# Talk:Continuous function

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## Old comment

I removed this:

There are two types of discontinuity. A removable discontinuity can be drawn as if it were continuous, should one redefine f(c). A nonremovable discontinuity cannot be drawn as a "continuous line", even if f(c) is redefined. For example: if f(x) = (x² - 1) / (x - 1); then, f(x) has a removable discontinuity at x = 1 (which can be removed by defining f(1) soas to equal the limit of f(x), which is 2). On the other hand, if f(x) = 1 / x; then, f(x) has a nonremovable discontinuity at x = 0. (see also: division by zero).

The two given functions are continuous. This seems to be more relevant to removable mathematical singularity. AxelBoldt 19:49 20 Jun 2003 (UTC)

## More continuity issues

If two functions f and g are continuous, then f + g and fg are continuous. If g(x) ≠ 0 for all x in the domain, then f/g is also continuous.

Even if g(x) is zero in a few points (a few == not all), (f/g)(x) can be defined for the points where g(x) ≠ 0. Won't this function be continuous too ?

The zeroes of g can be any closed set, however complicated. It is better to say something simple, and true. Charles Matthews 18:53, 18 May 2004 (UTC)

There seems to be some inconsistency between the definition given and properties on the one hand, and the examples on the other. Is the tangent function, which presumably quilifies as a trigonometric function, to be considered continuous? I'm not sure what to say about logarithm. Josh Cherry 02:25, 3 Dec 2004 (UTC)

I also noticed trouble with the tangent (and secant, cosecant, cotangent). Unfortunately, the page on trig functions deals with the domains of the inverse functions, but not of the direct ones. I have not checked the logarithm for domain - obviously we must exclude zero and we must refer to branches because the article purports to refer to the complex domain. Pdn 01:25, 26 July 2005 (UTC)

Yes, f/g is a continuous function, regardless g(x) is zero or not. It's just that it is not defined wherever g(x) = 0. And no, if f(x) = 1 and g(x) = x then f/g cannot be defined at g(x) = 0 in a way that would make the extended function a continuous function.

Let's see if cotangent is continuous. If it is not then there must exist an element p of its domain at which it is not continuous, that is, lim _{x -> p} cotan(x) != cotan(p). So, because such p does not exists (points of the form k*pi where k is an integer are not in the domain of function cotangent), cotangent is continuous. And so is tangent. — Preceding unsigned comment added by 76.171.57.52 (talkcontribs)

Please sign your posts with four tildes ~~~~. Also, I have moved these posts to the bottom. The items you are replying to are more than 10 years old, and your replies were interspersed into a very old conversation, breaking it up in a way that the original participants would not have intended. Sławomir Biały (talk) 14:11, 28 May 2015 (UTC)

## Continuity and the domain of the function

There appears to be some confusion between whether a function is continuous and whether it can be extended to the reals. Continuity depends only on the domain of the function, so 1/x is continuous: it is not defined at x=0 and so continuity there does not arise. The same is true for some of the trigonometric functions and real log. Complex log is not a function unless you take a principal branch. The question as to whether a function can be continuously extended to the reals is obviously related, but it is not the same. --JahJah 19:41, 18 August 2005 (UTC)

A quick followup to my own comment: where did the part about accumulation points of the domain come from in the definition? This is not in accord with current terminology. --JahJah 19:56, 18 August 2005 (UTC)

If the domain has no accumulation points, then every function on it is continuous; for example, every function whose domain is the set of all integers is continuous, because a variable x taking values in the set of all integers except c cannot approach c. Certainly it is in accord with standard usage, but the definition can be rephrased so that this need not be mentioned explicitly until we get into corollaries and such. Michael Hardy 21:02, 18 August 2005 (UTC)
... and now I've rephrased it. Michael Hardy 21:16, 18 August 2005 (UTC)
I still disagree with this. If you take the integers with the subspace topology from the reals, they have the discrete topology, and so every function defined on the integers is continuous. Continuity is an intrinsic property: it does not depend on the domain or codomain being included in any other space. This is the definition used for metric spaces (including the reals and subsets thereof) as well as for topological spaces in general. A discontinuity can mean several things, but bringing it into the definition like this is, in my opinion, a mistake in terms of exposition as well as being mathematically inaccurate. --JahJah 21:44, 18 August 2005 (UTC)
You're not reading carefully, I think. The article did not say, either before or after my rephrasing, that continuity depends on extendability to a continuous function at an accumulation point not in the domain. It said, about a point c that IS IN THE DOMAIN, that if it is an accumulation point, then a certain condition is necessary for continuity. Of course, that condition is still necessary if c is not an accumulation point, but then it holds vacuously, so it's not necessary to include it explicitly in the definition. Michael Hardy 22:07, 18 August 2005 (UTC)
Michael, my apologies: I was looking at an old version. I agree with the definition you have given, although I would still question putting accumulation points up front. Does this mean it is now OK for me to add rational functions and inverse trigonometric functions as examples of continuous functions without it being immediately reverted? I think it is important to convey to a novice that elementary functions are continuous. --JahJah 22:24, 18 August 2005 (UTC)

Sorry to be a pest on this, but I have found this a great source of confusion with students learning calculus. I am sure we all know what continuity means: the problem is explaining it to a Wikipedia user, who is presumably a novice. It is also clear from the page history and talk page that contributors are confused. In my opinion, continuity expresses the idea that a nearby point is mapped to a nearby point, and this idea gives enough for the definition for either metric spaces or topological spaces. Talking about points of accumulation in the domain when giving the definition does not help to convey this. Recent calculus textbooks avoid this by only defining continuity for functions defined on nontrivial intervals, but this seems unnecessarily restrictive. Similarly, talking about functions being continuous in their domain suggests the possibility that they may be continuous outside their domain, and leads to speculations that 1/x or trigonometric functions are not continuous. Older definitions of continuity have certainly encompassed this possibility, but this is not what the modern idea of continuity means. Perhaps the problem is that the idea of the domain of a function given by a formula is not clear (for example, many students think that x/x=1). I have no wish to argue about what continuity means, only to express it clearly and correctly to the intended audience. --JahJah 16:20, 19 August 2005 (UTC)

