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I have two closely related questions:


Firstly, suppose that I can find two hyperintegers $P$ and $Q$ s.t. $\frac{P}{Q}=\sqrt{2}$. Obviously, both $P$ and $Q$ lie in an extension of the integers. However, this does not make $\sqrt{2}$ rational because neither $P$ nor $Q$ is an integer.

This raises the question: are the rational, irrational, and integral qualities of infinities even worth discussing in the first place?

By simple algebra, $P=\sqrt{2}Q$; but, as stated $P$ and $Q$ are both hyperintegers - hence it is possible to obtain a hyperinteger which is a noninteger multiple of another hyperinteger. This seems far enough removed from the properties of the standard integers that the question of rationality effectively becomes irrelevant.


Secondly, suppose that I identify a unique infinite quantity $E$ as the upper bound of the natural numbers, and $\epsilon=1/E$. If...

$$\mathbb{N}^*=\bigcup_{k=0}^E \left\{a_k\right\}\quad:\quad a_k=a_{k-1}+1,\ a_0=0$$

...then presumably $\mathbb{N}^*\epsilon\approx[0,1]$. I would say that this leads to a contradiction by implying that $[0,1]$ is countable, except that I'm not sure if it makes sense to describe $\mathbb{N}^*$ as countable in the first place. Certainly, for any $n\in\mathbb{N}^*\cap\mathbb{N}$ the set...

$$N_n=\bigcup_{k=1}^n\left\{a_{k}\right\}$$

...is countable. But as long as $n$ is finite $N_n\epsilon\subset\text{monad}(0)$. In order to obtain any portion of $[0,1]$ which is not comprised solely of infinitesimals, I have that $n$ must be infinite and thus outside the natural numbers.

That $\mathbb{N}^*$ is defined by succession intuitively suggests that $\mathbb{N}^*$ is countable - however the fact that most elements of $\mathbb{N}^*$ are infinite means that an infinite number of successions is required to reach most elements of $\mathbb{N}^*$.

I thought about trying to prove that $\mathbb{N}^*$ is uncountable using something akin to diagonalization, but at a certain point I have to ask - does it even make sense to think about such sets as countable or uncountable to begin with?

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    $\begingroup$ Is it possible, though, to find these two hyperintegers? I always thought that the many standard proofs that $\sqrt2$ is irrational can be done internally in PA, which tend to hold in the usual hyperfinite systems (i.e. ultrapowers of $\Bbb N$). $\endgroup$ – Asaf Karagila Apr 2 at 14:49
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    $\begingroup$ Your definitions of $E$ and $\mathbb{N}^*$ make no sense. In particular, what do you mean by "the upper bound of the natural numbers"? If you're talking about a hyperreal field, there are many different upper bounds for the natural numbers (and no least upper bound). You also have not defined what $a_{k-1}$ means. $\endgroup$ – Eric Wofsey Apr 2 at 14:50
  • $\begingroup$ And why is there a unique quantity as an upper bound of $\Bbb N$? That's the whole point in the proof that an order complete field is $\Bbb R$, and so $\Bbb N$ has no least upper bound, it is either unbounded or there is a decreasing chain of upper bounds, without a least one. $\endgroup$ – Asaf Karagila Apr 2 at 14:51
  • $\begingroup$ @AsafKaragila I'm not entirely sure... finding these two hyperintegers would be time consuming. First I would need to find two divergent infinite sequences whose quotient as each tended to infinity converged to $\sqrt{2}$, then find a matching ultrafilter to construct the appropriate extension of the integers. $\endgroup$ – R. Burton Apr 2 at 14:53
  • $\begingroup$ Or, show that $\{m\mid\exists n\leq m: m^2=2n^2\}$ is empty, and note that ultrapowers are elementary... But you know, potato powasteyourtimeto... :-) $\endgroup$ – Asaf Karagila Apr 2 at 14:56
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For your first question, as noted in the comments there are no such $P,Q$. What we can find are hyperintegers $P,Q$ such that $P\over Q$ is in the "halo" around $\sqrt{2}$ - that is, $\vert {P\over Q}-\sqrt{2}\vert$ is infinitesimal - but that's very much not the same thing. This follows from the transfer principle, which you should make sure you understand right at the beginning of studying the hyperreals.

Your second question suffers from the same "halo-versus-equality" issue, but there it's inessential. Suppose $E$ is a nonstandard natural; then we do indeed get a surjection from $\mathbb{E}$, the set of hypernaturals $<E$, to $[0,1]$ gotten by sending $N$ to the real in whose halo $N\over E$ lies. However, $\mathbb{E}$ is never countable (indeed, that's a consequence of the existence of that surjection!), and more generally we can show that no internal set of hyperreals which is (externally) infinite is (externally) countable.


EDIT: Here's a proof of my bolded claim.

Suppose $A$ is an externally-infinite internal subset of $^*\mathbb{R}$. Internally, we can define the set $S_A$ of hypernaturals $n$ such that there is an injection from $n$ to $A$. Since $A$ is externally infinite, we know that $S_A$ contains every standard natural; by overspill, there is some nonstandard natural $N\in S_A$. But then we get a surjection from $S_A$ to $[0,1]$ per above, and so we know that $S_A$ is uncountable - and in fact $\vert S_A\vert\ge 2^{\aleph_0}$.

More generally, and avoiding the "internal/external" language: if $\mathfrak{Y}=\prod_{i\in I}\mathfrak{X}_i/\mathcal{U}$ is any ultrapower of structures (over an infinite index set, by a nonprincipal ultrafilter) then every infinite definable subset of $\mathfrak{Y}$ has cardinality $\ge 2^{\aleph_0}$. (A proof can be found, for instance, in Bell's book Models and ultraproducts.)

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  • $\begingroup$ Okay, I'm with you on the first part (in retrospect it should have been obvious), and the second part makes sense, but I'm not sure how I would prove it. Is there a theorem that I could look up? $\endgroup$ – R. Burton Apr 2 at 15:19
  • $\begingroup$ @R.Burton I added a proof. $\endgroup$ – Noah Schweber Apr 2 at 15:32
  • $\begingroup$ For the first part, you can also do what I suggested in the comments of the questions. Note that PA already proves that $\forall m\forall n\leq m(m\cdot m\neq n\cdot n+n\cdot n)$, and by elementarity, the same holds for hyperfinite integers as well. $\endgroup$ – Asaf Karagila Apr 2 at 15:41
  • $\begingroup$ @AsafKaragila Sure, but I think it's important to also be comfortable with the "transfer nuke" (which really is just elementarity higher up, after all). $\endgroup$ – Noah Schweber Apr 2 at 15:42

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