Talk:Continuous function

Counterexample for the equivalence of two definitions for topological spaces

The article originally stated:

• A function $f:X\rightarrow Y$ is continuous at a point $x\in X$ if and only if for any neighborhood of its image $f(x)\in Y$ the preimage is again a neighborhood of $x\in X$.

This definition can be shown to be equivalent to the same statement, with open neighborhoods instead of neighborhoods.

The same statement with open neighborhoods instead of neighborhoods would be:

• A function $f:X\rightarrow Y$ is continuous at a point $x\in X$ if and only if for any open neighborhood of its image $f(x)\in Y$ the preimage is again an open neighborhood of $x\in X$.

The last sentence then say these two statements are equivalent. That is false, here is a counterexample:

Let $(X,{\mathcal {T}}_{X})$ and $(Y,{\mathcal {T}}_{Y})$ be two topological spaces with $X=\{a,b,c\}$, ${\mathcal {T}}_{X}=\{\emptyset ,\{a,b,c\},\{a\}\}$, $Y=\{w,y,z\}$ and ${\mathcal {T}}_{Y}=\{\emptyset ,\{w,y,z\},\{w,y\}\}$.

Let $f\colon X\rightarrow Y$ be a function from the first topological space to the second defined by $f(x)={\begin{cases}w,&{\text{if }}x=a\\y,&{\text{if }}x=b\\z,&{\text{if }}x=c\end{cases}}$.

The only two neighborhoods of $f(a)$ are $\{w,y\}$ and $\{w,y,z\}$. $f^{-1}(\{w,y\})=\{a,b\}$ and $f^{-1}(\{w,y,z\})=\{a,b,c\}$. Both $\{a,b\}$ and $\{a,b,c\}$ are neighborhoods of $a$ so according to the first definition the function $f$ is continuous at $a$.

Now, the only two open neighborhoods of $f(a)$ are also $\{w,y\}$ and $\{w,y,z\}$. $f^{-1}(\{w,y\})=\{a,b\}$ and $\{a,b\}$ is not an open neighborhoods of $a$. Therefore, according to the second definition the function $f$ is NOT continuous at $a$.

Therefore, the two definitions can't be equivalent. Otherwise, the function $f$ would be continuous under the two definitions. Hence, I have deleted the sentence claiming that the two definitions are equivalent.

Pyrrhonist05 (talk) 17:42, 5 November 2017 (UTC)

True, the relevant neighborhoods of a need not be open, but relevant neighborhoods of f(a) may belong (WLOG) to any basis of the filter, in particular, be open. Thanks to Pyrrhonist05. Boris Tsirelson (talk) 21:12, 5 November 2017 (UTC)

I've moved this from the article. I had copy edited something similar out of the article several months ago. This appears to restate the definition trivially in several ways, but I don't think it makes the definition any easier to understand:

This definition can also be written symbolically as $\forall N\in N(f(x)),\ f^{-1}(N)\in N(x)$ where $N(x)$ is the neighborhood system of $x$.

Since neighborhoods systems are filters, they are upper sets. Therefore it is sufficient to say that for a function $f$ to be continuous at a point $x$ there must exist, for every neighborhood $N_{f(x)}$ of $f(x)$, a neighborhood $N_{x}$ of $x$ contained in the preimage $f^{-1}(N_{f(x)})$. Symbolically, it must be true that $\forall N_{f(x)}\in N(f(x)),\ \exists N_{x}\in N(x):N_{x}\subseteq f^{-1}(N_{f(x)})$.

In this second definition, it makes no difference if neighborhoods are restricted to open neighborhoods only. That is, it is equivalent to require that for every open neighborhood $U_{f(x)}$ of $f(x)$, there exist an open neighborhood $U_{x}$ of $x$ contained in the preimage $f^{-1}(U_{f(x)})$.

These last two definitions can also be restated to use images of the function $f$ rather than the preimages. That is, one may instead say that the function $f$ is continuous at $x$ if for every neighborhoods $N_{f(x)}$ of $f(x)$, there exist a neighborhood $N_{x}$ of $x$ such that its image $f(N_{x})$ is contained in $N_{f(x)}$. Again, restricting neighborhoods to open neighborhoods only doesn't make a difference.

