Definition of first order infinitesimal number using equivalence classes

99 Views Asked by At

I read about model construction in infinitesimal differential geometry (http://www.iam.fmph.uniba.sk/amuc/_vol-73/_no_2/_giordano/giordano.pdf , page 3-6)

I can't understand how equivalence class generated by $h(t)=t$ could be a first order infinitesimal number. How is it possible? What is main idea of this? How does it work?

$(1.1)$ $\forall h \in D: f(h) = f(0) + h \cdot f'(0)$

$(1.2)$ $\exists! m \in R: \forall h \in D: f(h) = f(0) + h \cdot m$

If we try to construct a model for $(1.2)$ a natural idea is to think our new numbers as equivalence classes $[h]$ of usual functions $h: R \to R$. In such a way we can hope both to include the real field using classes generated by constant functions, and that the class generated by $h(t)=t$ could be a first order infinitesimal number. To understand how to define this equivalence relation we can see $(1.1)$ in the following sense:

$(1.3)$ $$f(h(t)) \sim f(0) + h(t) \cdot f'(0)$$

If we think $h(t)$ "sufficiently similar to $t$", we can define $\sim$ so that $(1.3)$ is equivalent to:

$$\lim \limits_{t\to 0} \frac{f(h(t)) - f(0) - h(t) \cdot f'(0)}{t}=0$$

that is:

$(1.4)$ $$x \sim y :\iff \lim \limits_{t\to 0} \frac{x(t)-y(t)}{t} = 0$$

Thanks.