Dumb Question: Is the set $\Delta$ in Smooth Infinitesimal Analysis a subset of the set $\mathbb{R}$ of the real numbers?

Question

I am currently reading O'Connor's Introduction to Smooth Infinitesimal Analysis. In the paper, O'Connor defines $\Delta$ (or D in his paper) as the set of all numbers that are an element of R such that their squares are equal to zero (i.e., $\Delta = \{\delta \in R\text{ }| \delta^2=0\}$). So, the following question is probably quite stupid, but is $\Delta$ a subset of $R$ because if $\delta$ is an arbitrary element of $\Delta$, then it must be an element of $R$? Secondly, although I know the axioms in smooth infinitesimal analysis have been proved to be consistent (in topos logic, which I am not familiar with), but if the Law of Microcancellation holds and $\Delta \subseteq R$, then would the following proof be valid?

"Proof": Consider the function $f(x) = x^3$ (is $x^3$ smooth? I'm not sure but I believe it is.). Let us calculate $f'(x)$. By the Fundamental Fact about Derivatives,

$f(x+d) - f(x) = f'(x)d \text{ }$ for any $\text{ }d \in \Delta$. Then, $(x+d)^3 - x^3 = f'(x)d$, so $3x^2d+3xd^2+d^3 = f'(x)d$. Thus, since $d^2=0$, we have $3x^2d = f'(x)d$, so by the Principle of Microcancellation, we have $f'(x) = 3x^2$. Now, my main question lies around doing Microcancellation before considering that $d^2 = 0$. We would then have $f'(x) = 3x^2 + 3xd + d^2 = 3x^2 + 3xd$ because $d^2 = 0$. Then, $3x^2 = 3x^2 + 3xd$ for $x \in R$ and $d \in \Delta$ that is not equal to $0$.Let us take $x = 1$. Then, $3d = 0$, so $d = 0$, but this contradicts the fact that $d$ is an arbitrary element of $\Delta$.

Any advice on how to improve my question would be greatly appreciated.

Answer 1

$D$ is a subset of $R$ in smooth infinitessimal analysis. Note that it is usually typeset as $R$, not $\mathbb{R}$, because $R$ is the smooth real line which is being axiomatized as part of the theory. It is not just the set of real numbers $\mathbb{R}$ defined in set theory or similar. I'm not even sure if it corresponds to one of the usual definitions of real numbers in the topos models for SIA (although maybe it does).

The reason your microcancellation argument doesn't work is that it is being sloppy about variable binding. The principle says:

$$(∀d. a\cdot d = b \cdot d) → a = b$$

$d$ (the one being cancelled, anyway) is not in scope in the right hand equation. So:

$$(∀d. 3x^2d + 3xd^2 + d^3 = f'(x)d) → 3x^2 + 3xd + d^2 = f'(x)$$

is not an instance of this principle. The following would be:

$$(∀d_1. 3x^2d_1 + 3xd_0d_1 + d_0^2d_1 = f'(x)d_1) → 3x^2 + 3xd_0 + d^2_0 = f'(x)$$

Where you're cancelling one of two distinct variables of type $D$. But that wasn't the situation you were in. And also, it isn't necessarily the case that $∀d_0, d_1 \in D. d_0\cdot d_1 = 0$ as I recall, so you wouldn't be able to derive $f'(x) = 3x^2$.

Answer 2

To give a simpler version of Dan Doel's argument, the principle of microcancellation asserts that if for all such $d$ one has $ad=bd$ then $a=b$. However, the expressions $a$ and $b$ cannot contain the variable $d$ ! Therefore one cannot derive that $f'(x)=3x^2+3xd$ as you tried to do. Notice that since the background logic is intuitionistic, you should be careful to avoid claims such as "$d\in\Delta$ is not equal to zero".