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I have a very basic problem. I am confused about "continuous function" term.

What really is a continuous function? A function that is continuous for all of its domain or for all real numbers?

Let's say:

$\ln|x|$ - the graph clearly says it's continuous for all real numbers except for $0$ which is not part of the domain. So is this function continuous or not? I could say same about $\tan{x}$ or $\frac{x+1}{x}$

And also what about:

$\ln{x}$ - the graph clearly says it's continuous for all of its domain: $(0; \infty)$ - so is this $f$ continuous or not?

Thanks for clarification.

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    $\begingroup$ If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question? $\endgroup$ – Nap D. Lover Jan 29 at 21:16
  • $\begingroup$ Not really. Question is: Is function $f(x) = \ln{|x|}$ continuous? I assume it is? $\endgroup$ – weno Jan 29 at 21:35
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    $\begingroup$ continuous on what set? As others have pointed out: continuity is a property of functions either at points or on sets but not a property of the function alone without regard to the domain. $\log |x|$ is certainly continuous where it is defined. Is that clear? $\endgroup$ – Nap D. Lover Jan 29 at 21:38
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    $\begingroup$ functions are continuous at specific points. For example: $f(x) = (x-2)^2; x < 0$ and $f(x) = 1$ if $0 \le x < 1$ and $f(x) = \sqrt x$ for $x \ge 1$. Is continuous at $x = -7$ and $x = \pi$ but is not continuous and $x = 0$. To say a function is "continuous" is short hand to mean in is continuous on every point in it's domain. It makes utterly no sense to talk of a function being continuous on a point not in its domain. $\endgroup$ – fleablood Jan 29 at 22:49
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    $\begingroup$ That said some people will "abuse terminology" and say intuitive but wrong things such as "$f(x) = \frac 1x$ is discontinuous at $x = 0$" this is... well, it's simply wrong. $\endgroup$ – fleablood Jan 29 at 22:51
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Mathematicians (but not all calculus books) mean "continuous at every point of its domain" when they say a function is "continuous." The functions $f(x) = 1/x$ and $f(x)=\ln x$ are continuous functions.

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    $\begingroup$ Yes, but preferrably the domain should be specified if this is important. The functions $f : \mathbb{R}\setminus0\to\mathbb{R}$, $x\mapsto \tfrac1x$ and $g : ]0,\infty[\to\mathbb{R}$, $x\mapsto \ln x$ are continuous functions, and it's clear what's meant. $\endgroup$ – leftaroundabout Jan 30 at 11:07
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"Continuous" is not, in and of itself, a property of a function. You have to talk about being continuous at a given point, or on a collection of points as you have above.

It is generally safe to assume that if somebody leaves off the set, they intend to say that the function is continuous on its domain (as both of your examples are); but, I tend to believe that explicit is better than implicit.

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    $\begingroup$ This is just not true. You can define continuity in terms of continuity at a given point, but in many contexts, that is not a very natural definition. I also can't recall any important mathematical concept for which continuity at a single point (as opposed to not being continuous anywhere) changes much. $\endgroup$ – tomasz Jan 30 at 12:53
  • $\begingroup$ Once students get to a topology course, we do indeed teach them continuity as a property of a function, namely: A function is continuous if and only if the inverse image of each open subset of the range is an open subset of the domain. $\endgroup$ – Lee Mosher Jan 31 at 16:39
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The exact answer depends on your chosen definition of "function" (there is more than one). For most uses, a function is regarded as being continuous on an interval $(a,b)$ if for every number $c$ in $(a,b)$, $f(x)=\lim_{x\to c} f(x)$.

In your example $f(x)=\ln{x}$ is continuous on the interval $(0,\infty)$ and either undefined or complex/multivalued everywhere else, depending on whether you consider the codomain (range) of $f$ to include the complex numbers or not.

In other words, no function is ever just 'continuous' - it is continuous within an interval (which may or may not be its domain).

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    $\begingroup$ No. Not for most uses, a function is regarded as being continuous on an interval. $\endgroup$ – user370967 Jan 29 at 20:53
  • $\begingroup$ There is a case to be made for 'pac-man' continuity via quotient space. Also, not every function is a function of a single, real variable, so the notion of "interval" may not always apply. $\endgroup$ – R. Burton Jan 29 at 21:01
  • $\begingroup$ I have always seen continuity defined at a point and on a set of points. Every once in a while I saw a calculus textbook author assume continuity on an open interval to simplify the proof of a basic result. But in the context given I think it is exceedingly likely that "is the function continuous" implicitly refers either to "on $\mathbb R$" or "on its domain". I don't think the implicit meaning of "on an interval" is likely. $\endgroup$ – rschwieb Jan 30 at 16:04
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It is, but what you are looking for might be the notion of continous extension.

Here both $\frac{1}{⋅}$ and $\log$ are continuous in the sense that they are (pointwise) continuous on their respective domains, as is $$ \left| \begin{array}{lll} f : &\mathbb{R}^* ⟶ \mathbb{R}\\ & x \longmapsto x² \end{array} \right. $$ The difference is that there exists $g: \mathbb{R}→\mathbb{R}$ continuous such that $g(x)=f(x)$ for all $x∈\mathbb{R}^*$ — so $f$ has a continous extension on $\mathbb{R}$ — whereas $\frac{1}{⋅}$ has no such thing.

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