5
$\begingroup$

I've been trying to show that there exists a subset of the hyperreals which has order type $\omega_1$, that is, that there exists a subset $A \subseteq {}^*\mathbb{R}$ for which there is an order isomorphism from $\omega_1$ onto $A$.

Here the hyperreals are constructed using a free ultrafilter, $\mathscr{U}$, over $\omega$ - much like construction presented on the wikipedia page.

I know that one cannot find such a subset of $\mathbb{R}$, but I really just don't see how the situation is much better for the hyperreals. Initially I believed that one ought to be able to find a subset of the hyperreals simply because there were ''more'' hyperreals, but even that isn't (entirely) true because $\mathbb{R}$ and ${}^*\mathbb{R}$ have the same cardinality. I haven't really been able to even get off the ground and I was wondering if anyone could shed some light on how to construct such an embedding/subset of ${}^*\mathbb{R}$? Any feedback is much appreciated.

$\endgroup$
4
  • $\begingroup$ Can't we just construct a set by the following recursion: at $r_{\alpha+1}$ just pick a successor of $r_\alpha$, and at $r_\lambda$ we pick a successor of the $r_\beta$ for $\beta<\lambda$ (which will exist since $\{r_\beta:\beta<\lambda\}$ will be countable, and so we can diagonalise out). $\endgroup$
    – user104955
    Nov 24, 2014 at 15:12
  • $\begingroup$ This is interesting. I hadn't thought about it like that. I'm not quite sure what you mean by ''diagonalize out'' though. $\endgroup$ Nov 24, 2014 at 16:02
  • $\begingroup$ Let $r_0,...,r_n,...$ be a sequence of hyperreals. Then we define $r$ such that $r(n)$ is a real greater than $r_m(n)$ for $m\leq n$. Then, for all $m\geq n$, $r(m)> r_n(m)$. So $r$ is greater than all the hyperreals in the sequence. Does that sound right? $\endgroup$
    – user104955
    Nov 24, 2014 at 16:22
  • $\begingroup$ Oh I see. Yes this looks like it ought to work! $\endgroup$ Nov 24, 2014 at 17:42

2 Answers 2

5
$\begingroup$

What you want is the usual argument that the bounding number $\mathfrak{b}$ is uncountable. Given a countable family $\mathscr{F}=\{f_n:n\in\omega\}\subseteq{}^\omega\omega$, define

$$f:\omega\to\omega:n\mapsto 1+\max_{k\le n}f_k(n)\;;$$

then for each $n\in\omega$ we have $f(k)>f_n(k)$ for all $k\ge n$. Using this, it’s easy to construct recursively a family $\mathscr{F}=\{f_\xi:\xi<\omega_1\}\subseteq{}^\omega\omega$ such that $f_\xi<^*f_\eta$ whenever $\xi<\eta<\omega_1$, where for $f,g\in{}^\omega\omega$ we write $f<^*g$ iff there is an $n\in\omega$ such that $f(k)<g(k)$ for all $k\ge n$. The map

$$\varphi:\omega_1\to{}^*\Bbb R:\xi\mapsto [f_\xi]_{\mathscr{U}}$$

is then an order isomorphism (though there’s no reason to think that it’s continuous at limits).

$\endgroup$
3
  • $\begingroup$ There are plenty of reasons to think that: (1) Ignorance about order properties of the real numbers and the hyperreal numbers; (2) naiveté; (3) wishful thinking; (4) cranks might use these sort of facts to prove or disprove all sort of crazy things. The list can go on. $\endgroup$
    – Asaf Karagila
    Nov 24, 2014 at 20:53
  • $\begingroup$ @Asaf: I vote for (3). $\endgroup$ Nov 24, 2014 at 21:00
  • $\begingroup$ I agree, it's the best one. But it can be a curse too. $\endgroup$
    – Asaf Karagila
    Nov 24, 2014 at 21:01
2
$\begingroup$

Let $\alpha$ be some ordinal such that $f : \alpha \to {}^*\mathbb{R}$ is a bijection (i.e. a well-ordering of the hyperreal numbers). Let $I$ be the subset of $\alpha$ defined by $$ I = \{i \in \alpha : f(i) \ge f(j) \text{ for all } j \le i\} \quad (f(i) \ge f(j) \text{ as hyperreals.}) $$ It suffices to show that $I$ is uncountable, and we will have $\omega_1 \subset I$.

Suppose towards contradiction that $I$ is only countable. Then there is a countable set of hyperreal numbers $h_1, h_2, h_3, \ldots$ such that ALL hyperreal numbers $h$ are bounded by some $h_n$. Pick representative sequences \begin{align*} h_1 &= (r_{11}, r_{12} , r_{13}, \ldots) \\ h_2 &= (r_{21}, r_{22} , r_{23}, \ldots) \\ h_3 &= (r_{31}, r_{32} , r_{33}, \ldots) \\ \vdots \end{align*} And consider the hyperreal number $$ q := (1 + r_{11}, 1 + \max\{r_{12}, r_{22}\}, 1 + \max\{r_{13}, r_{23}, r_{33}\}, \ldots) $$ where $q_i = 1 + \max\limits_{1 \le j \le n} r_{ji}$. $q_i$ is real, so $q$ is hyperreal. But now for any $h_n$, the $i$th coordinate of of $q$ is strictly bigger than the $i$th coordinate of $h_n$ for all $i$ sufficiently large. So $q > h_n$ for all $n$, contradiction.

$\endgroup$
7
  • 1
    $\begingroup$ Why is there such a bijection? $\endgroup$
    – Asaf Karagila
    Nov 24, 2014 at 19:26
  • $\begingroup$ @Goos when you say ''i.e. a well-ordering of the hyperreal numbers'' do you mean a well ordering of the hyperreals induced by the map $f$? $\endgroup$ Nov 24, 2014 at 20:07
  • $\begingroup$ Regardless, it seems like the construction you use to draw a contradiction is exactly the one that gives you the subset you want. For example, let $r_0 = (0, 0, \ldots)$, let $r_{\alpha+1} = (r_\alpha(0) + 1, r_\alpha(1) + 1, \ldots)$ and for a limit ordinal $\lambda < \omega_1$, let $\{r_{\alpha_i}\}_{i \in \omega} \subseteq \{r_\alpha\}_{\alpha < \lambda}$ be a cofinal subsequence of length $\le \omega$ of the one you've constructed upto this point and then ''diagonalize'' this subsequence in exactly the way you did above to define $r_\lambda$. Does this work? $\endgroup$ Nov 24, 2014 at 20:15
  • $\begingroup$ @AndrewRoss Yes, that works also. And yes, I mean the order given by $f(a) < f(b) \iff a < b$. $\endgroup$ Nov 24, 2014 at 20:53
  • $\begingroup$ @AsafKaragila I was appealing to the well-ordering principle, but I should not have assumed that the ordinal in question was $\omega_1$. I shall edit. $\endgroup$ Nov 24, 2014 at 20:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .