How to prove that 2 points $\varepsilon$ and 2$\varepsilon$ are indistinguishable not only in sole case? (Smooth infinitesimal analysis)

Question

There are 2 points on the Real line: $A$ and $2A$. They are indistinguishable in sole case - if $A$ is $0$.
But how to prove that $\varepsilon$ and 2$\varepsilon$ are indistinguishable not only in sole case? ($\varepsilon$ is infinitesimal)

Thanks.

Answer 1

As you can find in e.g. Section 3 of O'Connor's SDG notes, the real line object $R$ of Smooth Infinitesimal Analysis satisfies the following: $$\forall x \in R.~x \neq 0 \rightarrow \exists y \in R. xy = 1$$

This means that if you can prove that some number $x$ is not equal to $0$, then you can find its multiplicative inverse $x^{-1}$ such that $x x^{-1} = 1$.

It follows that if $\varepsilon$ is a nilsquare infinitesimal, so that $\varepsilon^2 = 0$, then you cannot ever hope to prove that $\varepsilon \neq 0$. Why?

Assume you could somehow prove that $\varepsilon \neq 0$. Then, by the result above, you could find its multiplicative inverse $\varepsilon^{-1}$ satisfying $\varepsilon \varepsilon^{-1} = 1$. And once you found the multiplicative inverse, you could

• Start from the equation $\varepsilon \varepsilon = 0$;
• multiply both sides by $\varepsilon^{-1}$ to get $\varepsilon\varepsilon\varepsilon^{-1} = 0 \varepsilon^{-1}$;
• simplify using $\varepsilon\varepsilon^{-1} = 1$ to $\varepsilon = 0 \varepsilon^{-1}$;
• use the fact that $0x = 0$ holds for any $x$ to conclude that $\varepsilon = 0$.

So we'd now have that $\varepsilon \neq 0$ and $\varepsilon = 0$, a contradiction.

This means that you cannot distinguish any infinitesimal from zero: to distinguish an infinitesimal $\varepsilon$ from zero, you'd have to assume or prove that it's not zero, i.e. that $\varepsilon \neq 0$. But as we see above, $\varepsilon \neq 0$ would give rise to a contradiction!

In practice, this means that whenever you give a proof, construction or definition starting with "take an infinitesimal $\varepsilon^2 = 0$", your proof/construction/definition has to work even in the case that $\varepsilon = 0$. You can't ever rule this case out in any way.

Now, assume that you could somehow distinguish between $\varepsilon$ and $2\varepsilon$, i.e. you could somehow prove that $2\varepsilon \neq \varepsilon$. Then, since $x \neq y$ implies $x - c \neq y - c$, we'd have that $2\varepsilon - \varepsilon \neq \varepsilon - \varepsilon$, i.e. $\varepsilon \neq 0$. But that cannot be shown, as we've seen above.

Not only can you not distinguish infinitesimals from zero, but you cannot distinguish infinitesimals from each other at all. So whenever you give a proof, construction or definition starting with "take an infinitesimal $\varepsilon^2=0$", your proof/construction/definition has to work exactly the same way for every infinitesimal. You won't be able to make distinctions like "if I picked infinitesimal $a$, do something; if I picked infinitesimal $b$, do something else".

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aside: The key thing that makes this work is the use of intuitionistic logic. If we were using classical logic, by the law of excluded middle we'd have that "$\varepsilon = 0$ or $\varepsilon \neq 0$": the second case yields a contradiction, so we'd conclude that the first case, $\varepsilon = 0$ holds, and therefore we'd conclude that the only nilsquare number in a field is zero. That's why Smooth Infinitesimal Analysis cannot work using classical logic. Thanks to intuitionistic logic, we don't have the law of excluded middle, so it need not be the case that "$\varepsilon = 0$ or $\varepsilon \neq 0$" holds, and Smooth Infinitesimal Analysis keeps working.

As a side remark, I'd like to point out that Smooth Infinitesimal Analysis is a particularly difficult setting for mastering intuitionistic logic.

When one learns constructive mathematics that relies on intuitionistic logic, one finds that certain argument forms and constructions one is used to no longer work. This naturally takes some getting used to. However, there's at least no new stuff to keep track of: at least initially, you're still learning largely the same mathematics, you just have to work a bit harder to prove your theorems.

This happens in Smooth Infinitesimal Analysis too. But you also get something more. Smooth Infinitesimal Analysis has anti-classical axioms, forms of reasoning that wouldn't work at all in classical logic (they'd yield contradictions), but happen to work in intuitionistic logic. So when you learn intuitionistic logic and SIA at the same time, you're forced to grapple with the fact that your old familiar tools no longer work, and you also have to master new tools that look like they shouldn't work. Most people find managing these interactions fairly tough.