In real analysis Cauchy sequence for $0$ is $(1/2,1/4,1/8,...)$.

But in non-standard analysis (hyperreal numbers) this sequence is infinitesimal $\varepsilon$.

Since hyperreal numbers are extension of real numbers I can't understand how $(0,0,0,....)=(1/2,1/4,1/8,...)$ and $\varepsilon=(1/2,1/4,1/8,...)$ since $0 \neq \varepsilon$.

How is it possible?

Downvoters, please explain where did I go wrong?


4 Answers 4


In the usual construction of the real numbers from Cauchy sequences, two Cauchy sequences are considered equivalent iff the term-wise difference converges to $0$.

In the usual construction of the hyperreals from Cauchy sequences, two sequences are considered equivalent iff the term-wise difference converges to $0$, and also is actually equal to $0$ "often enough" (with a precise but non-constructive definition of "often enough").

So in the hyperreal case, equivalence is a stricter requirement. Which is to say, the actual equivalence classes are "smaller" and "more numerous". Sequences that are equivalent in the real case might not be in the hyperreal case. This is what you have come across, and it is not really strange at all.

  • $\begingroup$ I meant another thing. I can't understand how $(0,0,0,....)=(1/2,1/4,1/8,...)$ and $\varepsilon=(1/2,1/4,1/8,...)$ $\endgroup$
    – Mike_bb
    Commented Dec 24, 2023 at 16:18
  • 5
    $\begingroup$ No, this is exactly what you meant. The $=$ means different things in your two statements, because the equivalence classes involved are different. $\endgroup$
    – Arthur
    Commented Dec 24, 2023 at 16:28
  • $\begingroup$ We have only one class - $(1/2, 1/4, 1/8, ....)$ $(0,0,0,..)$ equal to $0$ $\endgroup$
    – Mike_bb
    Commented Dec 24, 2023 at 16:34
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    $\begingroup$ No. When the sequences are seen as real numbers you have one set of equivalence classes, and thus one meaning of $=$. And when the sequences are seen as hyperreal numbers you have a different set of equivalence classes, and thus a different meaning of $=$. $\endgroup$
    – Arthur
    Commented Dec 24, 2023 at 16:35

As Arthur's answer states, there is no contradiction here; you're just conflating two different notions of equivalence. The sense in which the sequence $(1,{1\over 2}, {1\over 4}, ...)$ "is" the number zero is quite different from the sense in which that sequence "is" a nonzero infinitesimal (and FWIW "represents" is better than "is" in this sort of context, since it engenders less of this sort of confusion).

To add to Arthur's answer, I'm going to give a very high-level and ahistorical framing of the two number system constructions which hopefully make the distinction between them more immediate.

Both standard and nonstandard (= hyperreal) analysis can be thought of as aspects of a single theme:

We can form interesting extensions of the field of rationals, $\mathbb{Q}$, by looking at substructures of quotients of the power structure $\mathbb{Q}^\mathbb{N}$.

Here $\mathbb{Q}^\mathbb{N}$ is the set of all infinite sequences of rationals, with the operations of addition and multiplication being defined componentwise. Many "basic properties" of these arithmetic operations persist from $\mathbb{Q}$ to $\mathbb{Q}^\mathbb{N}$, but some care is needed here (e.g. think about division). Quotients of $\mathbb{Q}^\mathbb{N}$ are the structures we get by "identifying" certain elements of $\mathbb{Q}^\mathbb{N}$ with each other; I'll use the notation "$A/{\sim}$" to describe the quotient of a set $A$ by an equivalence relation $\sim$.

OK, technically I should also talk about "structure-respecting" equivalence relations, i.e. congruences, since even though you sort of can look at quotient structures by non-congruential equivalence relations we really shouldn't. But this isn't a serious issue here; all the equivalence relations involved in this answer will be congruential and that won't be the important point.

The standard approach is to first form the substructure $\mathsf{CS}\subseteq\mathbb{Q}^\mathbb{N}$ of Cauchy sequences, and then use the equivalence relation $$\alpha\approx\beta\quad\iff\quad \forall \epsilon>0\exists N\forall n>N[\vert \alpha(n)-\beta(n)\vert<\epsilon].$$ For example, letting $\alpha(n)=2^{-n}$ and $\beta(n)=0$ we do indeed have $\alpha\approx\beta$. The corresponding quotient $\mathsf{CS}/{\approx}$ is the standard reals.