Also, we usually don't state things in the notation of symbolic logic on Wikipedia (note that we don't do this for the other definitions of continuity). So, I've removed it pending discussion. Sławomir Biały (talk) 14:11, 6 November 2017 (UTC)

I didn't reintroduced this paragraph because it made the definition clearer but because I think it gave valuable information. I agree with you that it don't help to understand the first definition but I find that the definitions it adds are not merely trivial restatements of the first one. If you already know what are neighborhoods systems and the fact that they are filters (and so upper sets), it's true that the definitions are not really involved. But if you don't really know about neighborhoods systems being filters I think it is not so clear that the definition can be restated this way and that they are equivalent. Also, the first time I stumbled across this section of the page, I didn't known neighborhoods systems, filter and upper sets and it makes me discover these concepts. Therefore, I find that there is an opportunities in this section to introduce theses concept to the reader. Additionally, you may say that because all theses definitions are all equivalent we could just give one and forgot about the others but as a mathematician you know that sometimes when we try to prove something we don't succeed by using the first definition but then by using only a subtly changed definitions we manage to do it. What I want to convey is that I think it is important to give several definitions, at least to let people know there isn't only one. At the same time, I'm not under the impression that this content clutter the section or make it less readable so I don't see any reason for not keeping it. As for the symbolic notation, we could just remove it if that is a problem. Pyrrhonist05 (talk) 18:36, 7 November 2017 (UTC)

@Sławomir Biały and D.Lazard: Any thoughts on this ?

To avoid being guilty of wp:original research here, it would help to have a source. - DVdm (talk) 10:08, 8 November 2017 (UTC)

I agree with Sławomir that this text is not useful for understanding the definition. It is also unnecessarily too WP:TECHNICAL, by using terms (filter and upper set), which are unknown by most users that know of continuous functions. Also, the last sentence consists of replacing a sentence like "this object has this property" by "there is an object having this property". This is never useful, and is confusing here.

By the way, this section is not an alternative definition, but the standard definition of the continuity at a point. I'll thus move and rename this section. D.Lazard (talk) 10:40, 8 November 2017 (UTC)

I never said that the content I reintroduced helped to understand the first definition, I said that it added valuable information. In my opinion, the definitions added are not the same as the first and are not merely simple restatement of it either. As for WP:TECHNICAL I think there is point where we just can't avoid to be technical. Should we remove the definitions using the hyperreals or oscillation because it is too technical ? Off course no ! If someone doesn't know filters and upper sets, I think it is fine since he or she can still understand and use the definitions and understand what is continuity without them. Pyrrhonist05 (talk) 15:24, 8 November 2017 (UTC)

Image in section "Continuous functions between topological spaces" is right

The image is reproduced here for clarity and completeness:

Continuity of a function at a point.

This image is referring to this definition of continuity at a point between two topological spaces:

• A function $f:X\rightarrow Y$ is continuous at $x\in X$ if for each neighborhood $V$ of $f(x)$ there exists a neighborhood $U$ of $x$ such that $f(U)\subseteq V$.

This definition is far from being unusual, I googled "topological space continuous function" and I found these web pages and PDF on the first result page:

• [1], page 4 definition 4.6.
• [2], page 1 definition 1.1.
• [3], page 1 Theorem 18.1, 4th item.
• [4], first paragraph

Now, one could argue that they are self-published works, so here a (more difficult to find) published work mentioning this definition:

• Willard, Stephen (2012). General Topology. Courier Corporation. p. 44. ISBN 9780486131788.

Also $U$ is not chosen after $V$, a $U$ such that $f(U)\subseteq V$ should just exists for (every) $V$ (neighborhoods) and the image just show that by picturing one such $U$ and illustrating what $f(U)\subseteq V$ means . One could say that the picture doesn't shows every $V$ (neighborhoods) but it's impossible to put everything in an image. An image is here to illustrate something so that it is better understood by schematizing things. Therefore the image is correct and should be restored because an image is always welcome to illustrate a point. Pyrrhonist05 (talk) 18:29, 8 November 2017 (UTC)

Incorrect definition of G^k continuity

The wiki claims that these are identical to their curvature continuity (C^k) counterparts, but they are not. The geometric continuity (G^k) conditions require continuity of the specified degree along their parameterization in addition to the continuity in the space in which they are embedded. See p. 185ff of Gerald Farin's "Curves and Surfaces for Computer Aided Geometric Design: A Practical Guide", 2nd ed. for details.

