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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.

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drhab
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    I think the question doesn't make sense. It sounds like asking if $\pi$ is even. – user2345215 Jan 01 '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 01 '15 at 15:45
  • For example, the definition of vertical asymptotes in real analysis: for $;x=k;$ to be a vert. asymp. of function $;f;$ it has to be that $; f;$ isn't defined at $;x=k;$ but it must be defined in some meighborhood (one sided or not) of that point, and also it must be at least one of the two sided limits of the function when $;x\to k;$ must be $;\infty;,;;-\infty;$ . – Timbuc Jan 01 '15 at 15:46
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    @user2345215 I think the question makes sense a lot. +1 – Timbuc Jan 01 '15 at 15:46
  • The books by Spivak, or Purcell-Varberg-Rigdon (and I can check others if wanted) either give specifically a definition from which can be understood that a point of non-definition can be considered a non-continuity point, or else use examples of this kind (e.g., the function $;1/x;$ not being continuous at $;x=0;$ because it isn't defined there). – Timbuc Jan 01 '15 at 15:56
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    @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 01 '15 at 16:33
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    @Siminore you are right: my above definition is incorrect, as the function can be perfectly well defined at $;x=k;$ yet one of the two one sided-limits there must be $;\pm\infty;$. Thank you. – Timbuc Jan 01 '15 at 16:38
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    Tibi gratulor, quod iam octoginta milia punctorum tulisti! :) – Joonas Ilmavirta Apr 27 '18 at 20:28
  • @JoonasIlmavirta Thank you very much for your kind words! "Favus mellis composita verba; dulcedo animae sanitas ossium" (Proverbs 16:24). – drhab Apr 28 '18 at 05:43

3 Answers3

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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.)

Hans Lundmark
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  • 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 01 '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 01 '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 01 '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 01 '15 at 16:36
  • @GPerez: Of course we can say something about the function's behavious near $0$; there's already plenty of terminology for that. For example, there's "pole" versus "essential singularity", or "can be extended continuously" versus "cannot". I don't see any need to define some new special meaning of the word "discontinuous". – Hans Lundmark Jan 01 '15 at 16:43
  • @Siminore there is no theory of discontinuous functions That can hardly be true; a lot of theories don't even need a topology. – GPerez Jan 01 '15 at 16:50
  • @HansLundmark I know there isn't a need, but the same lack of need applies to the initial question. As long we're dealing with a needless question, I wouldn't hesitate to give a needless answer. – GPerez Jan 01 '15 at 16:53
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    @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 01 '15 at 16:58
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    @Siminore Alright, alright, I guess I was a little bit nitpicky about your wording. I knew what you meant. – GPerez Jan 01 '15 at 17:07
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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.

ncmathsadist
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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.

Joonas Ilmavirta
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