Yes, I think that the problem is that the idea of the domain of a function given by a formula is not clear. When you say "tan (x) is continuous", most people will imagine that function which is periodic with huge jumps at π/2+kπ, and they will say that the thing in question can't be continuous. So, while the definion of continuity if of course about what happens in its domain, when going to concrete examples given by analyitic expression, never hurts to mention its domain to start with. Oleg Alexandrov 18:45, 19 August 2005 (UTC)
Maybe you are right. However, with the elementary functions listed, there should be no doubt about their domains. If someone imagines tan(x) to not be continuous, they have not quite got it: they are applying continuity to an extension. This misconception recurs in this page and the editing history, and including "in their domain of definition" may encourage it. --JahJah 19:31, 19 August 2005 (UTC)

Let us see if there are other opinions on the topic. Oleg Alexandrov 03:53, 20 August 2005 (UTC)

I'm not sure yet. "If someone imagines tan(x) to not be continuous, they have not quite got it." So we have to make it clear to them that they haven't understood it yet. I moved tan(x) etc. below the f(x) = 1/x example and added a sentence to that example stating explicitly that f(x) = 1/x is not continuous at x = 0 because that's not part of the domain. How does that sound? -- Jitse Niesen (talk) 12:11, 20 August 2005 (UTC)
Draft wording that conveys my point: It is important to remember that continuity depends only on the domain of the function. For example, if f(x)=1/x, the domain of f is R-{0}, and f is continuous. The question as to whether f is continuous at 0 does not arise, since 0 is not in the domain of f. The questions (a) is f continuous? (b) can f be extended to a continuous function on a larger domain? are different. --JahJah 12:49, 20 August 2005 (UTC)
Sounds good. Quick reaction: Perhaps add a function with a removable singularity? This would drive the point that there is a difference between (a) and (b) home. -- Jitse Niesen (talk) 13:51, 20 August 2005 (UTC)
Two examples. The simplest is x/x, but it is liable to misinterpretation given the current context. Another is (sin x)/x, but this is a bit more complicated. --JahJah 15:32, 20 August 2005 (UTC)

## Heine definition of continuity

It was pointed out by Jitse Niesen and Oleg Alexandrov that my changes regarding the Heine definition of continuity were erroneous (see User_talk:Kompik#Limits_and_AC). I tried to make (at least some of) changes which seemed to be appropriate according to this talk. Perhaps someone will add more on this topic. BTW analogous concept called Heine definition of limit appears in some textbooks. I'm not sure which one of these two notions is more common. --Kompik 07:08, 30 September 2005 (UTC)

## Absolute value function continuous at x=0 ?

This article claims that f(x) is continuous but not differentiable at x=0. How is f continuous if lh limit is not equal to rh limit? Once again, another erroneous statement arising from a subject filled with inaccuracies - real analysis. 158.35.225.229 15:25, 23 November 2005 (UTC)

But the limit on the left does equal the limit on the right: both are zero. -- Jitse Niesen (talk) 16:11, 23 November 2005 (UTC)
Jitse, if you want more of that, see Talk:Mathematical analysis :) Oleg Alexandrov (talk) 20:00, 23 November 2005 (UTC)
Sorry, I'm already engaged at Talk:Proof that 0.999... equals 1 :) -- Jitse Niesen (talk) 21:08, 23 November 2005 (UTC)

## Points are small

If between every two points you can draw another one, isn't every function discontinuous? Just curious... --anon

No, as long as your curve does not jump too much between those two points. Oleg Alexandrov (talk) 19:26, 6 December 2005 (UTC)

Suppose you have this function: _/\_ Is it continuous? The points in the top never jump too much. --anon

Yes, it is continuous, but not differentiable. Intuitively, a function is continuous if you can draw its graph without lifting the pencil from the paper. Oleg Alexandrov (talk) 18:59, 22 December 2005 (UTC)

## No, really, really amazingly small. You can't imagine how small points are.

Sorry about that. Rick Norwood 21:40, 23 February 2006 (UTC)

## An issue with epsilon-delta definition

In the paragraph that starts with "Alternatively written", what if one chooses a c such that c is in R (reals) but not in I (i.e. c is not in the domain of f)? Would "| f(x) - f(c) | < ɛ" then be a valid statement? --anon

Fixed now, thanks! Oleg Alexandrov (talk) 03:58, 13 March 2006 (UTC)

There are still a couple of problems, one mathematical and one typographical. The typographical problem is the fact that delta is so much smaller than epsilon. Looks funny and I don't know how to fix it.

The mathematical problem is this. Suppose I = the set of rational numbers and f(x) is 0 for all x in I. According to this definition, f is continuous at any rational number. I think you need to let I be an interval. Rick Norwood 15:19, 14 March 2006 (UTC)

In that case, f would be continuous on the space of rational numbers, but could still fail to be continuous at every rational number when viewed as a function defined on the larger space of all real numbers. Michael Hardy 22:10, 14 March 2006 (UTC)

I just realized that if the first sentence of the intro is taken literally, a constant function isn't continuous (because a chance in the input produces no change in the output. I'll going to try to fix that. Rick Norwood 21:49, 16 March 2006 (UTC)

## sup?

What is sup, for example on the main page: sup f(Y)=f(sup Y)

A search for sup in the wiki turns up nothing relevant. A search for sup at mathworld.com finds nothing. A page on TeX (http://cs.nyu.edu/~yap/student/LatexBasics.html) seems to indicate that this might be faulting typesetting.