You are right. Thank you. Now corrected. Boris Tsirelson (talk) 06:23, 19 January 2019 (UTC)

On the "Weierstrass and Jordan definition"

Shouldn't this be "uniformly continuous", and not just "continuous"? There is a discrepancy here between this page and the [uniform continuity page]. On this page (continuous function), it is simply called "continuous," but on the other (uniform continuity), it says "The difference between uniform continuity, versus ordinary continuity at every point, is that in uniform continuity the value of δ depends only on ε and not on the point in the domain. " I think something should be cleared up, here. — Preceding unsigned comment added by 2603:9001:2902:1300:849B:38FF:720A:4408 (talk) 05:36, 24 September 2019 (UTC)

There is no discrepancy, as, here, $x_{0}$ is defined before $\delta ,$ which implies that $\delta$ may depend on $x_{0}.$ D.Lazard (talk) 06:36, 24 September 2019 (UTC)

Definition in terms of limits of functions

I like this definition because its somewhat easier to check against, as the example of the signum function demonstrates. It also looks very similar to the definition of Scott continuity - in fact Scott continuity looks like a generalization of this definition of continuity. If this is in fact the case, could we state this connection somewhere? I imagine many people from a programming semantics background would be interested! Houseofwealth (talk) 01:11, 16 January 2020 (UTC)

Error in limit definition

Greetings maintainers of this page. In the section "Definition in terms of limits of functions" you give the standard definition as

$\displaystyle \lim _{x\to c}f(x)=f(c)$. This is correct. Two paras later you say:

"The formal definition of a limit implies that every function is continuous at every isolated point of its domain."

This does not follow. In fact in the Wikipedia article on limits of functions at https://en.wikipedia.org/wiki/Limit_of_a_function, in the section "More general subsets," it is explicitly stated that limits can only be defined at limit points. That's because if you try to define limits at isolated points, the limits are not unique. That latter point is not made explicitly in the article, but it's the reason that limits are only defined at limit points.

Therefore, continuity at isolated points does NOT follow from the limit definition; but rather only from the epsilon-delta definition.

99.171.179.102 (talk) 21:14, 20 February 2021 (UTC)

Please remove the "highschool definition"

"The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers. In this case, the previous two examples are not continuous ..."

and:

"... but is not continuous over the domain $\mathbb {R}$ because it is undefined at x = 0"

What is this noobish nonsense? $f(x)=1/x$ is a continuous function PERIOD - A function cannot be (dis-)continuous "over the domain $\mathbb {R}$" if it is not defined at every $x\in \mathbb {R}$. If something is defined in a wrong way in highschool (because it may be too complicated otherwise, I guess) this should not result in a Wikipedia article containing said wrong definition (as viable).

An example that is perfectly analogous to the above case: "The function $f(x)=x^{2}$, $x\in \mathbb {R}$ is not a continuous function over the domain $\mathbb {C}$ because it is undefined for $x\in \mathbb {C} /\mathbb {R}$."

I really hope it is clear that this is nonsense. --Felix Tritschler (talk) 12:56, 10 December 2021 (UTC)

I have clarified the definition(s) of (global) continuity. In particular, it depends on the context whether $f(x)=1/x$ is considered as a continuous function. D.Lazard (talk) 16:12, 10 December 2021 (UTC)

I'll cite somebody from above on a similar example:"If someone imagines tan(x) to not be continuous, they have not quite got it".

A function can only be continuous or discontinuous at points of its domain. "(1/x) ... is discontinuous at ${\displaystyle x=0}$" is wrong just the same way it is wrong that $f(x)=x^{2}$, $x\in \mathbb {R}$ is (not) continuous at e.g. $2+i.$

I know there are many who really want to cling to the idea of continuity of functions as whether they can be drawn without resetting the pen. But this is not in line with modern day mathematics. --Felix Tritschler (talk) 21:11, 12 December 2021 (UTC)

In the context of calculus, "function" means often "partial function", and the domain of a function is often unknown, or, at least, difficult to compute. For example, are you able to describe without a computer the domain of $f(x)=1/(x^{6}-4x^{5}+3x^{4}-7x^{3}+x^{2}-1)$? Another example is the complex function $f(z)=1/\Gamma (z),$ where $\Gamma$ denotes the gamma function; if you are able to describe its domain, congratulation, you will win the Millenium Prize for having solved the Riemann hypothesis. An opposite example is the function ${\textstyle f(x)={\frac {x^{2}-2x+1}{x^{2}-1}},}$ which is not defined for $x=1,$ but becomes continuous if one defines $f(1)=0.$

All these examples (but the last) are continuous in their domain, but cannot be prolongated in a function that is continuous in a larger domain. For this reason, modern mathematicians say that these functions are continuous except at the zeros of their denominators, and these zeros are called discontinuities or singularities, or, in these particular examples poles.

