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

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.

• I think the question doesn't make sense. It sounds like asking if $\pi$ is even. – user2345215 Jan 1 '15 at 15:43
• @drhab I'd say a function is discontinuous at z$\;z_0\;$ if it is defined in some (left, right or two-sided) neighbourhood of $\;z_0\;$ but either it isn't defined at $\;z_0\;$ or else at that point the function doesn't fulfill the condition of continuity. My reason to think this way is the following: – Timbuc Jan 1 '15 at 15:45
• @user2345215 I think the question makes sense a lot. +1 – Timbuc Jan 1 '15 at 15:46
• @Timbuc I think that even if $x=k$ belongs to the domain of $f$, the line $x=k$ can be a vertical asymptote. – Siminore Jan 1 '15 at 16:33
• Tibi gratulor, quod iam octoginta milia punctorum tulisti! :) – Joonas Ilmavirta Apr 27 '18 at 20:28

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.)

• I'm speaking a bit out of my league here, but maybe we could decide to say (formally) that $1/z$ has a discontinuity at $0$ if it is a discontinuity in $\Bbb C\cup \{\infty\}$ (and I think it isn't)? That way it's distinguishable from functions like $e^{1/z}$. – GPerez Jan 1 '15 at 16:11
• @GPerez: You have to say what the domain of your function is. In the question asked here, the domain is the punctured plane. You can define another function $g(z)=1/z$ from the Riemann sphere (including zero and infinity) to itself; this function $g$ is continuous (analytic, even) on the Riemann sphere, but that's a whole different question. – Hans Lundmark Jan 1 '15 at 16:17
• I'm aware of this, which is why I say it'd be a purely formal notion. Obviously if we only define continuity for $f:X\to Y$ at a point $x\in X$ as [$f$ is cont. at $x\in X$ if ...], then it makes no sense to speak of points not in $X$ anyway. I'm just saying, it's not uncommon to look at things in a "bigger" domain to decide things about the "smaller" one, and this could help here if only for having $something$ to say about $0$, albeit something ultimately useless. – GPerez Jan 1 '15 at 16:29
• Here in Italy, many colleagues teach that $f$ has a point of discontinuity at $x_0$ if it cannot be extended at $x_0$ continuously. My opinion is that this is fairly acceptable for functions of a single real variable, but it becomes useless in a general context. In higher mathematics continuity matters, and there is no theory of discontinuous functions. – Siminore Jan 1 '15 at 16:36
• @GPerez Come on, I mean that there is no branch of mathematics that studies only discontinuous functions. In general topology we hardly define or classify points of discontinuity. I guess it is a good chapter for calculus student, but not much more that this. – Siminore Jan 1 '15 at 16:58

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.

• @nc I think this is exactly drhab's point. – Timbuc Jan 1 '15 at 15:47
• This doesn't answer the question, which had to do with the word discontinuous. – Hans Lundmark Jan 1 '15 at 16:02
• More precisely, the question is: does "discontinuous at $x$" mean simply "not continuous at $x$" or rather "defined at $x$ but not continuous at $x$"? – Hans Lundmark Jan 1 '15 at 16:05
• A function can't have continuity or discontinuity where is is not defined. – ncmathsadist Jan 1 '15 at 20:36

Typically continuity or lack thereof is restricted to points on the domain where the function is defined, but it can also be reasonably extended to your problem. The question is not "is $f$ continuous at zero?" but "can $f$ be continuously extended to zero?", and this makes perfect sense.

Since we are talking about extension to a single point, the question is whether $f$ has a limit at that point. The answer depends on the target space. If you want $f$ to map to $\mathbb C$, there is no limit. If you want it to map to the Riemann sphere $\mathbb C\cup\{\infty\}$, then there is. In fact, the mapping $z\mapsto1/z$ is a bijection of the Riemann sphere to itself.