How to prove that hyper reals are not reals?

Question

I was told by my professors that infinitesimals are numbers which are smaller than all the reals. And I was also told that it is the sequence, $\{1,1/2,1/3,1/4,\dots\}$. And I am comfortable with those all. But finally they told that those are not reals. At this point I felt a bit of uncomfortable and wanted a proof of it. So I tried the proof of required thing using Archimedes property of reals. But I don't find a proof as I require. So I request you to help me with this. Thank you.

Answer 1

The short version is that the multiplicative inverse of an infinitesimal must be greater than every natural number: if $\alpha$ is positive but $<{1\over n}$ for each $n\in\mathbb{N}$, then $1\over\alpha$ exists in our "expanded number system" (since $\alpha\not=0$) and is $>n$ for every $n\in\mathbb{N}$ (since $\alpha<{1\over n}\implies {1\over\alpha}>{1\over({1\over n})}$).

But talking about infinitesimals can feel rather slippery; additionally, the comments you've quoted aren't really accurate. So below I've given a longer version of this answer which makes things more precise. In my opinion you should not read this until the previous paragraph makes sense. However, after you understand the basic idea in the previous paragraph, the following may be useful to you (especially the bit at the end about the hyperreals). I've tried to keep what follows as intuitive as I can while still introducing the key concepts; please let me know if there's a point you'd like me to simplify or elaborate on and I'll be happy to do so.

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The relevant type of structure subsuming the rationals, the reals, and the various versions of the hyperreals (see below) is the ordered field. Roughly, an ordered field is a structure where we have operations of addition, subtraction, multiplication, and division, an ordering $<$, and all the "basic properties" of these hold (e.g. if $x<y$ then $x+z<y+z$). The most common examples of ordered fields are $\mathbb{Q}$ and $\mathbb{R}$; another example is $\mathbb{Q}(\pi)$, intuitively consisting of everything you can build from the rationals and $\pi$ (e.g. it contains $\pi^2\over 17$ but does not contain $\sqrt{2}$). And even more exotic examples of ordered fields exist (see below).

• For the precise definition of ordered fields and a summary of their basic properties, see here.

Talking about ordered fields is a bit abstract, but it helps make the various concepts we're playing with here precise. Every ordered field $F$ "contains a copy of $\mathbb{Q}$" as follows: we identify $0$ with the additive identity $0_F$, $1$ with the multiplicative identity $1_F$, $n\in\mathbb{N}$ with $1_F+1_F+...+1_F$ ($n$ times), and take ratios and negatives to get everything else. For example, $-{2\over 3}$ gets identified with the additive inverse of ${1_F+1_F\over 1_F+1_F+1_F}$.

• In fact, this is unique in an appropriate sense: there is exactly one field homomorphism $\mathbb{Q}\rightarrow F$. So we often conflate $\mathbb{Q}$ itself with this "copy of $\mathbb{Q}$ inside $F$" and I will do so below.

Now given an ordered field $F$, an infinitesimal in $F$ is an $\alpha\in F$ such that $\alpha>0$ but $\alpha<q$ for all $q\in\mathbb{Q}$ with $q>0$. One consequence of having infinitesimals is that the Archimedean property must fail:

Suppose $F$ is an ordered field and $\alpha\in F$ is infinitesimal. Then there is some $\beta\in F$ such that $\beta>q$ for every $q\in\mathbb{Q}$.

The proof is exactly the argument way back in the first paragraph of this answer: consider the multiplicative inverse of an infinitesimal. In particular, once you know that $\mathbb{R}$ is Archimedean, you know that $\mathbb{R}$ has no infiniesimals.

Meanwhile, note that there are perfectly natural ordered fields which are non-Archimedean; for example, consider the rational functions in a single variable $x$, ordered by "behavior in the limit:" $$\mbox{$f<g$ $\quad\iff\quad$ $\lim_{x\rightarrow +\infty}f(x)<\lim_{x\rightarrow +\infty}g(x)$.}$$ Once we throw on an appropriate equivalence relation (e.g. we want to identify ${17x+42\over 17x+42}$ and $1$), we get an ordered field which is clearly non-Archimedean (since $x>n$ for each constant $n$).

And now we finally come to hyperreals. A hyperreal field is a particularly nice non-Archimedean ordered field. Essentially, a hyperreal field is a non-Archimedean ordered field which is similar to $\mathbb{R}$ in a very strong way (despite being non-Archimedean). The precise details are highly technical, but see the comments below. At this point it'd be great if I could give an example of a hyperreal field but unfortunately I can't: in a precise sense, no hyperreal field is "easy to describe." In particular, the non-Archimedean example above is not a hyperreal field.

Note the phrase "a hyperreal field" as opposed to "the hyperreal field." It is here that your professors have been imprecise: although you'll often hear people talk about "the hyperreals," there is no single "best" hyperreal field. The most common type of hyperreal field considered is built out of infinite sequences of real numbers in a manner vaguely similar (but only vaguely) to how we construct the reals from the rationals via Cauchy sequences: elements of such a hyperreal field are "named" by sequences of real numbers, vaguely similarly to how real numbers are "named" by sequences of rational numbers. If we build a hyperreal field $H$ in such a way, the sequence $$\alpha:\quad(1,{1\over 2}, {1\over 3},{1\over 4}, {1\over 5},...)$$ does indeed "name" an infinitesimal. However, so do lots of other sequences. For example, the sequence $$\beta:\quad(2,{2\over 2}, {2\over 3}, {2\over 4}, {2\over 5},...)$$ "names" an element of $H$ which $(i)$ is infinitesimal but $(ii)$ is bigger than $\beta$. This isn't something we can prove without diving into the details, but I can tell you the key observations which drive these facts and should make them sound plausible:

• The point driving $(i)$ is that each term of $\alpha$ and $\beta$ is positive, but for each positive rational $q$ "almost all" of the terms of $\alpha$ and of $\beta$ are $<q$.

• The point driving $(ii)$ is that each term of $\alpha$ is $<$ the corresponding term of $\beta$.