I also see this notation on the Bernstein_polynomial page, where it seems to be used as a set operator. A search of my memory on set operations only turns up "superset", which does not seem to shed any light in these contexts.

I've just added a link to it; see supremum. It's a universally standard usage in mathematics. Michael Hardy 18:00, 31 July 2006 (UTC)

## Notation

I was recently reading an (non-WP) article that uses the notation ${\displaystyle C^{(k,\alpha )}}$ with ${\displaystyle 0<\alpha <1}$ to denote some sort of continuity. The paper failed to describe the meaning of this notation. From the context, it was clear that it was kind-of-like ${\displaystyle C^{k}}$ viz. the set of k-times differentiable functions, but ... what's the precise definition? linas 15:06, 27 August 2006 (UTC)

See Hölder condition, though of course I can't guarantee that that's the definition used in your paper. -- Jitse Niesen (talk) 03:10, 28 August 2006 (UTC)

## "right" continuous

I've been searching all over the place for what the hell "right-continuous" means. Why isn't there something on this term here? Fresheneesz 01:41, 21 March 2007 (UTC)

Problem resolved. I created the section "directional continuity". Fresheneesz 03:09, 21 March 2007 (UTC)
Isn't this identical to semicontinuity? I've appended a note to see semi-continuity. --TedPavlic 20:43, 8 April 2007 (UTC)
It's close, but not the same. The function
${\displaystyle f(x)={\begin{cases}0&{\text{if }}x\leq 0\\{\frac {\sin x}{x}}&{\text{if }}x>0\end{cases}}}$
is right-continuous, not left-continuous, not upper semi-continuous and not lower semi-continuous. See topologist's sine curve for a picture of the function for x > 0. -- Jitse Niesen (talk) 03:38, 9 April 2007 (UTC)
The definition of directional continuity here only applies to functions of a single real variable. I assume it generalizes to Euclidean space by taking ${\displaystyle c_{i}. Does it also generalize to non-Euclidean lattices? LachlanA (talk) 17:59, 12 February 2008 (UTC)

## Merger

I think the page on Newton's notation for continuous should be merged into this one. I don't think there's enough to it to warrant a separate page. Perhaps it can be slipped into a paragraph on historical background.

Strongly agree. Rick Norwood 14:39, 29 July 2007 (UTC)
I merged that into fluxion, I thought it was better. Or one could merge it even to Method of Fluxions, where there is more context. Oleg Alexandrov (talk) 15:29, 29 July 2007 (UTC)

## Is this correct?

I have an issue with the second paragraph:

"An intuitive though imprecise (and inexact) idea of continuity is given by the common statement that a continuous function is a function whose graph can be drawn without lifting the chalk from the blackboard."

If you consider "drawable" curves of this form, at every point on the curve, you know in which "direction" to move the pen to continue drawing, i.e. you know the tangent at every point. So isn't the fact that you can draw a curve making a stronger assumption, namely that the curve is in fact differentiable (of course, therefore continuous).

Polyrhythm 06:05, 11 August 2007 (UTC)

The distinction is that you can draw a cusp (like a hat that comes to a point at the top) which is continuous but non-differentiable. Rick Norwood 12:44, 11 August 2007 (UTC)
Yep, your pen can change direction suddenly, so the function is not necessary differentiable. Oleg Alexandrov (talk) 16:44, 11 August 2007 (UTC)

I take issue with the tone of the sentence "An intuitive -- though imprecise, inexact, and incorrect -- idea", what the hell is this doing in an encyclopedia? It sounds like somebody is critisizing someone else's explanation but just leaving it there to be pompous. If the user who changed the sentence to this feels the analogy is so inexact, then they should either remove it or offer one they feel is more appropriate. Imagine the reaction of someone with no knowledge of maths reading this article and then reading this sentence, what are they to take from this? Is there any point in reading this analogy? Larryisgood (talk) 13:21, 7 September 2010 (UTC)

This analogy is commonly used in introducing the idea of continuous functions. Mathematical ideas are purely abstract, so analogies are imperfect. A physical triangle is an imprecise and inexact analogy of a mathematical triangle. But it would be hard to understand triangles without pictures. So a curve drawn with chalk (discontinuous at the atomic level) helps the student understand continuity. Rick Norwood (talk) 15:17, 7 September 2010 (UTC)
I think I may not have been clear. I'm fine with the analogy, it seems that the article undermines itself by including the sentence "An intuitive -- though imprecise, inexact, and incorrect -- idea". If it was up to me I'd just leave it as it is saying its an imprecise analogy, rather than tell explain to the reader the analogy after stating that it was incorrect.Larryisgood (talk) 11:12, 8 September 2010 (UTC)

## Heine or Heine's

Which one is correct? Heine definition or Heine's definition of limit/continuity. The same with Cauchy. (Not being a native speaker, I'm asking for advice.) --Kompik (talk) 19:53, 23 January 2008 (UTC)

Both are. In the first, Heine is an noun adjunct; in the second, Heine's is a possessive determiner. In sentences, the first form takes the definite article, the: the Heine definition, while the second does not; but in headings the definite article is omitted. EdC (talk) 00:12, 24 January 2008 (UTC)

They are used differently. If you refer to "the Cauchy criterion" you include the word "the"; if you refer to "Cauchy's criterion, then you don't. Michael Hardy (talk) 00:13, 24 January 2008 (UTC)

## Continuous functions as integrands

I removed the following text:

Let ${\displaystyle ~f~}$ and ${\displaystyle ~g~}$ be real continuous functions defined at the real segment ${\displaystyle [a,b]}$. Then, there exist ${\displaystyle ~y\in [a,b]~}$ such that
${\displaystyle \int _{a}^{b}f(x)g(x)~{\rm {d}}x=f(y)\int _{a}^{b}g(x)~{\rm {d}}x~.}$
In the case of a complex function, such a theorem would not be valid: for example, if ${\displaystyle f(x)=\exp({\rm {i}}x)}$, ${\displaystyle g(x)=1}$, ${\displaystyle a=-\pi }$, ${\displaystyle b=\pi }$, then the integral in the left hand side of the eqiation is zero, and that in the right hand side is equal to ${\displaystyle ~2\pi ~}$, but at the real axis, there is no point ${\displaystyle ~y~}$ such that ${\displaystyle \exp({\rm {i}}y)=0}$.