In summary, this is you who are not in line with modern mathematics (modern mathematics are mathematics of living mathematicians, not necessarily mathematics taught in elementary calsses). D.Lazard (talk) 22:37, 12 December 2021 (UTC)

I doubt that most modern mathematicians would agree with the statement "$f(x)=1/x$ ... is discontinuous at ${\displaystyle x=0}$" simply because it is plain wrong. It does not agree with the definitions of continuity given in this very article (- all definitions use the value of the function at the respective point, i.e. it has to be in the domain). I also doubt they treat the terms "singularity" and "discontinuity" as equivalent, in general. And if in certain areas of mathematics where mostly functions, the domain of which is difficult to evaluate, are considered and because of that, the term "continuous function" is used differently, then that should warrant a subheading within this article where this is specifically discussed. Also, could you please point me to review articles or similar sources where the term is used in the way you described it? --Felix Tritschler (talk) 18:23, 21 December 2021 (UTC)

Ok, apparently there have been some changes to the article in favor of my argument, thank you. There is still:

1. "$f(x)=1/x$ ... is discontinuous at ${\displaystyle x=0}$" - see above (or below)

2. On functions with singularities: "When they are continuous on their domain, one says, in some contexts, that they are continuous, although they are not continuous everywhere. In other contexts, mainly when one is interested with their behavior near the exceptional points, one says that they are discontinuous."

- Near poles these functions are continuous and, again, at the poles they are not defined, hence, neither continuous nor discontinuous.

And the first sentence (of 2.), I think, just adds confusion to the topic. Because it is spot on, let me repeat the quote from above:"If someone imagines tan(x) to not be continuous, they have not quite got it"

Continuity of functions is about wheter they "respect" the limes, not about drawing the graph of a function in one uninterrupted line. This needs to be a fundamental message of the article.

--Felix Tritschler (talk) 12:25, 8 January 2022 (UTC)

Since there has been no answer, can we/I please remove/change the parts of the article talking about discontinuities outside the domain of functions? This includes the graph of f(x)=1/x including the wrong statement "... but is discontinuous at $x=0$" - Or we could keep the graph but with a paragraph talking about the very common mistake of this and other functions with singularities being discontinuous there. E.g. the French article on this subject has such a paragraph. And, btw, neither the German, the French nor the Spanish article claim these functions have discontinuities. It is a widespread misconception and this article should help correcting it, not the opposite. --Felix Tritschler (talk) 15:14, 20 May 2022 (UTC)

One can find, in many reliable sources sentences such that The function $x\mapsto {\frac {1}{x}}$ is discontinuous at $x=0.$ It is not the role of Wikipedia to decide whether this formulation is wrong or not, but, as such a formulation is very common, Wikipedia must report it. I agree that the section deserves to be improved or clarified, and, as a first step in this direction, I have added a link to Partial function. As your peremptory assertions are unsourced, they are considered here as your own original research. So, please do not base edits on them, this is against one of the main Wikipedia policies (see WP:NOR). D.Lazard (talk) 16:55, 20 May 2022 (UTC)

What I said is not original research. I already mentioned the French article and found in Arens, Hettlich - Mathematik (2018), page 219, for exactly the same example function (translated from German):"One can only speak of continuity in places where a function is defined". And I asked before if you could point me to sources stating the opposite but there was no answer. So in which mathematics textbooks (or other reliable sources) is it written that "The function $x\mapsto {\frac {1}{x}}$ is discontinuous at $x=0$" (or similar, e.g. different function with singularity, with the same meaning)? --Felix Tritschler (talk) 16:27, 24 May 2022 (UTC)

The article Zeros and poles starts with "In complex analysis (a branch of mathematics), a pole is a certain type of singularity of a function, nearby which the function behaves relatively regularly", and it is later said that 0 is a pole of $1/x.$ You can find similar formulations in every textbooks of complex analysis. So the fact that in some contexts 0 is a singularity of $1/x$ is a sourced fact. Your peremptory assertion that this is wrong is thus your own opinion that is not a consensus of the mathematical community. D.Lazard (talk) 18:17, 24 May 2022 (UTC)

Yes, but this was never in question. The question was whether a pole/singularity is a discontinuity. I say, it is not, you say it is. You said in the post before your last one, that:"One can find, in many reliable sources sentences such that The function $x\mapsto {\frac {1}{x}}$ is discontinuous at $x=0.$" - Where are they? I offered 2 sources that this is wrong, regarding even exactly the same example. --Felix Tritschler (talk) 08:35, 27 May 2022 (UTC)

