# Point of discontinuity

I have a function: $$f(x) = x$$ Defined over the domain $\mathbb{R} \backslash 0$. Is it correct to say that:

The function is continuous, but it has a point of discontinuity at $x=0$?

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No it doesn't. Why would $f(x)=x$ have a discontinuity? – nbubis Jan 21 '13 at 7:56
I can't imagine a better example, therefore I said there's f(x)=x with the domain R without 0. – TomDavies92 Jan 21 '13 at 7:57
I think it is a valid a good question to clarify subtleties regarding fundamental definitions. The choice of function is very good since it concentrates on the issue at hand: the question whether 'not defined at a point' means 'not continuous there'. – Ittay Weiss Jan 21 '13 at 8:04
@TomDavies92 - Edited the English, I think it makes more sense now. – nbubis Jan 21 '13 at 8:12

The function is continuous for all points where it is defined, which according to you is the set $\mathbb R - \{0\}$. It has no points of discontinuity. A point $x$ is a point of discontinuity for a function $f:D\to \mathbb R$ if the function is defined at that point but its value there is not the same as the limit. When $f$ is not defined at $x$ at all then $x$ can't be considered a point of discontinuity. Think of it this way: the function $f(x)=x^2$, defined on all of the real numbers, is not defined for $x$="the moon". Does it mean that $f$ is discontinuous at the moon?

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Ittay Weiss said "When $f$ is not defined at $x$ at all then $x$ can't be considered a point of discontinuity." What about $f(x) = 1/x$? $0$ is not defined at $f$, nevertheless, as long as I know $0$ is considered a point of discontinuity for the function $f(x) = 1/x$. It is enough for the function $f$ to be defined on some neighborhood of $x_0$ even though it is not defined at $x_0$. I think this point is a point of discontinuity since $\lim_{x\to 0+} = \lim_{x\to 0-} = 0$, but $f(0)\neq 0$ (undefined by definition). – fade2black Aug 20 '13 at 22:34
It is true that $f(x)=1/x$ is not continuous at $x=0,$ since it isn't defined there. However, this does not mean that $f$ has a discontinuity at $x=0.$ See here, especially: "The term removable discontinuity is sometimes used by abuse of terminology...." – Cameron Buie Aug 21 '13 at 4:51
Then what about the example $1/x$, where $0$ is not in the domain of $f$? My text book on calculus says that the point 0 is a non-removable point of discontinuity (Calculus, 9th Ed., Ron Larson, pg. 71, Example 1 (a), function $f(x)=1/x$). – fade2black Aug 21 '13 at 8:15
@fade2black: Again, that is not a discontinuity. It is what is known as a pole. Many textbooks would call that a discontinuity, but again, this is an abuse of terminology. Continuity and discontinuity are defined only for points of the function's domain. I suspect that your textbook did not give a precise definition of what continuity at a point means, because otherwise, they couldn't claim that $0$ was a point of discontinuity of $f(x)=1/x$. – Cameron Buie Aug 21 '13 at 13:56

No, the continuity or discontinuity of a function at a point is only defined if the point is in the domain. The function is continuous at every point of its domain, which was stipulated to be $\mathbb R\setminus \{0\}$. It is not defined at $0$.

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What about limit points? I heared, that the point of discontinuity is also a limit point, that does not belong to the domain. Is it true? – TomDavies92 Jan 21 '13 at 7:58
I do not understand the question. – Jonas Meyer Jan 21 '13 at 8:01
@TomDavies92: It is depend on what kind of discontinuty you faced at that point. – S. Snape Jan 21 '13 at 8:02
@TomDavies92: But perhaps you are thinking about something like the following. If $f:\mathbb R\setminus\{0\}\to\mathbb R$ is a function, then when is it possible to extend $f$ by defining a value $f(0)$ in such a way to make the extended function continuous at $0$? This could be answered by checking whether or not $\lim\limits_{x\to 0}f(x)$ exists. – Jonas Meyer Jan 21 '13 at 8:04

If $x=0$ is not part of the domain of the function, there is no sense in talking about the properties of the function at that point - continuity or anything else.

It happens that the domain on which this function is defined is a part of a larger domain - it is a fundamental issue in mathematics to identify the possibility of extending functions to such larger domains in "good" ways - either preserving useful properties like continuity, or acquiring new ones - like connectedness, compactness or roots to specific equations.

Your $f$ could be extended to $\mathbb R$ by defining $f(0)=\pi$. If you want to preserve continuity, however, you need to define $f(0)=0$.

You might consider the function $g(x)$ defined on the same domain with $g(x)=-x$ when $x$ is negative, and $g(x)=x$ when $x$ is positive. Defining $g(0) = 0$ keeps $g$ continuous, but it is no longer differentiable over the whole domain of definition. Sometimes there is a trade-off between the properties you want and the domain you choose.

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First, I agree with most other answers: What you say about your $f$ is not totally right but $f$ is continuous where is defined.

I would like to add, that probably some confusion is caused by the fact that there is an obvious extension to a superset of its domain of definition. (Here, set $f(0)=0$, obviously.)

Indeed, the problem if a continuous function defined on some set has a continuous extension to a larger set is very important in some places of mathematics (e.g. if a continuous linear operator on a Banach space can be extended to a larger Banach space in which the former one is embedded...). Another example is the theorem that a continuous function defined on a dense set of a metric space has a continuous extension to the whole space.

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