But what is a continuous function?












6












$begingroup$


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.






real-analysis limits continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
    $endgroup$
    – LoveTooNap29
    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






  • 3




    $begingroup$
    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?
    $endgroup$
    – LoveTooNap29
    Jan 29 at 21:38






  • 2




    $begingroup$
    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






  • 3




    $begingroup$
    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
















6












$begingroup$


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.






real-analysis limits continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
    $endgroup$
    – LoveTooNap29
    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






  • 3




    $begingroup$
    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?
    $endgroup$
    – LoveTooNap29
    Jan 29 at 21:38






  • 2




    $begingroup$
    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






  • 3




    $begingroup$
    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














6












6








6


1



$begingroup$


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.






real-analysis limits continuity







share|cite|improve this question











$endgroup$




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.






real-analysis limits continuity




real-analysis limits continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 7:46









YuiTo Cheng

2,0342636




2,0342636










asked Jan 29 at 20:09









wenoweno

29411




29411








  • 1




    $begingroup$
    If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
    $endgroup$
    – LoveTooNap29
    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






  • 3




    $begingroup$
    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?
    $endgroup$
    – LoveTooNap29
    Jan 29 at 21:38






  • 2




    $begingroup$
    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






  • 3




    $begingroup$
    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














  • 1




    $begingroup$
    If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
    $endgroup$
    – LoveTooNap29
    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






  • 3




    $begingroup$
    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?
    $endgroup$
    – LoveTooNap29
    Jan 29 at 21:38






  • 2




    $begingroup$
    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






  • 3




    $begingroup$
    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








1




1




$begingroup$
If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
$endgroup$
– LoveTooNap29
Jan 29 at 21:16




$begingroup$
If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
$endgroup$
– LoveTooNap29
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




$begingroup$
Not really. Question is: Is function $f(x) = ln{|x|}$ continuous? I assume it is?
$endgroup$
– weno
Jan 29 at 21:35




3




3




$begingroup$
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?
$endgroup$
– LoveTooNap29
Jan 29 at 21:38




$begingroup$
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?
$endgroup$
– LoveTooNap29
Jan 29 at 21:38




2




2




$begingroup$
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




$begingroup$
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




3




3




$begingroup$
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




$begingroup$
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










4 Answers
4






active

oldest

votes


















22












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Yes, but preferrably the domain should be specified if this is important. The functions $f : mathbb{R}setminus0tomathbb{R}$, $xmapsto tfrac1x$ and $g : ]0,infty[tomathbb{R}$, $xmapsto ln x$ are continuous functions, and it's clear what's meant.
    $endgroup$
    – leftaroundabout
    Jan 30 at 11:07





















13












$begingroup$

"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.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    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



















0












$begingroup$

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 $(a,b)$, $f(x)=lim_{xto c} f(x)$.



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).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    No. Not for most uses, a function is regarded as being continuous on an interval.
    $endgroup$
    – Math_QED
    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





















0












$begingroup$

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.






share|cite|improve this answer









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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    22












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      Yes, but preferrably the domain should be specified if this is important. The functions $f : mathbb{R}setminus0tomathbb{R}$, $xmapsto tfrac1x$ and $g : ]0,infty[tomathbb{R}$, $xmapsto ln x$ are continuous functions, and it's clear what's meant.
      $endgroup$
      – leftaroundabout
      Jan 30 at 11:07


















    22












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      Yes, but preferrably the domain should be specified if this is important. The functions $f : mathbb{R}setminus0tomathbb{R}$, $xmapsto tfrac1x$ and $g : ]0,infty[tomathbb{R}$, $xmapsto ln x$ are continuous functions, and it's clear what's meant.
      $endgroup$
      – leftaroundabout
      Jan 30 at 11:07
















    22












    22








    22





    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 29 at 21:53









    Ted ShifrinTed Shifrin

    64.2k44692




    64.2k44692








    • 3




      $begingroup$
      Yes, but preferrably the domain should be specified if this is important. The functions $f : mathbb{R}setminus0tomathbb{R}$, $xmapsto tfrac1x$ and $g : ]0,infty[tomathbb{R}$, $xmapsto ln x$ are continuous functions, and it's clear what's meant.
      $endgroup$
      – leftaroundabout
      Jan 30 at 11:07
















    • 3




      $begingroup$
      Yes, but preferrably the domain should be specified if this is important. The functions $f : mathbb{R}setminus0tomathbb{R}$, $xmapsto tfrac1x$ and $g : ]0,infty[tomathbb{R}$, $xmapsto ln x$ are continuous functions, and it's clear what's meant.
      $endgroup$
      – leftaroundabout
      Jan 30 at 11:07










    3




    3




    $begingroup$
    Yes, but preferrably the domain should be specified if this is important. The functions $f : mathbb{R}setminus0tomathbb{R}$, $xmapsto tfrac1x$ and $g : ]0,infty[tomathbb{R}$, $xmapsto ln x$ are continuous functions, and it's clear what's meant.
    $endgroup$
    – leftaroundabout
    Jan 30 at 11:07






    $begingroup$
    Yes, but preferrably the domain should be specified if this is important. The functions $f : mathbb{R}setminus0tomathbb{R}$, $xmapsto tfrac1x$ and $g : ]0,infty[tomathbb{R}$, $xmapsto ln x$ are continuous functions, and it's clear what's meant.
    $endgroup$
    – leftaroundabout
    Jan 30 at 11:07













    13












    $begingroup$

    "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.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      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
















    13












    $begingroup$

    "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.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      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














    13












    13








    13





    $begingroup$

    "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.






    share|cite|improve this answer









    $endgroup$



    "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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 29 at 20:19









    Nick PetersonNick Peterson

    26.8k23962




    26.8k23962








    • 2




      $begingroup$
      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














    • 2




      $begingroup$
      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








    2




    2




    $begingroup$
    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




    $begingroup$
    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




    $begingroup$
    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











    0












    $begingroup$

    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 $(a,b)$, $f(x)=lim_{xto c} f(x)$.



    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).






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      No. Not for most uses, a function is regarded as being continuous on an interval.
      $endgroup$
      – Math_QED
      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


















    0












    $begingroup$

    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 $(a,b)$, $f(x)=lim_{xto c} f(x)$.



    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).






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      No. Not for most uses, a function is regarded as being continuous on an interval.
      $endgroup$
      – Math_QED
      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
















    0












    0








    0





    $begingroup$

    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 $(a,b)$, $f(x)=lim_{xto c} f(x)$.



    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).






    share|cite|improve this answer









    $endgroup$



    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 $(a,b)$, $f(x)=lim_{xto c} f(x)$.



    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).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 29 at 20:22









    R. BurtonR. Burton

    633110




    633110








    • 1




      $begingroup$
      No. Not for most uses, a function is regarded as being continuous on an interval.
      $endgroup$
      – Math_QED
      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
















    • 1




      $begingroup$
      No. Not for most uses, a function is regarded as being continuous on an interval.
      $endgroup$
      – Math_QED
      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










    1




    1




    $begingroup$
    No. Not for most uses, a function is regarded as being continuous on an interval.
    $endgroup$
    – Math_QED
    Jan 29 at 20:53




    $begingroup$
    No. Not for most uses, a function is regarded as being continuous on an interval.
    $endgroup$
    – Math_QED
    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




    $begingroup$
    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






    $begingroup$
    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













    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 31 at 16:21









        EvpokEvpok

        511417




        511417






























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