Talk:Continuous function

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[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] No, really, really amazingly small. You can't imagine how small points are.

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

[edit] 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)

[edit] 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)

[edit] Notation

I was recently reading an (non-WP) article that uses the notation C(k,α) with 0 < α < 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 Ck 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)
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