I think this theorem is not important enough to be included on this page. The intermediate and extremal value theorems are taught in every analysis course, while I don't remember seeing this theorem. -- Jitse Niesen (talk) 12:35, 17 March 2008 (UTC)

## Why not use this cleaner, more elegant notation for continuity?

Suppose we have a function f from X to Y. We say f is continuous at x for some ${\displaystyle x\in X}$ if for any open set ${\displaystyle V\subseteq Y}$ containing f(x), there is an open set ${\displaystyle U\subseteq X}$ containing x such that ${\displaystyle f(U)\subseteq V}$. 128.210.146.26 (talk) 19:00, 3 October 2008 (UTC)

That's not a notation; that's an equivalent definition. Which definition should be used depends on context. If one is addressing the broadest possible audience one should start with one that that broad audience will understand. Michael Hardy (talk) 19:43, 3 October 2008 (UTC)

This definition is used in the article. Or do you mean, why not use this definition from the start? Well, I think it's more abstract and thus harder to understand, that's why. -- Jitse Niesen (talk) 20:13, 3 October 2008 (UTC)

The problem with using the topological definition of continuity is that you then have to explain what an open set is, and that's too abstract for beginners. Not that epsilon delta is easy. Rick Norwood (talk) 12:15, 4 October 2008 (UTC)

In the context of functions from reals to reals it is sufficient to talk about intervals. Most people - also calculus beginners - understand "open interval". Consider the following:
A real-valued function f is continuous at the point c if
- f is defined on some open interval (a,b) containing c, and
- every open interval containing f(c) contains the image of some open interval containing c.
In this definition, the "image" of a neighbourhood N, f(N), contains the values f(x) for all points x in N.
A function that is continuous at every point in some set is said to be continuous on or in the set. A function that is continuous on its domain is called just continuous.
This removes some of the the mental and typographical clutter that beginners experience with the usual epsilon-delta formulation. Epsilon and delta are used only to define open intervals (although implicitly, through the expression |xc| < δ or c − δ < x < c + δ). The circumstance that these intervals become symmetric about c resp. f(c) has no consequence for the continuity and need not be emulated. For beginners, "open interval containing c" is much easier to grasp than |xc| < δ, and "set S contains set T" is much easier than "[formula for x is in S] whenever [formula for x is in T] for every x in the domain of f". The epsilon-delta formulation has three nested quantifiers: "for every epsilon, there is a delta, such that for every x in ...". The more modern definition do not entirely remove these quantifiers, but express them using sets as vehicles. The sets reify the relationships in a way that most people are better prepared (even though often still not well prepared) to handle.
The intervals need not even be open, if instead it is said that they must contain whatever as interior points or non-boundary points. One can rely on the everyday understanding of "interior" here, and bypass the definitions of a topology, etc. Another possibility is to invent a concept "surrounds", like "In the following definition we shall say that and interval surrounds a point x if it extends on both sides of x, i.e., it contains x but not as an endpoint." Or, what do ya think about this:
A real-valued function f is continuous at the point c if
- f is defined on a neighbourhood of c, and
- every neighbourhood of f(c) contains the image of some neighbourhood of c.
For the purposes of this definition it is enough to take "neighbourhood of a point" to mean some interval that has the point in its interior (not at a boundary). The image f(J) of some set J consists of the values f(x) of all points x of J.
Another variation rewrites the second condition to
- every neighbourhood of f(c) contains the values f(x) corresponding to all points x in some neighbourhood of c.
With either variant, maybe the definition could be accompanied with something like (please improve):
Less formally, the effect of this definition is that if one considers the values f(x) as x slides through its possible values, x cannot reach c without first reaching other points in the neighbourhood of c. These points have values f(x) "near" f(c) -- in a neighbourhood of f(c). The definition specifies "every neighbourhood of f(c)" so that this can be the case for every degree of "near". If you want a stricter definition of "near", consider a smaller neighbourhood, and this definition guarantees that f(x) still must reach some values in this smaller neighbourhood before reaching f(c) proper. No matter the strength of your magnifying glass, you will not see any sudden jumps to f(c). The approach is infinitely gradual.
Cacadril (talk) 20:25, 19 October 2009 (UTC)

## last example

is it correct that the last example [ f(x)=0 for x element Q and f(x)=1 otherwise ] is continuous at x=0 ? Danielschreiber (talk) 20:18, 26 November 2008 (UTC)

You've misstated the example. The last example is "f(x)=x for x in Q and f(x)=0 otherwise", which is indeed continuous at 0. But your "f(x)=0 for x in Q and f(x)=1 otherwise" is nowhere continuous. --Zundark (talk) 21:40, 26 November 2008 (UTC)

## Twice continuous

when people say twice continous, what does it mean? the first derivative is continuous, or the second derivative is contiuous? This should be mentioned in the article. Jackzhp (talk) 03:14, 22 February 2009 (UTC)

I'm a mathematician, and I've never run arcoss the phrase "twice continuous". Can you cite an example. Rick Norwood (talk) 13:43, 22 February 2009 (UTC)
I guess "twice continuous" actually is an inexact recollection of the first part of "twice continuously differentiable". It means having a continuous second derivative. Cacadril (talk) 16:08, 19 October 2009 (UTC)

