Denote by $\Bbb R$ the real line and by $\Bbb R^*$ the hyperreal line. For any real numbers $x < y < z$ and infinitesimal $\epsilon$ the following holds: \begin{equation} \forall a,b,c \in \Bbb R:~~~x + a\cdot \epsilon<y+b\cdot \epsilon<z +c\cdot \epsilon \end{equation} This, together with the ordering of $\Bbb R$ being a subset of the ordering of $\Bbb R^*$, makes me think that there is an analogy between the hyperreal line and the open long line, understood as an ordered countable infinity of real lines.

However, the hyperreal line contains at least an uncountable infinity of real lines, one for each real number. Then there are the infinite hyperreals. So the topology is not the same.

What is the topology of the hyperreal line?

  • $\begingroup$ There is of course an order topology. But it is not especially useful; its notion of convergence is not really the notion of convergence that we would "want". On the other hand the "hyperreal metric" (i.e. the natural extension of $d(x,y)=|x-y|$) does not form a metric, so it does not give us a topology. $\endgroup$
    – Ian
    Commented May 13, 2015 at 19:22
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    $\begingroup$ Note that, in addition to any infinitesimal hyperreal $\epsilon$, you also have $\sqrt\epsilon$, $\epsilon^2$, $\ln\epsilon$, and all of their inverses. EDIT: Fairly certain that $\ln\epsilon$ is negatively infinite, though. $\endgroup$ Commented May 13, 2015 at 19:23
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    $\begingroup$ @Ian What do you mean by that remark about convergence? (I'm not even sure how convergence would work in $\mathbb R^*$. $(\frac1n)_{n\in\mathbb Z^+}$ doesn't converge to anything, for example, because every element in that sequence is greater than every infinitesimal number, and there's neither a greatest infinitesimal nor a smallest real. The hyperreals aren't complete.) $\endgroup$ Commented May 13, 2015 at 19:27
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    $\begingroup$ @AndreaDiBiagio It's $\mathbb N$ (the nonnegative integers)… plus infinitely many points at $+\infty$. More specifically, they are all infinite hyperreals $N$ such that $\lfloor N\rfloor=N$. (This makes sense due to the transfer principle.) With the ultrafilter construction, they are all sequences of integers (under the ultrafilter). $\endgroup$ Commented May 15, 2015 at 14:40
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    $\begingroup$ The hypernaturals actually satisfy Peano's (first order) Axioms, IIRC. The induction principle — that if $\phi(x)$ is a first-order property, $\phi(0)$ holds, and $\phi(x)\implies\phi(x+1)$, then $\phi(x)$ holds for all $x$ — works because of the transfer principle. Also, there are uncountably many hypernaturals, because $f:\mathbb R\to\mathbb N^*$ defined by $f(x)=\lfloor Nx\rfloor$ is one-to-one. (Where $N$ is infinite.) $\endgroup$ Commented May 15, 2015 at 16:07

2 Answers 2


I believe there isn't a single thing we can point to as "the" right topology on the hyperreals, but there are a few natural candidates:

  • The interval topology was discussed in the comments.

  • We can take as basic opens the intervals $(r-\epsilon, r+\epsilon)$ with $r\in {}^*\mathbb{R}$ and $\epsilon\in\mathbb{R}_{>0}.$ Unions of such intervals are called real open sets; unfortunately the real open sets are not closed under intersection, and the induced topology is not well-behaved.

  • We can replace $(r-\epsilon, r+\epsilon)$ with $((r-\epsilon, r+\epsilon))=\{x: x$ is well inside $(r-\epsilon, r+\epsilon)\}.$ (This just means that we demand $x$ not be infinitesimally close to $r-\epsilon$ or $r+\epsilon$, in addition to $x\in(r-\epsilon, r+\epsilon)$.) These are the $S$-neighborhoods, and they induce the $S$-topology. This seems to be the nicest standard topology on the hyperreals.

See Robert Goldblatt's book, esp. chapters 10 and 11.

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    $\begingroup$ Is the $S$-topology Hausdorff? $\endgroup$ Commented May 15, 2015 at 16:13
  • $\begingroup$ @columbus8myhw No. You can't separate $0$ from $\epsilon \approx 0$. $\endgroup$
    – wlad
    Commented May 19, 2015 at 11:10

The "sameness" of the standard and nonstandard models is about elementary equivalence, not homeomorphism; by the transfer principle, $\mathbb{R}$ and ${}^\star \mathbb{R}$ have exactly the same internal properties.

That is, the most well-behaved topology on ${}^\star \mathbb{R}$ is the internal topology given by the transfer of the usual topology on $\mathbb{R}$, but it's only well-behaved internally. For example, it only has the internal least upper bound property: every bounded, nonempty, internal subset has a least upper bound. External subsets can go either way.

Whether or not that internal topology is an external topology is a nontrivial question I'm not prepared to answer; even at the most naive level of taking some formulation of topology verbatim, the answer depends on which formulation of topology you use!

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    $\begingroup$ Could you please define "internal" and "external" please? I am no expert in non-standard analysis. Also, could you clarify as to how does your post answer my question? $\endgroup$
    – Andrea
    Commented Feb 2, 2017 at 21:19

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