Discontinuous function on $\Delta$ in Smooth Infinitesimal Analysis

Question

As is known, there isn't discontinuous functions in Smooth Infinitesimal Analysis.

I tried to define discontinuous function on $\Delta$:

$f(x) = \begin{cases} 1, & \text{if $x$ = $0$} \\ 0, & \text{if $x$ $\in$ $\Delta$ \ {0}} \end{cases}$

Where did I go wrong?

Thanks.

Answer 1

Intuitionistic logic does not allow you to carve up a set into a union of "one point" and "all the rest": that requires excluded middle.

The good news is that there is a map $f$ that satisfies your definition in SDG. But its domain is not $\Delta$; instead, it's $\{0\} \cup (\Delta \setminus \{0\})$.

We can show that $\Delta \setminus \{0\}$ is empty, since $x \in \Delta \setminus \{0\}$ precisely if $x \in \Delta$ and $x \neq 0$, but no infinitesimal satisfies $x \neq 0$.

So the domain of definition of your function $f$ is the object $\{0\}$. And of course $f$ is continuous on $\{0\}$.

Now, for the bad news: you'll keep being confused about synthetic differential geometry in very elementary ways until you decide to put in hard work and learn intuitionistic logic properly. And while I'm normally very happy to answer elementary queries about SDG, we're reaching a point where, without progress, asking questions like these becomes a bad use of your (and my) time. My recommendation is to pick up the logicmatters.net study guide, fill in any gaps you can identify in your logic background, learn enough intuitionistic logic in enough detail that you can follow constructive treatises on undergraduate topics such as basic Bishop-style analysis, and then return to SDG. This is not fun, but it's much easier in the long term: there are currently no resources that successfully teach SDG to somebody who doesn't already know intuitionistic logic. I'm not saying such resources are impossible (they're not), but as of 2023, they simply don't exist.