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Thanks to Z. A. K. answer I edited my question

In this lecture Synthetic Differential Geometry they have the following

Definition 4.3. An infinitesimal on $R$ is any nilsquare element of $R$, i.e. $x^{2}=0 .$ We denote the collection of infinitesimals on $R$ by $\Delta:=\left\{x \in R \mid x^{2}=0\right\}$.

Axiom 4.1. (Kock-Lawvere) For any mapping $g: \Delta \longrightarrow R$ there exists a unique $b$ in $R$ such that for all $\varepsilon$ in $\Delta$ we have $g(\varepsilon)=g(0)+b \varepsilon$.

Theorem 4.2. $ \Delta \neq\{0\} \text {. } $

the proof of the theorem goes like this :

$g(\varepsilon)=g(0)+\varepsilon b$ for $b=0$, since $g(\varepsilon)=g(0)=0=0+\varepsilon 0$, but also for $b=1$, since $g(\varepsilon)=g(0)+\varepsilon=\varepsilon$. So this $b$ is not unique (because $0 \neq 1$ ), which contradicts the Kock-Lawvere axiom. So $\Delta$ cannot coincide with $\{0\}$.

Why is that if $\Delta \neq\{0\}$ would solve the contradiction in the proof?

amilton moreira
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1 Answers1

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Let me restate the axiom for you.

Kock-Lawvere axiom: for any map $g: \Delta \rightarrow R$, we can find precisely one element $b_g \in R$ that satisfies the following:

  • for every $\varepsilon \in \Delta$, the equality $g(\varepsilon) = g(0) + b_g\varepsilon$ holds.

What happens if we assume $\Delta = \{0\}$? Well, under that assumption we would know that

  1. Every $\varepsilon \in \Delta$ satisfies $\varepsilon = 0$.
  2. Accordingly, for any map $g: \Delta \rightarrow R$ and any $\varepsilon \in \Delta$, we have $g(\varepsilon) = g(0)$.

But this would mean that for any map $g: \Delta \rightarrow R$ we can in fact find two different numbers, $b_g=0$ and $b_g=1$, so that for every for every $\varepsilon \in \Delta$, the equality $g(\varepsilon) = g(0) + b_g\varepsilon$ holds. If $b_g=0$ then we have equalities $g(\varepsilon) = g(0) + 0 \varepsilon = g(0)$, if $b_g=1$, we would also have $g(\varepsilon) = g(0) + 1 \varepsilon = g(0) + 1 \times 0 = g(0)$. This contradicts the Kock-Lawvere axiom, which asserts that we can find exactly one number with this property.

Since assuming $\Delta = \{0\}$ leads to a contradiction, our assumption must have been wrong, and so we can conclude $\Delta \neq \{0\}$.

edit: Note that if we do not assume $\Delta = \{0\}$, then neither of 1,2 above hold, and we cannot get a contradiction. Since it is not the case that $\Delta$ consists of only one element, it's perfectly fine and non-contradictory that $g(d) = g(0) + 0d$ and $g(d) = g(0) + 1d$ both work for some specific element $d \in \Delta$, as long as they don't both work for all elements $\varepsilon \in \Delta$, the K-L axiom is not violated.

Z. A. K.
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  • Thanks to your answer I edited my question – amilton moreira Nov 14 '21 at 12:21
  • @amiltonmoreira See the edit on how $\Delta \neq {0}$ avoids the same conclusion. – Z. A. K. Nov 14 '21 at 12:31
  • what you mean is this assuming $\Delta={0 }$, than $b=0$ and $b=1$ would satisfy $g(0)=g(0)+0\dot b$. than since $\Delta$ consist of only the element $0$ that we would have two elements $b$ for such that for all $\epsilon \in \Delta g(\epsilon)=g(0) +b\epsilon$ . In the case $\Delta \neq {0}$ we have only the element $b=1$. Is this what you mean? – amilton moreira Nov 14 '21 at 12:41
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    No. For each function $g$, we only ever have one value $b_g$ such that for all $\varepsilon \in \Delta$, $g(\varepsilon) = g(0) + b_g \varepsilon$. The one value that works depends on the function $g$ - the value that works may be $0$ (e.g. for $g(x) = x^2$), may be $1$ (e.g. for $g(x) = x$), may be $\pi$, may be $e^e - 2$, or any other value, really. But if $\Delta= {0}$ was true, then multiple values would work for one function $g$ and all $\varepsilon \in \Delta$. Since according to K-L, only one value ever works for a fixed $g$, $\Delta= {0}$ is not true, and so $\Delta \neq {0}$. – Z. A. K. Nov 14 '21 at 12:47
  • A. K what I meant in my comment is this suppose $\Delta= {0}$ than $b=1,b=2,b=0...$ would satisfy $g(\epsilon)=g(0)+\epsilon \dot b$ because $\epsilon=0$ so for all $\epsilon in \Delta$ we have more than on $b$ such that for all $\epsilon \in \Delta , g(\epsilon)=g(0)+\epsilon \dot b$ because $\Delta$ consist only of the elemennt ${0}$ – amilton moreira Nov 14 '21 at 13:16
  • That's right, @amiltonmoreira. That's the essence of this proof. – Z. A. K. Nov 14 '21 at 13:20
  • But for example suppose $\Delta={0,c}$ in this case the only $b$ that would work is $b=1$ because for example for $b=0$ would work for $\epsilon=0$ but not for $\epsilon=c$ so we would have only the value $b=1$ that would satisfy $g(\epsilon)=g(0)+b \dot \epsilon$ for all $\epsilon \Delta$ – amilton moreira Nov 14 '21 at 13:22
  • @Z.A.K. If we prove that "every $\varepsilon \in \Delta$ satisfies $\varepsilon=0$" leads us to contradiction then why can't we conclude that there is only one $\varepsilon \in \Delta$ satisfies $\varepsilon=0$? Thanks. – Mike_bb Oct 21 '23 at 02:18
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    @Mike_bb: I'm not sure I understand your question. There is exactly one $\varepsilon \in \Delta$ that satisfies $\varepsilon = 0$, but as far as I can tell that fact has nothing to do with this proof and this question. And in general you certainly can't conclude "there is exactly one $x \in S$ that has property $P$" from "it is not the case that all $x \in S$ have property $P$", why could you? And $\Delta \neq {0}$ in particular is a theorem of SDG, with the proof given above. – Z. A. K. Oct 24 '23 at 19:26
  • @Z.A.K. big thanks! Your explanations and answers are very useful! – Mike_bb Oct 25 '23 at 05:19