This keeps getting better. Now you added complex analysis in the section "Real functions"-->"Definition"- again, without source (and with typos/missing articles). And the Cauchy Prinicipal Value part, you added as well, does not at all show that functions with singularities are discontinuous there - this concept allows for integration over singularities - why is this in the introductory section of the article "Continuous Function"? The part on partial functions also does not argue for functions with poles being discontinuous there, is phrased vaguely and also has a typo: "In other contexts, mainly when one is interested WITH their behavior near the exceptional points, one says that they are discontinuous."--> (I said this above already, btw):"near the exceptional points" they are continuous anyways (!), it is nowhere to find in the article "Partial Functions" that they are discontinuous at the poles and, so far, there are no sources backing it. - And again, where are your "many reliable sources sentences such that "The function $x\mapsto {\frac {1}{x}}$ is discontinuous at $x=0.$"? Even if you come up with them (which I doubt for this example), I already gave two sources saying that this is wrong and I'm quite sure I can bring more. --Felix Tritschler (talk) 09:15, 2 June 2022 (UTC)

Again, there is no answer and no sources for your claim, so, so far, this is original research. I have provided two sources backing my position. So I repeat what I wrote above:"Can we/I please remove/change the parts of the article talking about discontinuities outside the domain of functions? This includes the graph of f(x)=1/x including the wrong statement "... but is discontinuous at $x=0$" - Or we could keep the graph but with a paragraph talking about the very common mistake of this and other functions with singularities being discontinuous there. E.g. the French article on this subject has such a paragraph. And, btw, neither the German, the French nor the Spanish article claim these functions have discontinuities. It is a widespread misconception and this article should help correcting it, not the opposite".

In addition, the complex analysis-part you added in the section "Real functions" should be removed for an obvious reason. --Felix Tritschler (talk) 12:23, 17 July 2022 (UTC)

Umm, how to put it politely: that's not correct. $f(x)=1/x$ is a standard textbook example of a function that is not continuous at x=0. It appears in pretty much all standard introductory pre-calculus textbooks used in high-schools throughout the land. Its a standard example, precisely because, to the young student, it looks like a nice smooth function, and so whenever some kid starts getting glib about continuity, you ask them to do a delta-epsilon proof at x=0, and they can't. Try it yourself. The entire raison d'etre of theorems and proofs is that, by being forced to write a detailed proof, you avoid glib but incorrect claims about this-and-such being true. You have to actually prove it. Try it. You'll fail. There is no proof of continuity at x=0. 67.198.37.16 (talk) 06:25, 29 December 2023 (UTC)

You are mistaken. For the epsilon-delta criterion (or the test for continuity via limits) at a position, the function has to be defined there. This is not the case for $f(x)=1/x$ at x=0 (and poles of functions in general). I went over this numerous times above - did you read that at all? Note that D.Lazard, in an attempt to justify his erratic notion of continuity, produced citation 7 (currently): Strang, Gilbert (1991). Calculus, which says:

"Objection: The definition makes $f(x)=1/x$ a continuous function! It is not defined at x=0, so its continuity can't fail. The logic requires us to accept this, but we don't have to like it."

Isn't that interesting? He, himself, gave a source that directly contradicted his view. Further, it's not about whether "we like it" or not, as it says in the source. It's about what is correct. And correct is that a function is continuous if it is continuous at every position of its domain - which is the case for $f(x)=1/x$. And a discontinuous function has at least one discontinuity - but it can only be discontinuous where it is defined. It's part of the definitions of both the limits- and the delta-epsilon-criterion. If some school- and undergrad books ignore these definitions and arrive at the conclusion that $f(x)=1/x$ is a discontinuous function, then they are wrong in this regard, which (being wrong) isn't that special for such books.

Btw I found that D.Lazard also wrote inaccurate content in the Wikipedia article on the square root, where he implies that the square root of e.g. 4 is {-2, 2}. He also contributed to the article "Multa-valued function", which doesn't exist in Wikipedias of other languages - simply because it doesn't exist at all. Apparently not enough mathmaticians/scientists who care for accurate content care enough for it to contribute to Wikipedia pages. This is very problematic because Wikpedia has such a wide reach. --Felix Tritschler (talk) 00:21, 12 April 2024 (UTC)

Please just use direct language instead of sarcasm or rhetorical flourishes, and try to stay polite and assume good faith. –jacobolus (t) 01:27, 12 April 2024 (UTC)

Could you please point me to where was I less polite and more sarcastic than "Umm, how to put it politely:"?

And regarding assuming good faith: In the original dispute from 2021/2022 with D.Lazard, he just stopped replying to my arguments that he could not counter and ignored the source I gave to back my standpoint - and his own source (Strang, Gilbert (1991)) contradicts his own standpoint. Yet, he didn't revert his changes (e.g. where now there is complex number content under the heading "Real Functions") to the article. --Felix Tritschler (talk) 11:03, 12 April 2024 (UTC)

I stopped replying, because I had replied before, and I do not like to waste my time to repeat the same thing again and again. About self-reverting my edits, I would do it if there were a consensus for that. In this case, you alone is certainly not sufficient for a consensus. D.Lazard (talk) 11:41, 12 April 2024 (UTC)

Yes, you replied, but you didn't address any of my objections. So it had the same effect as no reply. --Felix Tritschler (talk) 11:03, 15 April 2024 (UTC)

Parts such as "did you read that at all?", "an attempt to justify his erratic notion", "Isn't that interesting?", and "Apparently not enough mathmaticians/scientists who care for accurate content care enough for it to contribute to Wikipedia pages." seem entirely gratuitous to me. –jacobolus (t) 12:26, 12 April 2024 (UTC)

More productively, if you want to find sources, one keyword to try is "infinite discontinuity", which will turn up many textbooks, papers, university course notes, etc. (though sometimes this means something else). –jacobolus (t) 13:00, 12 April 2024 (UTC)

Suggestion to change the article based on the discussion 'Please remove the "highschool definition" '

Since there has been no reaction, specifically no sources backing that functions are discontinuous at singularities etc. (see above), I am suggesting here anew to change the article accordingly, that is, to clarify that functions are neither continuous nor discontinuous at singularities or at any other points outside their domain and, hence, that functions like $x\mapsto {\frac {1}{x}}$ or the tangent function are continuous functions (A function that is continuous at every point in its domain is a continuous function). In the above discussion I have given a source from a mathematics textbook (and the French article on continuous functions). Btw, these issues with the article have been addressed before, in the now archived discussion "Continuity and the domain of the function" where it is correctly stated that "elementary functions are continuous" but either the article was never changed in this regard or these changes were reverted. (Plus, there have been very questionable recent additions and changes to the article like content on complex analysis in the section "Real functions".) I would encourage other users, especially the contributors of the now archived discussion, to get involved. Thank you in advance. --Felix Tritschler (talk) 11:50, 6 August 2022 (UTC)

You are proposing something that is contradicted by pretty much every textbook on pre-calculus used throughout the land. The reason this is a "high-school definition" is because it is used in high-schools. It is what students see, it is what is taught by convention. It appears on exams and in homework problems. 67.198.37.16 (talk) 06:31, 29 December 2023 (UTC)

Please have a look at my reply to your other comment (IP address 67.198.37.16 from 29.Dec.2023). --Felix Tritschler (talk) 11:05, 12 April 2024 (UTC)

Definition in terms of neighborhoods has issues

The definition begins with: "A neighborhood of a point c is a set that contains, at least, all points within some fixed distance of c." This is true for a metric space but there are topological spaces that are not metrizable.

The definition continues: "Intuitively, a function is continuous at a point c if the range of f over the neighborhood of c shrinks to a single point as the width of the neighborhood around c shrinks to zero." Again, the "width" is only well defined in a metric space.

It ends with "This definition only requires that the domain and the codomain are topological spaces and is thus the most general definition." This may be the case if we replace "neighborhood" by "open set" but given that neighborhoods are introduced as "points within some fixed distance", it is plainly wrong. 120.51.141.38 (talk) 05:51, 17 April 2023 (UTC)

As this section is a subsection of § Real functions, there is no problem, except in the last paragraph, which I have fixed by mentioning that this is a generalization of the concept of neighborhood that is used there. D.Lazard (talk) 09:01, 17 April 2023 (UTC)

1/x is discontinuous at 0?

I know that some calculus textbooks does say that $x\mapsto 1/x$ is discontinuous at 0 since it is not defined there. But this is a kind of old way to describe the matter and the map $x\mapsto 1/x$ becomes continuous (even meromorphic) after defining $1/0=\infty$. If we are to keep this example in the article I think it's best to replace the function by a map

$f(x)={\begin{cases}1/x&x\neq 0\\0&x=0\end{cases}}$

which is also defined at 0. 慈居 (talk) 08:42, 28 October 2023 (UTC)

And by the way, I'm here because this confusing statement has arrived at Korean Wikipedia by translation. 慈居 (talk) 08:47, 28 October 2023 (UTC)

The article says Many commonly encountered functions are partial functions that have a domain formed by all real numbers, except some isolated points. Examples are the functions $x\mapsto 1/x$ and $x\mapsto \tan x.$ When they are continuous on their domain, one says, in some contexts, that they are continuous, although they are not continuous everywhere. In other contexts, mainly when one is interested in their behavior near the exceptional points, one says they are discontinuous. [...] For example, the functions $x\mapsto 1/x$ and $x\mapsto \sin {\frac {1}{x}}$ are discontinuous at 0, and remain discontinuous whichever value is chosen for defining them at 0.. I think that this answers your concern. Also, when not otherwise specified, $x\mapsto 1/x$ is a (partial) function from the reals to the reals or from the complexes to the complexes. Its extension as a function on the projectively extended real line or the Riemann sphere is continuous everywhere, but it is another function. D.Lazard (talk) 12:11, 28 October 2023 (UTC)

