Suppose we construct the hyperreals by fixing a free ultrafilter $\mathcal F$ – formalizing the idea of "large subsets of $\mathbb N$" – and defining an equivalence relation between two real sequences $\{r_k\}_{k=1}^\infty$ and $\{s_k\}_{k=1}^\infty$, denoted as $\langle r_k \rangle$ and $\langle s_k \rangle$ as

$$\langle r_k \rangle \sim \langle s_k \rangle \iff \{k \in \mathbb N : r_k = s_k \} \in \mathcal F$$

Each hyperreal is hence an equivalence class as defined by the relation above.

Can we define a real valued norm $\|\cdot\| : {^*\mathbb R} \to \mathbb R$, without extending it to a hyperreal norm $\|\cdot\| : {^*\mathbb R} \to {^*\mathbb R}$?

One problem that emerges is that the norm must satisfy $\|x\| > 0$ whenever $x \neq 0$ which covers infinitesimal $x$. Hence the intuitive norm $\operatorname{sh}(|x|)$, where $\operatorname{sh}(|x|)$ denotes the shadow or standard part of $|x|$, doesn't work.

  • $\begingroup$ I don't recall encountering the "shadow" terminology before. Where did you see it? $\endgroup$ – Andrés E. Caicedo Aug 28 '18 at 11:53
  • $\begingroup$ @Andrés: In the shadows! :) $\endgroup$ – Asaf Karagila Aug 28 '18 at 12:57

No. Because cofinality matters.

The cofinality of any ultrapower of $\Bbb N$ by a free ultrafilter is uncountable. Therefore, the cofinality of ${}^*\Bbb R$ is uncountable as well. But $\Bbb R$ has only countable cofinality.

Any norm would have to be order preserving, in which case you just cannot do it.

Of course, the hyperreals form a vector space over $\Bbb R$ of dimension $2^{\aleph_0}$, so we can find an $\Bbb R$-linear isomorphism with some normed space, say $\ell^2$, but that's not what I think you want to do.

  • 1
    $\begingroup$ Why would a norm be order preserving? $\endgroup$ – nombre Aug 28 '18 at 20:18
  • $\begingroup$ Well, it obviously doesn't have to be. But it seems to me that somehow the OP wants a norm that is order preserving. If not, I also suggested a solution for that. $\endgroup$ – Asaf Karagila Aug 28 '18 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.