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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.

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  • $\begingroup$ 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
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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.

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    $\begingroup$ 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

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