This question already has an answer here:

In our booklet it is written that :

$$A\ function\ is\ continuous\ at\ every\ isolated\ point.$$

MY doubt:-

Let us consider an example:

Let $f:N\rightarrow R$ $\ $ such that $$f(x)=x$$

As $\ N=\begin{Bmatrix} 1,2,3,... \end{Bmatrix}$ and $1,2,3,..$ are all isolated points i.e $1,2,3... $ are not the limit point of set $N$

Now according to the above statement the function is continuous at every isolated point

But according to the definition of continuity, the continuity at point $a$ is$$\lim_{x\to a}f(x)=f(a)$$

Now take any number from set $N$ , for example take$\ 2\ $then$$f(2)=2$$and $$\lim_{x\to 2}f(x)=not \ possible\ to\ determine\ or\ cannot \ be\ evaluate $$

More pecisely $2$ is not limit point, so limit at $2$ cannot be calculated

So, the function is not continuous at all isolated point in this example.

How the function can be continuous at all isolated points? Can anyone tell me?


marked as duplicate by Eric Wofsey, JMP, Claude Leibovici, M. Vinay, MickG Jun 6 '16 at 8:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Whats is a sequence $(x_n)\in \Bbb{N}$ such that $x_n \to 2$, for example? Think about that. $\endgroup$ – Irddo Jun 6 '16 at 4:56

This is because the main different between the definition of limit in a point and continuity in a point is the inclusion (in the last case) of the distance zero in the domain, i.e. $|x-c|<\delta$ for continuity and $0<|x-c|<\delta$ for limit.

Definition of limit of a function in a point (to exist a limit the point must be a limit point):

$$\forall\varepsilon>0,\exists\delta>0,\forall x\in\mathcal D:0<|x-c|<\delta\implies|f(x)-L|<\varepsilon$$

where $\mathcal D$ is the domain of the function. Notice that $c$ not need belong to the domain of $f$. Now definition of continuity:

$$\forall\varepsilon>0,\exists\delta>0,\forall x\in\mathcal D:|x-c|<\delta\implies|f(x)-f(c)|<\varepsilon$$

Notice that here we need that $c\in\mathcal D$ and for $x=c$ the definition is trivially true that is what happen in an isolated point.

Trying to answer the comment we can define the limit of a function in some point using sequences, this is named the sequential characterization of the limit: if for any sequence in the domain of the function that converges to some point $c$ (maybe in the domain or not) the images of the function are converging to some point $L$ in the codomain (maybe not in the range of the function) then we says that $L$ is the limit of the function at $c$.

Symbolically if

$$(\forall (x_n)\in\mathcal D: (x_n)\to c\land x_j\neq c,\,\forall j\in\Bbb N\implies f(x_n)\to L) \iff\lim_{x\to c}f(x)=L$$

Notice that for any $(x_n)\to c$ where exist some finite number of $x_j=c$ then we can quit these points of the sequence and produce a convergent subsequence $(x'_n)\to c$ that hold the condition of distance different to zero for the $\delta,\varepsilon$ definition of limit.

  • $\begingroup$ u gave the nice explanation on the inclusion and exclusion of zero. i agree with your solution. but still i have one question that is: can we say that the limit of a function at 2 is equal to 2 $\endgroup$ – Girish Kumar Chandora Jun 7 '16 at 4:46
  • $\begingroup$ @girishkumarchandora it depends: if the function at $x=2$ have a limit of $2$ then yes... but this is a rare coincidence. The limit of any function in a point of it domain is a point in the codomain. Maybe you are confused due to the implicit fact that in the real line any point can be approached from anywhere, i.e. we approach to $x=2$ with points that are not $2$ and we see what happen then to the values of $f(x)$: we can see if converging to some $x$ then the images are converging to some point in the codomain. $\endgroup$ – Masacroso Jun 7 '16 at 6:50
  • $\begingroup$ i didn't get it what u want to say.... i just want to know that it is true or wrong: limit at 2 is 2 $\endgroup$ – Girish Kumar Chandora Jun 7 '16 at 7:51
  • $\begingroup$ For the function $f:\Bbb R\to\Bbb R$ such that $f(x)=x$ then we have that $\lim_{x\to c}f(x)=f(c)=c$, so yes $\lim_{x\to 2}x=2$. $\endgroup$ – Masacroso Jun 7 '16 at 7:54
  • $\begingroup$ iam considering the above function $f:N\rightarrow R$ $\endgroup$ – Girish Kumar Chandora Jun 7 '16 at 8:02

By definition, a function $f$ is continuous at a point $p$ if, as you get near $p$, $f($points near p$)$ approaches $f(p)$.

For isolated points, there are no points near $p$, so the statement is trivially true!

It's like saying: if there were unicorns, I would be green. The statement is always true if there are no unicorns, as the precondition is never satisfied.

  • $\begingroup$ if we are unable to move at point $p$ from both sides then we can say that the function is continuous at point $p$ ?. If yes then what would be the limit of function $\endgroup$ – Girish Kumar Chandora Jun 6 '16 at 5:08
  • $\begingroup$ Yes, correct. No need to define limit as the prior is not satisfied $\endgroup$ – user341502 Jun 18 '16 at 20:00

Well, your mistake is really conceptual. It all comes back to the following:

$$p\implies q$$

is true, if $p$ is false.

So, if there are no sequence $x_n\in N$ converging to $2$, then, $$x_n\to 2\implies f(x_n)\to 2$$ is still correct.

Also, you're missing a fact. Although $2$ is not a limit point, there are sequences that converge to $2$. For example, $2,2,2,2,2,\ldots$ converges to $2$. More precisely, any sequence that constantly becomes $2$ after some point converges to $2$, and these are the only sequences in that are converging to $2$ in $N$.

  • $\begingroup$ explain briefly , i didn't get it $\endgroup$ – Girish Kumar Chandora Jun 6 '16 at 5:02
  • $\begingroup$ $f$ is continuous at $a$ if for any sequence $x_n$ in $N$; $$x_n\to a\implies f(x_n)\to f(a)$$ holds. Now a statement $p\implies q$ holds iff whenever $p$ is true, $q$ is also true. So, if $p$ is false, $p\implies q$ is still holds. Thus, if there are no sequences converging to $2$, the first part of the $\implies$ is false. Thus, the statement still holds and $f$ is continuous. $\endgroup$ – Emre Jun 6 '16 at 5:11

One can interpret $$\lim_{x\to a}f(x)=f(a)$$

to be true vacuously, since $f$ is not defined in a sufficiently small deleted neighbourhood of $a$. Alternatively, you could consider the standard $\epsilon-\delta$ definition of continuity and note that it is satisfied.

  • $\begingroup$ please explain briefly , i didn't get it $\endgroup$ – Girish Kumar Chandora Jun 6 '16 at 5:01
  • $\begingroup$ math.stackexchange.com/questions/70736/… if you don't understand what the vacuous truth is. $\endgroup$ – rb612 Jun 6 '16 at 5:05
  • $\begingroup$ But isn't $\lim{x\rightarrow a}f(x)$ usually defined only where $a$ is a limit point? And so if $a$ is an isolated point, then this expression is undefined. $\endgroup$ – dtcm840 Oct 30 '18 at 1:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.