finite, actual infinity, potential infinity

Hypernaturals are uncountable because they are bigger than all the nats and so can’t be counted.

This isn't the condition for countability. For instance, consider the ordering ${<}_{ext}$ of $N\cup \left\{\infty \right\}$ where when $n\in N$ then $n{<}_{ext}\infty$. This ordering has $\infty$ bigger than all the nats, but it's still countable because you have a bijection $N\to N\cup \left\{\infty \right\}$ given by $0\to \infty$, $n+1\to n$.

Also countability of the hypernaturals is a subtle concept because of the transfer principle. If you start with some model $M$ of set theory with natural numbers $N_M$ and use an ultrafilter to extend it to a model $M^{\ast }$ with natural numbers $N^{\ast }$, then you have three notions of countability of a set $S$:

1. $M$ contains a bijection between $S$ and $N_M$,
2. $M^{\ast }$ contains a bijection between $S$ and $N_{M^{\ast }}$ (which is equal to $N^{\ast }$),
3. The ambient set theory contains a bijection between $S$ and $N$.

Tautologically, the hypernaturals will be countable in the second sense, because it is simply seeking a bijection between the hypernaturals and themselves. I'm not sure whether they can be countable in the third sense, but if $N\subset N_M$^[1] then intuitively it seems to me that they won't be countable in the third sense, but the naturals won't be countable in the third sense either, so that doesn't necessarily seem like a problem or a natural thing to ask about.

Whether cardinality of continuum is equivalent to continuum hypothesis

Not sure what you mean here.

1. ^{^}

   Is it even possible for $N_M⊊N$? I'd think not because but I'm not 100% sure.

↑ comment by Alok Singh (OldManNick) · 2024-05-12T10:51:58.077Z · LW(p) · GW(p)

Gonna sleep bc 3 am but will respond later. Also the remark that hyperfinite can mean smaller than a nonstandard natural just seems false, where did you get that idea from?

Replies from: tailcalled

↑ comment by tailcalled · 2024-05-12T11:30:01.204Z · LW(p) · GW(p)

Also the remark that hyperfinite can mean smaller than a nonstandard natural just seems false, where did you get that idea from?

When I look up the definition of hyperfinite, it's usually defined as being in bijection with the hypernaturals up to a (sometimes either standard or nonstandard, but given the context of your OP I assumed you mean only nonstandard) natural $n$. If the set is in bijection with the numbers up to $n$, then it would seem to have cardinality less than $n+1$^[1].

1. ^{^}

   Obviously this doesn't hold for transfinite sizes, but we're merely considering hyperfinite sizes, so it should hold there.

Replies from: OldManNick

↑ comment by Alok Singh (OldManNick) · 2024-05-13T07:13:16.780Z · LW(p) · GW(p)

From Goldblatt, since {0..H} is internal.

Replies from: tailcalled

↑ comment by tailcalled · 2024-05-13T07:23:12.881Z · LW(p) · GW(p)

Ok, so this sounds like it talks about cardinality in the sense of 1 or 3, rather than in the sense of 2. I guess I default to 2 because it's more intuitive due to the transfer property, but maybe 1 or 3 are more desirable due to being mathematically richer.

I thought about this since. Bigger is not the right word. Complicated maybe? Like how the unit interval contains non-measurable sub intervals, or a compact set contains non-compact subsets.

Could you give a sneak peak on how Sylvia /Alain use these uniform distributions?