Is function $f:\mathbb C-\{0\}\rightarrow\mathbb C$ prescribed by $z\rightarrow \large \frac{1}{z}$ by definition discontinuous at $0$?

Question

Is function $f:\mathbb C-\{0\}\rightarrow\mathbb C$ prescribed by $z\rightarrow \large{\frac{1}{z}}$ by definition discontinuous at $0$?

Personally I would say: "no". In my view a function can only be (dis)continuous at $z$ if $z$ belongs to its domain.

But I have heard other sounds, that made me curious.

This question was inspired by comments/answers on this question.

Answer 1

For a function $f$ to be continuous at a point $a$, you must have $a\in\text{dom}(f)$. The function you cite is continuous on the punctured plane.

Answer 2

I suspect that there is no universal agreement among different sources. But for example Rudin's Principles (p. 94) says "If $x$ is a point in the domain of the function $f$ at which $f$ is not continuous, we say that $f$ is discontinuous at $x$, or that $f$ has a discontinuity at $x$". He doesn't mention anything about points not in the domain of $f$, but this omission sort of implies that for such points neither of the terms continuous or discontinuous should be applied.

I think this practice makes a lot of sense, since your example function is continuous (being continuous at all points in its domain), and allowing continuous functions to have discontinuities would be strange, wouldn't it? (Singularity is a better word in such a case.)