## Closed sets

If f is a continuous function, is the image of f on a closed set closed? —Preceding unsigned comment added by Standard Oil (talkcontribs) 12:46, 8 April 2009 (UTC)

No. Consider the canonical injection of a non-closed subset of a topological space. It is continuous, the whole space of the domain is (always) closed, but by construction the image is not closed. Cheers, Stca74 (talk) 15:14, 8 April 2009 (UTC)
So you mean find a non-closed subset S of a space (X,T), get its subspace topology Ts, and because (S,Ts) is a closed subset of itself, the continuous function f(x)=x gives the non-closed image set S in T? Thanks, saved me the trouble of trying to prove something that's wrong :O.Standard Oil (talk) 05:58, 9 April 2009 (UTC)

## Examples from the real world and physics

I question some of the assertions made in the third paragraph of this article. How is it known that the height of the flower is a continuous function of time? Perhaps it may be assumed to be for the sake of simplicity in the macroscopic world, but that is another issue.

My understanding is that the physicists learnt the hard way (through erring on the assumption of continuity wrt the problem of black body radiation (which in part led to quantum theory where this so-called dictum of physics breaks down completely (does it not?))). Please disabuse me of any misunderstandings I may have about this issue.Aliotra (talk) 18:18, 16 September 2009 (UTC)

Okay, I see my error; it says "classical" physics.Aliotra (talk) 15:51, 17 September 2009 (UTC)

## Missing notions of continuity

Several "professional" notions of continuity should at least be mentioned in passing. Approximate continuity as treated in Federer: Geometric Measure Theory, Point 2.9.12 is one such example. —Preceding unsigned comment added by 141.35.15.251 (talk) 21:37, 7 April 2010 (UTC)

## Redundance

There's quite some stuff twice on the page, even literally quoted (e.g. the definition for continuity between topological space). So I've added a cleanup suggestion. -- Roman3 (talk) 12:57, 17 June 2010 (UTC)

## Definition in terms of sequence

Given ${\displaystyle a_{n}\rightarrow a,b_{n}\rightarrow b}$, and ${\displaystyle f\left(a_{n}\right)\geq f\left(b_{n}\right)}$, if we have ${\displaystyle f\left(a\right)\geq f\left(b\right)}$, then ${\displaystyle f\left(x\right)}$ is semi continuous and hemi continuous??. Jackzhp (talk) 15:24, 10 August 2010 (UTC)

## Continuous binary relations??

I removed the following bit of text, since I'm not sure this makes sense in this generality:

A binary relation R on A is continuous if R(a, b) whenever there are sequences (ak)i and (bk)i in A, which converge to a and b respectively, for which R(akbk) for all k. Clearly, if one treats R as a characteristic function in two variables, this definition of continuous is identical to that for continuous functions.

Jakob.scholbach (talk) 22:16, 23 August 2011 (UTC)

## A discrete function is never continuous.

(However, if one assumes a discrete set as the domain of function M, for instance the set of points of time at 4:00 PM on business days, then M becomes continuous function, as every function whose domain is a discrete subset of reals is.)

...no. There is a continuous function that's equal to this function at those points, but it's not the same function. The function described is not continuous because the limit does not exist at any point. Twin Bird (talk) 20:59, 30 August 2011 (UTC)

You're using an insufficiently general notion of continuity that doesn't work at isolated points. Go learn some point-set topology. Sniffnoy (talk) 01:22, 31 August 2011 (UTC)

## Is 1/x continuous at 0?

I have read the "Definition in terms of limits" and it seems to be lot of confusion there. First it says that continuity is defined only for points c in the domain (nothing about limit points). Then it defines the continuity by lim f(x)=f(c) AND lim f(x) exists. Then it says that a function defined on integers is continuous (contradicting the previous conditions).

Probably this part about "Definition in terms of limits" should be either removed or reformulated so that it is compatible with the usual topological definition. However, it is really confusing, because 1/x is a continuous function, but you always find in some books that "it has a discontinuity in 0" (although 0 is not in the domain). So, what to do? Franp9am (talk) 23:22, 4 November 2011 (UTC)

In analysis, a function is continuous at a point c if both f(c) and the limit of f(x) as x goes to c are defined and equal. f(x) = 1/x has an asymptote at x = 0. It is discontinuous at x = 0 because f(0) is not defined and it has no limit at x = 0 because the left hand limit is negative infinity and the right hand limit is positive infinity. It is, of course, continuous at every point in its domain, and books disagree as to whether or not to call a point not in the domain of a function a discontinuity of that function. In topology, a function is continuous if the inverse image of every open set is open. Under this definition, f(x) = 1/x is still not continuous at x = 0, because 0 is not in the domain of f. If we add to the real line a point at infinity, and define f(x) = 1/x for x not equal to zero or infinity, f(0) = infinity, and f(infinity) = 0, then we get a continuous function from S1 to S1, but that's a different function. As for a function defined on integers being continuous, it is continuous at every integer n, because f(n) exists, and the limit as x goes to n of f(x) exists and equals f(n), because for every epsilon greater than zero, we can let delta = .1 (for example), and x within .1 of n but not equal to n is FALSE, and both FALSE implies TRUE and FALSE implies FALSE have a truth value of TRUE. This is, however, too teachnical for this article. Rick Norwood (talk) 13:22, 5 November 2011 (UTC)
I agree, but what to do? It looks confusing for me to say that "-1 is a point of discontinuity of the real function sqrt{x}".. Would it not be better to formulate it like "For a limit point of the domain c, we define f to be continuous at c iff..." and "A function is called continuous, if it is continuous at each point of its domain. However, some authors [reference] define continuity by the condition that f is continuous at each limit point of the domain.", or something like that? Franp9am (talk) 18:03, 5 November 2011 (UTC)

This sounds plausable, but is OR unless you have a book that says this. The book I teach calculus out of, Thomas's Calculus, is conflicted on the subject. If, however, you assume that you have to say something about all the points on the real line, then yes, the square root function is discontinuous at -1 because it is not continuous there. I certainly don't want my students saying, for example, y = tan(x) is continuous everywhere, just because it is continuous everywhere in its domain. I think your limit point idea is a good one if you can find a source. Rick Norwood (talk) 11:46, 6 November 2011 (UTC)

f(x) = 1/x is clearly not continuous at x=0 since a) x is undefined at x=0, and b) lim x->0+ f(x) \= lim x->0- f(x). The question should be, "is f(x) continuous." There is an ambiguity here over what domain do we mean when we say a function is continuous? For f(x)=1/x, do we mean its domain (R\0) or R? To the lay reader, it is not clear why we would choose R\0 rather than R. In fact, it seams like cheating since we are saying "it is continuous except for when it is not".

• At least 3 sources say that for f(a) to be continuous at point a, it must be defined at point a [1] [2] [3], implying that we can't hand-wave away undefined parts of the domain.
• At least one source says that lim x-> a f(a) must exist [4], again implying that we cant hand-wave it away.
• At least one source says that "A rational function is continuous EXCEPT where the denominator is zero" [5], again implying that we cannot hand-wave away parts of the domain.
• Another source explicitly states that f(x)=1/x "... is not a continuous function since its domain is not an interval. It has a single point of discontinuity, namely x = 0..." [6].

I've made an edit to the page to indicate that the domain in question matters when examining a function. jthillik (talk) 21:50, 2 September 2016 (UTC)

References

1. ^ "Continuous Functions" (PDF). p. 2. Retrieved 2016-09-02. f is defined at a
2. ^ "Lecture 5: Continuous Functions" (PDF). p. 1. Retrieved 2016-09-02. f(a) is defined (a is in the domain of f).
3. ^ "Continuity and Differentiability". UBC Calculus Online Course Notes. Retrieved 2016-09-02. Since the function is not defined at these points, it cannot be continuous.
4. ^ "Lecture 5: Continuous Functions" (PDF). p. 1. Retrieved 2016-09-02. lim x→a f(x) exists
5. ^ "Continuous Functions" (PDF). p. 6. Retrieved 2016-09-02. A rational function is continuous except where the denominator is 0.
6. ^ Speck, Jared (2014). "Continuity and Discontinuity" (PDF). MIT Math. p. 3. Retrieved 2016-09-02. Example 5. The function 1/x is continuous on (0, ∞) and on (−∞, 0), i.e., for x > 0 and for x < 0, in other words, at every point in its domain. However, it is not a continuous function since its domain is not an interval. It has a single point of discontinuity, namely x = 0, and it has an infinite discontinuity there.

## Most recent change.

I personally disagree with the most recent change to the history section, which removes the statement that Cauchy's definition of continuous was very closely related to the limit definition of continuity used today. Instead it has now states that Cauchy's definition of continuity was very closely related to the infinitesimal definition.

My real issue is that Tkuvho takes out a sourced statement, and as I pointed out on Tkuvho's talk page before he removed this the statement, the comment that Cauchy based continuity on limits can be sourced from several different sources. Thenub314 (talk) 20:17, 15 December 2011 (UTC)

I am certainly aware of such claims in the literature. Such claims are out of date. They have been refuted in recent scholarly literature. Moreover they are patently false. This is not the first time I urge you to take a look at Cauchy page 34. Instead of being a wikipedia bureaucrat, try to be a mathematician. Tkuvho (talk) 21:09, 15 December 2011 (UTC)
Several of the references I mentioned to you were published within the last few years. How are they are out of date? History is a bit different than mathematics, it doesn't take one paper of someone who disagrees to refute something and make the point of view invalid. It is not a counter example, it is just an alternate analysis. I will take a look at Cauchy page 34, but it really won't change matters. It is not my opinion that is important in any way, but what people publishing on the subject are saying. Thenub314 (talk) 22:58, 15 December 2011 (UTC)
The book by Boyer was reprinted a couple of years ago, but it is mostly a reprint of the old edition written in the 1940s, and therefore reproduces the same errors. Tkuvho (talk) 09:32, 16 December 2011 (UTC)
I didn't say Boyer was recent, I said several of the references I mentioned to you are recent. For example Schubring was published in 2005 and makes a very detailed analysis supporting the idea that Cauchy's definition of infinitesimal was based in limits. V. Katz book is also quite recently, and says precisely the same thing. Thenub314 (talk) 16:32, 16 December 2011 (UTC)

## "Oscillatory" definition of continuity and the Weinstrauss function

I've never heard of oscillation being zero, in order for continuity to hold... But how does this work with the Weinstrauss function which is continuous everywhere and differentiable nowhere.... an example of the oscillation definition to this problem would provide a lot of insight to this article... as the Weinstrauss function (using ordinary language) seems to have LOADS of oscillation (but perhaps it has oscillation 0, in the mathematical sense). 99.149.190.128 (talk) 04:31, 17 February 2012 (UTC)

Think of it this way: if the input changes by an infinitesimal, the output may change by an incomparably bigger infinitesimal, but still only an infinitesimal. The oscillation at that point is the standard part of that infinitesimal, and therefore zero. Tkuvho (talk) 09:17, 17 February 2012 (UTC)

## Flower example is incorrect

"As an example, consider the function h(t), which describes the height of a growing flower at time t...[this] function is continuous". This ignores the fact that a flower is composed of (discrete) molecules, and thus its growth will not be continuous but rather "molecule-by-molecule" (roughly speaking). JDiala (talk) 01:24, 15 June 2015 (UTC)

That's kinda pedantic, but if you want to look at it that way, consider how a flower grows and how its molecules are added. If a molecule were to be added to the very top (like in crystal growth), going from being not part of the flower to being part of it on the surface, you might have a point, but plant growth does not really happen like that – molecules are added internally, and the existing "top" gets pushed a little, which would be a continuous movement (many molecules moving, pushing the top molecules along, or one molecule overtaking the previous "top" molecule. So to say the example is incorrect does not really hold water, ever for the most pedantic reasoning. If you feel there is a more intuitive simple example that might be accessible to everyone, feel free to make a suggestion. —Quondum 02:16, 15 June 2015 (UTC)
I see your point, but a much more straightforward example would be the position of some particle at time t; this is clear since space, and therefore the concept of "position", is continuous(ie. instant teleportation is impossible). JDiala (talk) 06:21, 15 June 2015 (UTC)
I don't think "position of a particle" is a good example of a function of time, because it is not properly defined. We could say "the height of a ball" or "the distance a particle has travelled". Sławomir Biały (talk)

According to quantum mechanics, even the position of a particle is not continuous. Continuity is a property of a mathematically defined function. The article should not say any physical process is continuous, but should rather give an example of a physical process that is modeled by a continuous function, and one that is modeled by a discontinuous function. Good examples might be the position of a particle as a function of time for continuity and the date as a function of time for discontinuity, but it should be clear that "continuous" and "discontinuous" apply to the function, not the physical process. Rick Norwood (talk) 11:42, 15 June 2015 (UTC)

https://xkcd.com/1240/ Sławomir Biały (talk) 12:01, 15 June 2015 (UTC
: ) Rick Norwood (talk) 12:51, 15 June 2015 (UTC)
The center of mass of a rigid body (can QM be applied to that?)? I would find it hard to believe that there exist no examples of actual continuity in the universe, particularly since spacetime is modeled as a differentiable (pseudo-Riemannian) manifold. JDiala (talk) 12:57, 15 June 2015 (UTC)
I was wondering whether QM was going to be dragged into this. Because the theory is not fully known (or at least, people make the weirdest assertions about it such as that it provides examples of discontinuity ), we cannot make definitive statements about physics. From this perspective of this article, one should accept a classical model, or find something else. Oh, and the centre of mass is no different from any particle in QM; only the scale is different. —Quondum 13:44, 15 June 2015 (UTC)
Well, I think all of this is clearly being pedantic way beyond what an example is meant to illustrate. One might as well say that a model is also not continuous, because models are made out of sentences in the propositional calculus. In short, no example will ever be completely satisfactory under all conceivable interpretations. That does not mean that we should do away with examples that relate mathematical concepts to everyday things people are familiar with. Obviously, we don't want to regard flower growth as a mathematical definition of continuity, nor do we want to extrapolate it to absurd degrees, but it is usefully illustrative, particularly in contrast to an obviously discontinuous process. The objections raised are just overreaching, in a "prove the teacher wrong" spirit, but do not address whether it is actually a helpful example of continuity. They are overreaching pedantry. Sławomir Biały (talk) 15:04, 15 June 2015 (UTC)
That's a fine response; I agree with you then. JDiala (talk) 02:37, 16 June 2015 (UTC)

## Should we rephrase the limit definition of continuity?

In the paragraph 'Definition in terms of limits of functions' (subsection 'Definition') we essentially say this:

${\displaystyle {\displaystyle \left\{{\text{Function }}f:{\mathcal {D}}\rightarrow \mathbb {R} {\text{ is continuous at }}c\in {\mathcal {D}}\right\}\iff \lim _{x\to c}f(x)=f(c).}}$

We later explain that the above statement also works when c is not a limit point of the domain D, since the condition given is 'vacuously' satisfied; but is that really so? If c is not a limit point, then, according to most of the definitions given by textbooks (including the one given by Wikipedia), the limit of f at c is left undefined (c is required to be a limit point prior to the definition of the limit), the notation ${\displaystyle {\textstyle \lim _{x\to c}f(x)}}$ is vague, and no one can tell weather the equality ${\displaystyle {\textstyle \lim _{x\to c}f(x)=f(c)}}$ holds or not, hence continuity is not well-defined this way.

Apart from these logic issues, we'd better avoid that kind of tricky phrasings and use simpler language and terminology if possible. So, I suggest changing the statement of the definition, presenting explicitly the two cases (c is a limit point, c is not a limit point).

By the way, another article, 'Limit of a function', addresses the same definition (section 'Relationship to continuity') and has also an obscure way of expressing these two cases.

--SiriusGR (talk) 17:36, 11 October 2015 (UTC)

I think it would be clearest just to say that ${\displaystyle f}$ is continuous at a point ${\displaystyle p}$ of the domain if, for every ${\displaystyle \epsilon >0}$, there is a ${\displaystyle \delta >0}$ such that ${\displaystyle |f(x)-f(p)|<\epsilon }$ for all ${\displaystyle x}$ in the domain of ${\displaystyle f}$ such that ${\displaystyle |x-p|<\delta }$. We could add that, in the case where ${\displaystyle p}$ is also a limit point of the domain that this coincides with ${\displaystyle \lim _{x\to p}f(x)=f(p)}$ but that, when ${\displaystyle p}$ is not a limit point, the condition then holds vacuously. 00:14, 12 October 2015 (UTC)
So you suggest merging paragraphs 'Definition in terms of limits of functions' and 'Weierstrass definition (epsilon–delta) of continuous functions', in my opinion though we should keep the structure of the article and give each one of the definitions separately. This article may be also read by somebody who is not particularly interested in, nor familiar with, rigorous calculus (a chemistry student perhaps); the limit definition should be more handy to them. So, I was thinking of rewriting the passage like this...

The function f is said to be continuous at some point c of its domain just when one of the following statements holds:
(1) The limit of f(x) as x approaches c through the domain of f exists and is equal to f(c). In mathematical notation, this is written as
${\displaystyle \lim _{x\to c}{f(x)}=f(c).}$
(2) c is not a limit point of the domain of f; that is the notion of ${\displaystyle {\textstyle \lim _{x\to c}f(x)}}$ is undefined.
In detail statement (1) means three conditions: First, f has to be defined at c. Second, the limit on the left hand side of that equation has to exist. Third, the value of this limit must equal f(c).
Note that if the point c in the domain of f is not a limit point of the domain, then, by statement (2), f is taken to be trivially continuous, since x cannot approach c through values not equal to c. Thus, for example, a function whose domain is the set of all integers is continuous at every point of its domain.
The function f is said to be continuous if it is continuous at every point of its domain; otherwise, it is discontinuous.

--SiriusGR (talk) 22:36, 12 October 2015 (UTC)
A definition like this lacks motivation. Are there sources that define continuity at a point in this way? Standard introductory texts in analysis (e.g., Bartle, Rudin, Apostol) all define continuity at a point via epsilon-delta, and then observe that this happens to agree with ${\displaystyle \lim _{x\to p}f(x)=f(p)}$ if p is also a limit point. I don't see any reason why the article should depart from the usual structure, especially not to provide a kludgy definition of our own. 23:14, 12 October 2015 (UTC)
I have to admit that the new definition in the article is better and much clearer than my improvisations. I've just noticed that Greek Wikipedia resolves the problem in a similar way: it gives the definition first, and then explains that the definition is not sufficient, because it cannot be applied to the case of isolated points. Thank you.
--SiriusGR (talk) 16:31, 13 October 2015 (UTC)

## Small oscillation versus absence of jumps

A recent edit changed the first sentence from this

In mathematics, a continuous function is, roughly speaking, a function for which small changes in the input result in small changes in the output.

to this

In mathematics, a continuous function is, roughly speaking, a function for which small changes in the input result in absence of jumps in the output.

I don't really agree that this is more correct than what was there before, and it is arguably less clear. One characterization of continuity is that the oscillation of the function can be made arbitrarily small by making the domain sufficiently small. That is to say, the change in the output value of a function is small provided the change in the input value is sufficiently small. This is more or less what is directly embodied in the familiar epsilon delta definition of a function (depending on the details, one gets different notions of continuity, e.g., uniform continuity or pointwise continuity). That definition does not mention "jumps".

Furthermore, the case of the Thomae function shows that a function can have small (!) jumps if we make small changes in the input, and still be continuous at a point.

On the issue of clarity, what could it mean to say that small changes in the input "result in the absence of jumps"? Were the jumps there before we made the small change? That makes no sense mathematically, and is very misleading. 23:07, 19 December 2015 (UTC)

I agree. Moreover, could we say that the function sin(1/x) jumps at the origin? D.Lazard (talk) 10:25, 20 December 2015 (UTC)

There are four major kinds of discontinuities: removable singularities, where the graph of a function has a hole in it that can be filled in to make a continuous function; jump discontinuities, usually found in functions defined piecewise; asymptotic singularities, where the function "goes to infinity" and we cannot travel along to graph to get from one side of the asymptote to the other, and essential singularities, like the topologist's sine curve mentioned above, that goes all over the place near the singularity. (In complex variables, the types of discontinuities are even more striking.) Only one of these has a "jump". The earlier definition better captures all cases. Rick Norwood (talk) 12:56, 20 December 2015 (UTC)

## preimage of closed function is closed.

This article states, and I quote: "This is equivalent to the condition that the preimages of the closed sets (which are the complements of the open subsets) in Y are closed in X." The article open and closed maps makes a similar assertion. The talk page Talk:open and closed maps seems to provide a simple counter-example. I feel like I am missing something ... what am I missing?

Here consider this: f: [0,1] -> (0,1] where f(x) = x (for x != 0) and f(0) = 1/2. Clearly, f is not continuous, if one uses the definition that preimages of open set must be open (since the preimage of (1/4, 3/4) is not open). Yet, from what I can tell, the preimage of any closed set is closed. Right?

The talk page Talk:open and closed maps gives a hand-waving argument about "passing through complements" but his does not work: consider the closed set V=[1/8,1/4] u [3/4,1] in the target space. Then it's complement (in the target space) is C(V) = (0,1/8) u (1/4,3/4) and is clearly open. Next, f-1(C(V))= (0,1/8) u (1/4,3/4) u {0} = [0,1/8) u (1/4,3/4) is clearly NOT open!

This seems to imply that the "equivalent" definition of continuity in terms of preimages of closed sets cannot possibly be right. Am I missing something? Misunderstanding something? 67.198.37.16 (talk) 17:33, 30 June 2016 (UTC)

The preimage of (0, 1/4] (which is closed in the image of f) is (0, 1/4], which is not closed in [0, 1]. If one consider the same function as a function from [0, 1] to [0, 1], things become clearer, as the preimage of [0, 1/4] is (0, 1/4]. D.Lazard (talk) 17:55, 30 June 2016 (UTC)
Never mind. I see the loophole. In the target space, the interval (0,1/4] must be considered to be closed yet its preimage is not closed, therefore f is not continuous. This is subtle -- a confusion about the definition of open and closed in the target space vs the definition in the standard topology. Dohh. 67.198.37.16 (talk) 18:09, 30 June 2016 (UTC)
Oops thank you, yes, I came to that realization at the same time, and see you just answered just now also. Thanks. 67.198.37.16 (talk) 18:11, 30 June 2016 (UTC)