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In Bell's A primer of infinitesimal analysis, three microneighbourhoods of zero are defined:

  • $M_1 = \{ \epsilon \ |\ \epsilon^2 = 0 \}.$
  • $M_2 = \{ \epsilon \ |\ \lnot \epsilon \ne 0 \}.$
  • $M_3 = [0, 0].$

It is immediate that $M_1 \subseteq M_2 \subseteq M_3.$ However it seems to me that actually $M_2 = M_3.$ Is my proof right?

If $\epsilon \in M_3,$ but $\epsilon \ne 0,$ then $\epsilon > 0$ or $\epsilon < 0.$ The first case contradicts $\epsilon \le 0$ and the second $\epsilon \ge 0.$ Thus $\lnot \epsilon \ne 0.$

The dichotomy axiom appears as one of the order axioms of $\mathrm{BSIA}$ in page 105. But inequality is just denial-equality, so maybe it is not true in other versions of $\mathrm{SIA}?$ It seems to be a form of Markov's priciple.

I'm trying to understand why the two microneighbourhoods are defined if they seem to coincide.

Dominique
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Keplerto
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1 Answers1

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Here is a more detailed version of your argument as I now understand it. Let $\epsilon\in M_3$. If $\epsilon\not=0$ then either $\epsilon>0$ or $\epsilon<0$. Either way contradicts the definition of $M_3$. This therefore establishes constructively $\neg \epsilon\not=0$. In other words, $\epsilon\in M_2$.

The subsequent discussion in Bell suggests that he would not be too bothered if it turned out that $M_2=M_3$; in fact, what he immediately says is that it is possible (i.e., there is a model such) that all three of $M_1$, $M_2$, and $M_3$ are the same! Of course that's weaker than having a proof of $M_2=M_3$ from the axioms.

Bell only says that this is possible but does not give any reference. I tried looking this up in books by Kock as well as Moerdijk and Reyes, but so far no luck. You may want to try this at MathOverflow where heavyweights like Mike Shulman and Andrej Bauer should be able to answer this instantly.

Mikhail Katz
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  • We write $x \ne y$ for $\lnot x = y;$ this may be read as $x$ and $y$ are distinct or distinguishable.

    < So triple negation can be converted to a single one, right?

     – Keplerto Aug 01 '25 at 11:01
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    Also, I was attempting a direct proof, I did not assume $\varepsilon \notin M_2.$ From your answer, however, I get that maybe it is rare to have apartness that is stable (under double negation). I know that's the case in constructive math because Markov's principle is not accepted, but is the same true in $\mathrm{SIA}?$ – Keplerto Aug 01 '25 at 11:48
  • @Keplerto, then where did your condition $\varepsilon\not=0$ come from? – Mikhail Katz Aug 04 '25 at 05:55
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    I assume $\varepsilon \ne 0$ in order to reach a contradiction, isn't that valid? How else would a proof of $\lnot \varepsilon \ne 0$ go, in principle? – Keplerto Aug 04 '25 at 10:57
  • @MikhailKatz $M1$ and $M2$ aren't the same because $M1$ contains only $\varepsilon$ that satisfy $\varepsilon^2=0$ but $M2$ contains not only $\varepsilon$ that satisfy $\varepsilon^2=0$, – Mike_bb Aug 05 '25 at 06:28
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    I didn't say they are the same. I said that Bell said that it is possible for them to be the same. I assume he means that there are models where they are the same. As you point out, in these models you couldn't have for example nilpotent infinitesimals of order 3 (as you can have in other models). – Mikhail Katz Aug 05 '25 at 06:29
  • @MikhailKatz You are right. I agree. – Mike_bb Aug 05 '25 at 06:31