of $P$ which is also in $M$). I claim that $\bigcup G$ is a function $\omega\to 2$ and that it is not contained in $M$. (so identifying $\Bbb R$ with $2^\omega$ I'm claiming that in $M[G]$ we have an extra real number that wasn't in $M$)

(this is called a Cohen's real because that's how Cohen showed that $\mathsf{CH}$ can fail in $\mathsf{ZFC}$: by adding $\aleph_2$ Cohen reals)

Well it's a filter so we can definitely take the union of things in it

And this is $\mathsf{ZF}^-$ so we have no powerset?

Right, since it is a filter the union is a function

@AkivaWeinberger Yes, it's not needed, but you can think about a model of full ZFC if you prefer

Ah

Wait hold on

Isn't the set of reals dense?

what do you mean with dense?

Its closure is the set of all partial colorings

By real I mean a complete coloring

I think so

Oh

wait

Finite partial functions

yes, that's important

so we don't have the reals in the poset in the first place

OK, I missed that the first time

Yeah stuff is making more sense now

no problem

Ok so now to show that $\bigcup G$ is a total function on $\omega$ we can look at $D_n=\{f\in P\mid n\in\mathrm{dom}(p)\}$, show that this is dense in $P$ (this is very easy), hence it must meet $G$ and so $n\in\mathrm{dom}(\bigcup G)$ for every $n\in\omega$

Mhm

It has to have something that has $n$ because the set of things that have $n$ is dense

Do you see why $D_n$ is dense?

because the closure of $D_n$ is the set of subcolorings of things in $D_n$

and if you have a coloring with $n$ uncolored

Uhmmm I don't find the colouring analogy very intuitive but I think so

you can just color $n$ one of the colors

and $D_n$ will include that

and so our original coloring is the subcoloring of that that ignores $n$

It's just that for every partial $f:\omega\to 2$ either it already has $n$ in its domain, or you can extend it to $f\cup\{(n,0)\}$

Pretty sure you're saying the same thing in a different language :P

We are

$f=(f\cup\{(n,0)\})\big|_{\operatorname{dom}f}$

Right so $\bigcup G$ colors everything (has everything in its domain)

Alright, so we have that $\bigcup G$ is a real number (I'm going to use real number to mean "function $\omega\to 2$")

Sure

We only need to show that $\bigcup G$ is not already in $M$

Oh

Are you gonna flip all the digits

Erm, look at $1-\bigcup G$

Wait never mind

That doesn't make sense

Give me a sec

Hm OK how about this

Suppose $\bigcup G$ is in $M$

We look at all the things that differ from $\bigcup G$ in at least one place

This is dense

and $G$ can't intersect it

The only question is, why can't we do this when $\bigcup G$ isn't in $M$

Hmmm what I had in mind is very similar: So consider $f:\omega\to 2$ with $f\in M$ and let $D_f$ be the set of functions in $P$ that differ from $f$ somewhere. $$D_f=\{g\in P\mid \exists n\in\mathrm{dom}(g)(g(n)\neq f(n))\}$$ this set is dense in $P$ (and is in $M$), so $G$ must meet it, hence it differs from $f$ somwhere. Since $f$ was arbitrary we're done.

@AkivaWeinberger I'm not convinced this set is in $M$ if $\bigcup G$ isn't

@AkivaWeinberger Which should answer this

I see

So basically

that differs from all sequences of 0s and 1s in M at at least one place

And

if M is countable

I see how you can construct this explicitly

We have countably many such sequences in $M$, so label them $f_0,f_1,f_2,\dots$

and then diagonalize:

indeed, that's the way to go

define a sequence $g$ such that $g(n)=1-f_n(n)$

Now there are a lot of technical details to work out before forcing really works and we can say that there is this "$M[G]$" which is also a model of set theory and contains both $M$ and $G$ (hence also $\bigcup G$), but that's the main idea

I mean

This is literally Cantor's diagonal argument

Clearly if M is countable then we're missing stuff

@AkivaWeinberger What do you mean?

If $M$ is countable, then $\Bbb R\setminus M$ is nonempty

That's delicate

I mean, when I talk about $\Bbb R$ I really mean "the set that $M$ believes to be the powerset of $\omega$"

Would you write that $\Bbb R^M$?

