Talk:Classification of discontinuities

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Better to classify like this:

==1) Removable

==2) Essential ... which have the following sub classifications....

=2a) (Finite) Jump =2b) Infinite (Jump) =2c) Oscillatory

==1) means... limit exists.... but not equal to function value

==2) means....limit does not exist

=2a) limit does not exist because the left and right limits (which exist and are finite) are not equal =2b) limit does not exist because left and/or right limit is +/- infinity =2c) limit does not exist because of oscillatory action near point --anon

I don't think that a finite jump discontinuity is an "essential discontinuity". Do you have references for that? Oleg Alexandrov (talk) 13:59, 15 May 2006 (UTC)
(Finite) Jump Discontinuities are essential because the criteria to determine if a discontinuity is essential or not is whether the limit exists at the point. Of a function has a finite jump, the limit does not exist there (left and right limits are unequal), thus there exists an essential discontinuity there.
here is the first link I clicked after searching for essential disc http://oregonstate.edu/instruct/mth251/cq/Stage4/Lesson/jumps.html
Unfortunately, this use of essential clashes with the use of essential singularity in complex analysis. The same goes for the article, by the way. Another possible point of confusion with the classification in the article is that "not removable" and "non-removable" are not the same. -- Jitse Niesen (talk) 02:55, 17 May 2006 (UTC)

Confusion

This article was confusing continuous functions (which are continuous on their domains) with continuous funciton on the real numbers. So technically all the example functions are continuous if they are not defined at x=1. I have edited the examples so that they are all total functions, removing this confusion. I would like to make the jump discontinuity half-continuous but then I would have to edit the image. —The preceding unsigned comment was added by 2006 80.57.33.218 (talkcontribs) .

You are right of course. I did not think of that when i added the examples. Oleg Alexandrov (talk)
I would like to add that a jump discontinuity can be defined as a place where the left and right limits both separately exist, but are not equal. This is quite different from an 'essential' singularity, as the function Exp(-1/x^2) certainly has an essential singularity at the origin when thought of as a complex function, but I would argue that it has a removable discontinuity when thought of as a function of real x only. It is certainly discontinuous at x = 0, simply because it is not defined there. We cannot say that it is continuous there, so it's not continuous (discontinuous) there. I understand where the other editors are coming from when they say that discontinuity can only be defined at a point in the domain of the original function, but I think that this is splitting hairs a bit too much and can be confusing to the uninitiated reader. It is incorrect to classify a point at which the function is not defined, but for which the limit exists, as a 'removable singularity', as this term means that the limit exists from all COMPLEX directions. The function f(x) = \sqrt(x) for x > 0 and 0 for x < 0 is certainly undefined at zero, and the limit certainly exists (when thought of for only real x), but defining f(0) = 0 will not remove the singularity at x = 0. The function is still 'singular', but it will now be continuous. This is why I prefer the term 'removable discontinuity'. Any comments on this take? (from Jonathan Osborne)— Preceding unsigned comment added by Physicself (talkcontribs) 01:08, 4 May 2012 (UTC)

Removal of isolated point case

I removed the following paragraph by Patrick

Note that a function is continuous if it is continuous in all the points of its domain. If a continuous function is not defined in some isolated point, we can similarly distinguish the three cases. While above "removable discontinuity" means that we can make the function continuous at the point by changing the function value, if the function is not defined yet the corresponding case means that we can make the function continuous at the point by defining the function value.

The issue of whether the function is actually defined at ${\displaystyle x_{0}}$ is relevant only for removable discontinuities, and that case is already treated by the remark added by Patrick at that case. Repeating the same thing again in a separate paragraph does not make sense to me. Oleg Alexandrov (talk) 04:51, 28 May 2007 (UTC)

Never mind. I think Patrick has a point, although it took me a long time to understand what the point was from the text above. I tried to add something to that extent in the article, but stated differently. Oleg Alexandrov (talk) 05:16, 28 May 2007 (UTC)
After having slept on it, I believe if a function has a removable discontinuity at a point, it has to have a discontinuity there, so it has to be defined there. As such, it does not make sense to talk about things if the function is not defined at the given point, per the anon remark in the previous section. For that reason I removed again the case when the function is not defined at a point but is defined around it. Oleg Alexandrov (talk) 15:28, 28 May 2007 (UTC)
I agree that "make the function continuous at the point by defining the function value." was not correct, that should have been "defining the function value such that the function is continuous at this point". That is an interesting property (e.g. of (sin x)/x at x=0) that could be mentioned on this or another page.--Patrick 14:16, 29 May 2007 (UTC)

The last paragraph under the first section used to read like this:

The term removable discontinuity is sometimes (improperly) used for cases in which the limits in both directions exist and are equal, while the function is undefined at the point ${\displaystyle x_{0}.}$[1] This use is improper because, according to the precise definition of continuity of a function, the function is then actually continuous.

I changed the last sentence of this paragraph to read as follows:

The term removable discontinuity is sometimes (improperly) used for cases in which the limits in both directions exist and are equal, while the function is undefined at the point ${\displaystyle x_{0}.}$[2] This use is improper because the precise definition of continuity of a function applies only to points which are in the domain of the function.

