Discontinuous function in Smooth infinitesimal analysis I read that there isn't discontinuous functions in Smooth infinitesimal analysis. But I tried to define discontinuous function ($\varepsilon$ is infinitesimal):
$f(x) = 
\begin{cases}
1,  & \text{if $x$ = $\varepsilon$} \\
0, & \text{if $x$ $\neq$ $\varepsilon$}
\end{cases}$
What is mistake in my definition?
Thanks.
 A: tl;dr It is not the case that all $x \in R$ satisfy either $x = \varepsilon$ or $x \neq \varepsilon$, so your function is not well-defined.
Let $R$ denote the real line of Smooth Infinitesimal Analysis. What you read is true: in Smooth Infinitesimal Analysis all functions from $R$ to itself are continuous and in fact differentiable everywhere. This is an easy consequence of the Kock-Lawvere axiom.
Now, the function $f$ you defined is a perfectly well-defined discontinuous function -- from the set $S = \{ x \in R \:|\: x = \varepsilon \vee x \neq \varepsilon \}$ to $R$.
However, $(\forall x \in R. x = \varepsilon \vee x \neq \varepsilon)$ is not a theorem of Smooth Infinitesimal Analysis for any $\varepsilon$: in fact, its negation is a theorem! This is fine, since Smooth Infinitesimal Analysis uses intuitionistic logic, where not all instances of the law of excluded middle ($A \vee \neg A$) are taken as axioms.
Since the negation of $(\forall x \in R. x = \varepsilon \vee x \neq \varepsilon)$ holds, we have $S \neq R$. So your function has domain $S$, and not domain $R$, and consequently its existence does not contradict the fact that all functions from $R$ are continuous.