Oh, that definition for partial functions is even a special case of continuous functions in the normal sense, by adding a new isolated point to the codomain. This is indeed enough to make the statement make sense. But as for my home wiki, I'd rather not introduce this definition and only discuss continuity at points in the domain, cause that seems enough. 慈居 (talk) 13:48, 28 October 2023 (UTC)

First, I doubt that the consensus is that, when not otherwise specified, $x\mapsto 1/x$ is assumed to be a partial function. But even if, that does not produce a discontinuity at x=0: Neither via the epsilon delta criterion, nor via limits is it possible to determine a discontinuity at x=0. As a partial function, x=0 still does not belong to the domain of this partial function and the criteria for discontinuity can only be applied at points of the domain of a (partial) function.

Please have a look at the source "Strang, Gilbert (1991). Calculus. SIAM. p. 702. ISBN 0961408820", where it says:

"The definition makes f(x)=1/x a continuous function! It is not defined at x=0, so its continuity can't fail." --Felix Tritschler (talk) 11:33, 12 April 2024 (UTC)

The meaning of the expression ⁠$x\mapsto 1/x$⁠, without further explicit detail, depends on context and is somewhat fuzzy and not always consistent. In complex analysis, dealing with arbitrary rational functions, it is most convenient to be working with the projectively extended complex numbers, ⁠${\hat {\mathbb {C} }}=\mathbb {C} \cup \{\infty \}$⁠, so naming this function ⁠$f$⁠ we have ⁠$f(0)=\infty$⁠ and is continuous there. In applications working with real-variable calculus, it's most common in practice to consider partial functions from ⁠$\mathbb {R} \to \mathbb {R}$⁠, so that ⁠$f(0)$⁠ is considered undefined and is a discontinuity because ⁠$\textstyle \lim _{x\to 0^{+}}f(x)\neq \lim _{x\to 0^{-}}f(x)$⁠. In computer floating point arithmetic the number system includes signed zero (representing either 0 or possibly underflow) and signed infinity (representing either infinite quantities or overflow), so ⁠$f(0)=\infty$⁠ and ⁠$f(-0)=-\infty$⁠, both well defined but clearly a discontinuity. Some authors, especially of introductory textbooks, work only with functions per se, with the domain defined to exclude values where the function is undefined. For these authors, ⁠$f(0)$⁠ is undefined and ⁠$\textstyle \lim _{x\to 0^{+}}f(x)=+\infty$⁠, ⁠$\textstyle \lim _{x\to 0^{-}}f(x)=-\infty$⁠, but the function's domain is said to be ⁠$(-\infty ,0)\cup (0,\infty )$⁠ (as is the function's image) and the function is said to be continuous for points in this (itself non-continuous) domain. –jacobolus (t) 17:19, 12 April 2024 (UTC)

As a real valued function of a real variable, isn't $1/x$ most commonly a function $\mathbb {R} \setminus \{0\}\to \mathbb {R}$? 慈居 (talk) 17:41, 12 April 2024 (UTC)

My impression is that defining continuity at an undefined point is rather rare. And the only reference that I know that does so, defines every function $\mathbb {R} \setminus \{0\}\to \mathbb {R}$ to be discontinuous at 0, because it is not defined at 0, which has nothing to do with analysis at all. 慈居 (talk) 17:58, 12 April 2024 (UTC)

Continuity has to do with whether the limit from the left is the same as the limit from the right and whether these equal the value of the function. In the case where the function is undefined at a particular point, it is indeed usually taken to be discontinuous there, but sometimes there is a removable singularity which can be trivially fixed by just defining the function at that point to be equal to its limit (e.g. starting from ⁠$x\mapsto \sin x/x$⁠ we can explicitly define the value at ⁠$0{\vphantom {)}}$⁠ to obtain the continuous sinc function even though ordinarily division by zero is undefined), while other times there is some kind of "jump" where the limits from right and left do not agree, or a place where the function "blows up" and tends to infinity (from one or both sides, with the same or differing signs from above and below), or even an essential singularity where the behavior is more pathological. –jacobolus (t) 20:39, 12 April 2024 (UTC)

Such classification is made because we often consider a function from and to a subset of $\mathbb {C}$, but we also wish to see their behavior as a map from and to a subset of ${\widehat {\mathbb {C} }}$. But actually there is only one concept of "holomorphy" for a fixed domain and codomain. Holomorphy is defined for points in (the interior of) the domain and I think continuity should be too. 慈居 (talk) 05:53, 13 April 2024 (UTC)