As I understand it, the concept of continuity is defined only for points in the domain of the function, so for points outside the domain of the function it doesn't make sense to say that the function is either continuous or discontinuous. I changed the wording of the sentence to make this clearer, I think, and also because it is not necessarily true that a function is continuous if the limits in both directions exist and are equal and the function is undefined at the point. For example, consider the function ${\displaystyle f\colon \mathbb {R} \setminus \{0\}\to \mathbb {R} }$ defined by

${\displaystyle f(x)={\begin{cases}x,&{\mbox{if }}x{\mbox{ is rational}};\\0,&{\mbox{if }}x{\mbox{ is irrational}}.\end{cases}}}$

Then ${\displaystyle \lim _{x\to 0^{+}}f(x)=\lim _{x\to 0^{-}}f(x)=0}$, but the function is continuous nowhere in its domain. —Bkell (talk) 05:45, 10 October 2007 (UTC)

A common way of making functions total is to turn the codomain into a pointed set by adding one new element standing for "undefined", often represented by the symbol ⊥. If the codomain C is a topological space, then so is the disjoint union C ∪ {⊥} of C and {⊥} viewed as terminal object of Top. Then the "reparable holes" as for sin(x)/x become removable discontinuities. Specialized to the case of functions defined on real numbers, it is possible to define an operation that takes a function f to the function g defined by:
${\displaystyle g(x)={\begin{cases}y,&{\mbox{if }}\lim _{\xi \to x^{-}}f(\xi )=y=\lim _{\xi \to x^{+}}f(\xi );\\f(x),&{\mbox{if no two-sided limit exists}}.\end{cases}}}$
This operation will remove removable discontinuities and fix reparable holes. One would expect such an operation to have been discussed in the literature – although I'm unaware of any instances.  --Lambiam 11:25, 10 October 2007 (UTC)

Calling the use of the term removable discontinuity improper is POV. The use of the word discontinuity to refer to points outside the domain of a function is well-established. James Stewart gives the following definition:

If f is defined near a (in other words, f is defined on an open interval containing a, except perhaps at a), we say that f is discontinuous at a (or f has a discontinuity at a) if f is not continuous at a.

David Radcliffe (talk) 20:52, 3 May 2008 (UTC)

In my opinion it is improper to use a term for instances that are not covered by its definition. The function defined by f(x) = (sin x)/x is not defined in any open interval containing 0. Therefore the definition does not apply, and it is improper to state that the function has a removable discontinuity for x = 0. Why is that POV?  --Lambiam 12:22, 4 May 2008 (UTC)

I Googled "discontinuity" to see if there is any consensus on how it is defined, precisely because I am teaching a class using James Stewart's Calculus textbook right now, and the above definition is driving me crazy. Why? Because with that definition, a continuous function can have points at which it is discontinuous (for example, the function ${\displaystyle f(x)=1/x}$, defined for ${\displaystyle x\neq 0}$, is continuous, but it has a discontinuity at ${\displaystyle x=0}$, which is not in the domain of ${\displaystyle f}$). This is just plain confusing. I remember being confused by this issue when I was learning about continuity, and I refuse to subject my students to the same confusing terminology. Instead, I am going to borrow the complex analysis term "singularity" to describe the case (as in Stewart's definition above) where the function is defined in a deleted neighborhood of a point, but not at the point itself. I think the term "discontinuous" should be synonymous with "not continuous" -- this seems to be the common usage. Thus, as others have noted above, since continuity is only defined at points in the domain of the function, it should also only make sense to talk about "discontinuity" at points in the domain. 67.139.106.10 (talk) 02:35, 27 June 2011 (UTC)

A constructive suggestion

I agree with the above criticism of Stewart's text. Defining discontinuity of f at x for x not in the domain of f seem to indicate a lack of coherent language that allows to express properties of f without confusion. Rather than saying that function f, undefined at x, has a non-removable discontinuity at x, it would be more appropriate, and consistent with set theory, to say that there is no continuous extension of f over x. The latter means that whatever value will be assigned to x, the extended function will not be continuous at x.

To also cover the case when x is in the domain Dm(f) of f, one can say that:

"function f restricted to Dm(f)\{x} has no continuous extension over x."

This carries all the information of the phrase "f has a non-removable discontinuity at x" without confusion when f is a continuous function.

Removable discontinuity may be similarly replaced with the existence of a continuous extension.

I can speculate that someone (a colleague of mine suggested Richard Courant as the inventor of the above taxonomy of discontinuous at x with no regard whether x is in Dm(f) or not) interpreted "whatever value" as "not defined value", and then decided to include "undefined value" just to make it more general. — Preceding unsigned comment added by 76.171.57.52 (talk) 06:55, 29 May 2015 (UTC)

Can we define "x = a is a discontinuity of f(x)" by negating "f(x) is continuous at x = a" ?

The current article has examples of discontinuities, but no definition of the general phenomenon. Do we obtain the definition of "x = a is discontinuity of f(x)" by negating the definition of "f(x) is continuous at x = a"? If so, that would be a start in settling the controversies that arise about values x =a that are not in the domain of a function f(x).

We could analyze whether the definition of "f(x) is continuous at x = a" can only be satisfied by values of x in the domain of f. That would make any value not in the domain of f(x) automatically a discontinuity of f(x). However, I don't know if that conclusion would be harmonious with the assertion that discontinuities must fall into one of the three categories illustrated in the current article. For example, if we define ${\displaystyle f(x)={\sqrt {x(x-2)}}}$ to have a domain consisting of only those real numbers where of ${\displaystyle {\sqrt {(x)(x-2)}}}$ is a real number then which type of discontinuity would ${\displaystyle x=-1.3}$ be?

There is also the hair-splitting question of whether a discontinuity of a function of a real variable must be a real number. For example, my brother's red socks are not in the domain of a function of a real variable, but presumably my brother's red socks are not a discontinuity of such a function. Perhaps (in the case of real valued functions of a a real variable) the technicality is that "f(x) is continuous at x = a" is only defined for real numbers "a", so its negation is likewise only defined for real numbers "a".

Tashiro~enwiki (talk) 02:53, 1 March 2017 (UTC)

1. ^ See, for example, the last sentence in the definition given at Mathwords.[1]
2. ^ See, for example, the last sentence in the definition given at Mathwords.[2]