The nonstandard approach is quite different. We skip the "substructure of" step, at the cost of performing a much more complicated quotient step. Specifically, start with a nonprincipal ultrafilter $\mathcal{U}$ on $\mathbb{N}$, and define the equivalence relation $$\alpha\sim_\mathcal{U}\beta\quad\iff\quad\{i: \alpha(i)=\beta(i)\}\in\mathcal{U}.$$ We then look at the structure $\mathbb{Q}^\mathbb{N}/\sim_\mathcal{U}$, which is our field of hyperreal numbers (note that it depends, in principal at least, on the choice of $\mathcal{U}$; we shouldn't say things like "the hyperreals").

The point is that the equivalence relations $\approx$ and $\sim_\mathcal{U}$ (fixing a choice of $\mathcal{U}$ for simplicity) are just different equivalence relations (and on different sets, for that matter, but that's a less substantive point here). For example, letting $\alpha$ and $\beta$ be defined as two paragraphs prior we have $$\alpha\approx\beta\quad\mbox{but}\quad\alpha\not\sim_\mathcal{U}\beta,$$ and there's no tension here at all. The only possible confusion comes from the conflation of two different notions of equivalence.

Incidentally, the whole "find interesting extensions of a structure $\mathcal{A}$ by looking at quotients of its power structures" theme is developed more systematically in universal algebra (see especially the $\mathsf{HSP}$ theorem), but that's a ways down the road.


Here is a slightly more elementary version of Noah Schweber's answer. Let's start with the ring $CS$ of Cauchy sequences (of, say, rational numbers). Let $I\subseteq CS$ be the set of sequences tending to zero, i.e., null sequences. Then the quotient $CS/I$ (collapsing all null sequences to $0$) is a field, namely the real numbers $\mathbb R$. But we could refine the implied equivalence relation by declaring that a null sequence should be collapsed only if almost all of its terms (i.e., all but finitely many terms) are $0$. Then the quotient will be a ring $R$ with zero divisors, where one already observes the phenomenon that some null sequences generate nonzero elements of $R$. To obtain a tool useful in analysis, we need to introduce additional collapsing in a certain consistent way, and form the quotient field which will give the hyperreal field $\mathbb R^\ast$, a proper ordered field extension of $\mathbb R$.

The hyperreals are useful in studying metric spaces and topology, among many other fields. For example, a set $K\subseteq \mathbb R$ is compact if and only if every element of $K^\ast$ is infinitely close to an element of $K$, giving a transparent criterion for compactness that does not involve either open covers or finite subcovers.

Furthermore, the hyperreals satisfy the internal least upper bound property, which is the only version that is reasonable to expect to hold. Similarly, internal completeness holds.


The hyperreals are not a metric space because they are not metrizable. Since Cauchy sequences are defined using metric properties we lose the ability to use them in the hyperreals.

The main problem is that $\epsilon$ is infinitely close to $0$ so we can't separate them meaning the space is not Hausdorff. So if we try to define a metric we get $d(0,\epsilon)=0$ but since $\epsilon \neq 0$ we've violated positivity.

They still carry the order topology however and I think there is a nice analogy with the field of rational functions. In that field the element $x^{-1}$ is infinitesimal and $x$ is infinite. So while I can clear say that for all real $r$ that $x^{-1} < r < x$ there is always this gap we have to jump to get from the finite to the infinite.

This also makes it clear that the hyppereals do not have the least upper bound property. For example $(2,4,8,16,\dots)$ is bounded above by $x$ but there is no least upper bound because every polynomial with an $x$ term is also an upper bound and we can't pick a least among them. So we've lost completeness as well.

These complications can be addressed in way that conforms with our intuition about infinitesimals though by being careful to restrict our attention to the most well-behaved hyperreals. For example we might not really care that $1+2\epsilon$ isn't exactly $1 + 3\epsilon$ or are happy to say that $1 + \frac1\epsilon$ and $2 + \frac1{\epsilon^2}$ are both $\infty$. Thankfully the order properties alone do give us significant power when suitably generalized so we can make sense of the theory.

  • 2
    $\begingroup$ While a true statement about the hyperreals, this doesn't actually address the question. $\endgroup$ Commented Dec 25, 2023 at 4:43

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