Still, $\Bbb R\setminus\Bbb R^M$ is nonempty

$M$ also believes that this set is uncountable, just because there is no set in $M$ that $M$ believes to be a bijection between $\Bbb R^M$ and $\omega^M$ ($\omega^M=\omega$ as long as $M$ is transitive though)

so our Cohen real is just an element of that

so now you're trying to construct $M[G]$

which I guess means $\Bbb R^{M[G]}\supseteq\Bbb R^M\cup(\bigcup G)$

Right, but the point is that $V$ and $M$ might believe $\mathsf{CH}$ to be true, but then you add $\aleph_2$ (according to $M$) Cohen reals to $M$ to get this $M[G]$ where $\mathsf{CH}$ fails

I don't know how you'd construct $M[G]$ though

@AlessandroCodenotti I see

And what I just did assumed $M$ was countable

If $M$ isn't countable I dunno why $G$ should exist

I think existence of a generic $G$ is independent in that case

Everyone assumes the ground model ($M$) is transitive and countable when dealing with forcing :P

How can you add $\aleph_2$ things to it if you need it to be always countable

$\aleph_2$ things from the point of view of $M$

Ah

$M$ and $V$ wildly disagree on what $\aleph_2$ means since $V$ believes $M$ to be countable

I see the goal, I don't see how you'd get there

This is another technical point that needs to be addressed, we added $\aleph_2$ things from the point of view of $M$, but why is that still $\aleph_2$ things in $M[G]$?

Mhm

(there is an important theorem saying that you can do forcing without collapsing cardinals or something similar, this is stuff I haven't really seen yet)

If it's transitive, doesn't it have the "right" ordinals?

Wait

No of course not

Wait now I'm confused again

OK hold on so why does $\omega^M=\omega$

Everything which is an ordinal in $M$ will be an ordinal in $V$ ("$x$ is an ordinal" is a $\Sigma_0$-formula so it's absolute between transitive classes. "$x$ is $\omega$" is also absolute)

But $\mathsf{Ord}^M$ could smaller than $\mathsf{Ord}$, for example if $M=V_\kappa$ for a strongly inaccessible $\kappa$

What if $M$ is a countable model though

So $M$'s version of $\omega_1$ will be a countable ordinal?

It should still agree with $V$ on what is $\omega$

Oh wait what's the name of the weird ordinal

The Church–Kleene ordinal

@AkivaWeinberger countable from the point of view of $V$, but $M$ believes it to be uncountable

I mean, the whole of $\mathsf{Ord}^M$ will be countable according to $V$, but $M$ even thinks that it is a proper class!

So I don't 100% remember what that is, but I'm guessing $M$'s version of $\omega_1$ will be some countable ordinal larger than the Church–Kleene ordinal

@AlessandroCodenotti Yeah yeah

I don't know how the Curch-Kleene ordinal works to be honest

$\omega_1^{\rm CK}\le\omega_1^M<\omega_1$

That's my conjecture

It's kinda like the smallest "noncomputable" ordinal in a sense

but the word they use is "nonrecursive"

Anyway $\phi(x)$ saying "$x$ is $\omega_1$" looks like it can be written as a $\Pi_1^{\mathsf{ZF}}$-formula to me, so it is downward absolute, but not upward absolute, which makes sense, since $M\subset V$ and the thing $M$ believes to be $\omega_1$ is not the thing $V$ believes to be $\omega_1$

"Downward absolute"?

If $M_0\subseteq M_1$ are transitive classes and $\phi(x_0,\ldots,x_n)$ is an $\in$-formula then it is $M_0$-$M_1$-downward absolute if $\phi^{M_1}(a_0,\ldots,a_n)\rightarrow\phi^{M_0}(a_0,\ldots,a_n)$ for all $a_0,\ldots,a_n\in M_0$

upward absolute means that $M_0$ and $M_1$ are swapped in the implication

(if $M_1=V$ one usually writes $M_0$-downward absolute and if $M_0$ is clear from context also just "downward absolute")

So you're saying that, if $\omega_1\in M$, then $\omega_1=(\omega_1)^M$?

yes

Don't see why that would be but OK

It's not too hard to see after seeing absoluteness for $\Sigma_0$ formulas

The proper statement is that "if $M$ is a nonempty transitive class, then every $\Sigma_0$-formula is $M$-absolute" (so absolute between $M$ and $V$)

Which one's $\Sigma_0$ again

And the idea here is to look at the collection of all $M$-absolute formulas, which clearly contains the atomic formulas since they're unaffected by relativization to $M$ and then show that it has enough closure properties to contain all $\Sigma_0$ formulas

@AkivaWeinberger smallest collection of formulas containing the atomic formulas and closed under $\neg$, $\rightarrow$ and bounded existential quantifiers ($\exists x_1\in x_2$)

(of course this also gives closure under $\land$,$\lor$ and bounded universal quantifiers)

Yeah 'cause $\rightarrow$ and $\lnot$ are a complete set

yep

I asked a related question in the logic room chat.stackexchange.com/rooms/44058/logic

Do mirrors confuse radar

dunno

Anyway I need to go and learn some type theory now