[Retric]

How would you turn f(x != 1) = 0, f(1) = (has no value) to be continuous?

Because it sounds like you’re simply redefining the terms. At which point you might as well be using “Spork”. Because, defining a new system has zero impact on a different system.

[Ericson2314]

1. You are talking about partial functions, that is a separate concern.

2. Behold https://en.wikipedia.org/wiki/Discrete_space . Topologies define continuinity, and here is a discrete topology.

3. With e.g. probability measures / expected values, which unify "discrete" and "continuous" statistics, you'll notice that there's lots of rules that are trivially obeyed in the discrete case, but take some care in the "continuous" case. For example, not ever set can have a measure in the latter but can in the former. This directly relates to discrete things being trivial to deem continuous. It's also a useful to define coarser topologies / event sigma-algebras in the finite case to better understand the issues are the unavoidable in the infinite cases. We only make the discrete discontinuous in that last "artificial" exercise.

[GolDDranks]

I believe that the function you just defined is already continuous. It's not defined at 1 so checking its continuity doesn't make sense there, and at every other point it's clearly continuous because it's constant. Points that are arbitrarily close to 1 are still continuous because the intervals around 1 are open.

[Retric]

In calculus f(x) = x/x - 1 is not continuous at 1 based on the initial definitions. Moving to fixed-point arithmetic the idea of limits gets odd, but the function still needs a definition for that input.

[GolDDranks]

I would say the continuity isn't just defined at that point... I very much agree that one can't affirmatively say that the function is continuous at that point. Btw. I guess you meant to say f(x) = (x - 1)/(x - 1).

However, let's ignore that; the problem disappears if we define f(x ) = 1 at x = 1, and f(x) = 0 elsewhere.

In standard analysis that function would be discontinuous at that point.

I'm not sure about implications in the OP's kind of analysis. Would the existence of m(1) imply that there exist some smallest input that is larger than 1, that would make difference in output? Same for the largest input that is smaller than 1.

[tel]

You handle continuity in discrete spaces through topology. A function is continuous for a (really, two) given topology if the inverse images of open sets are open sets.