In the original source which introduced the name "holomorphic function", Briot & Bouquet's Théorie des fonctions elliptiques (2nd edition, 1875), it is taken for granted that we can take the reciprocal of a function which has a pole at a point and get a zero there, or vice versa. That is, the pole of the function is at least implicitly considered to be part of the domain (or something close enough to the domain that it can counts as part of the domain of the reciprocal function). –jacobolus (t) 06:09, 13 April 2024 (UTC)

I see it but I didn't get what your point is. Another way to say this would be that a function holomorphic at points except some isolated poles has an analytic continuation to a domain that contains those poles, but valued in ${\widehat {\mathbb {C} }}$. The reciprocal function also has a continuation to a domain containing the infinity, and the two maps can be composed freely. 慈居 (talk) 06:23, 13 April 2024 (UTC)

++ I'm fine with assuming that the domain already contains the poles, or not assuming it, which varies among different authors. I'm only concerned with defining continuity only at points in the domain. 慈居 (talk) 06:47, 13 April 2024 (UTC)

The current article text seems quite clear and explicit, discussing both points of view. Are you just wondering about finding sources for each version? For example, a 2021 version of introductory calculus notes from Harvard have "The function ⁠$f(x)=1/x$⁠ is continuous except at ⁠$x=0$⁠. It is a prototype with a pole discontinuity at ⁠$x=0$⁠. One can draw a vertical asymptote. The division by zero kills continuity." By comparison a 2023 version of introductory calculus lecture notes from Columbia have "However, this new definition doesn’t necessarily correlate with our intuition of which functions are continuous. For example, think about the function ⁠$f(x)={\tfrac {1}{x}}$⁠. This clearly has a discontinuity at ⁠$0$⁠: the limit ⁠$\textstyle \lim _{x\to 0}{\tfrac {1}{x}}$⁠ does not exist, and neither does ⁠$f(0)$⁠, so we certainly can’t say that they’re equal if they’re not even defined. However, at every other point f(x) is continuous, and since 0 is not in its domain this means f(x) is continuous everywhere in its domain and therefore continuous!" –jacobolus (t) 07:10, 13 April 2024 (UTC)

Uhm, I'm fine with discussing both points of view, but to me the current article looks "A, but some authors B" and I prefer "B, but some A". 慈居 (talk) 08:21, 13 April 2024 (UTC)

Repeating D.Lazard from above, the current article says explicitly:

"Many commonly encountered functions are partial functions that have a domain formed by all real numbers, except some isolated points. Examples are the functions ${\textstyle x\mapsto {\frac {1}{x}}}$ and $x\mapsto \tan x.$ When they are continuous on their domain, one says, in some contexts, that they are continuous, although they are not continuous everywhere. In other contexts, mainly when one is interested in their behavior near the exceptional points, one says they are discontinuous."

What would be your preferred wording? –jacobolus (t) 16:01, 13 April 2024 (UTC)

If I may try (but in better English than mine): Many commonly encountered functions have a domain formed by all real numbers, except some isolated points. Examples are the functions $\textstyle x\mapsto {\frac {1}{x}}$ and $x\mapsto \tan x$. They are continuous functions, since each of them is continuous at every point in the domain. They are not continuous nor discontinous at the exceptional points, since those points do not belong to the domain. In other contexts, mainly when one is interested in their behavior near the exceptional points, one says that they are discontinuous at those points, and therefore not continuous functions on the entire real line. 慈居 (talk) 16:20, 13 April 2024 (UTC)

One of my reasons is that this should be at least more compatible with the definition for topological spaces. There is no apparent natural parent spaces asggined to arbitrary spaces, so continuity of maps between spaces is only defined at points in the domain. 慈居 (talk) 08:27, 13 April 2024 (UTC)

Then perhaps better use the standard definition (as in Starng's Calculus) by default in the article? As for the other one, we could also mention that, citing a reference; I've seen such a definition, but not in an English reference, so I'm not going to add one myself. 慈居 (talk) 17:37, 12 April 2024 (UTC)

In both the real number system and the complex number system, infinity is not a number. In particular, the real number system obeys the Archimedean axiom. which says that for any two positive real numbers, some integer multiple of the smaller is bigger than the larger. If infinity were allowed as a real number, the Archimedean axiom would not hold.

What is called calculus at a lower level is, at a higher level, called either Real Analysis or Complex Analysis, the name indicating which number system is used. Rick Norwood (talk) 11:32, 28 October 2023 (UTC)