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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?

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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
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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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 $N$.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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