Fitzpatricks Continuity Criterion an Isolated Point Cannot Be a Limit Point

2. continuity at isolated point

continuity at isolated point

Solution 1

This is because the main difference 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.

The definition of the limit of a function at 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 $c$ doesn't need to belong to the domain of $f$. Now the definition of continuity of a function at a point is:

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

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.

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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 functional limit: if for any sequence $(x_n)_n$ in the domain of the function that converges to some point $c$ (maybe in the domain or not) the sequence $(f(x_n))_n$ converge 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 $f$ at $c$.

Symbolically if

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

Notice that if $(x_n)_n\to c$ and there is some finite number of $x_j=c$ then we can quit these points of the sequence and produce a subsequence $(x'_n)_n\to c$ such that $x'_n\neq c$ for all $n\in \mathbb N$, then this subsequence hold the condition $|x'_n-c|>0$ that is required in the $\delta,\varepsilon$-definition of the functional limit.

Solution 2

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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Comments

• In our booklet it is written that :

  A function is continuous at every isolated point.

  MY doubt:-

  Let us consider an example: let $f:\mathbb N \rightarrow \mathbb{R} $ such that $f(x)=x$. As $\ \mathbb N =\{1,2,3,...\}$ and $1,2,3,..$ are all isolated points i.e $1,2,3... $ are not limit points of $\mathbb 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 $\mathbb N $ , for example take $ 2$ then $f(2)=2$ and $$\lim_{x\to 2}f(x)=\text{not possible to determine or cannot be evaluate }$$

  More precisely $2$ is not a limit point, so limit at $2$ cannot be calculated. So, the function is not continuous at all isolated points in this example.

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

  • Whats is a sequence $(x_n)\in \Bbb{N}$ such that $x_n \to 2$, for example? Think about that.

• please explain briefly , i didn't get it

• explain briefly , i didn't get it

• 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

• $f$ 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.

• 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

• @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.

• 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

• 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 $\lim_{x\to 2}x=2$.

• iam considering the above function $f:N\rightarrow R$

• it is true or wrong for above function that limit at 2 is 2

• it is true or wrong for above function that limit at 2 is 2

• Oh, sorry @girishkumarchandora, I was not paying attention. Yes, it is true that the function $f:\Bbb N\to \Bbb R$ s.t. $f(n)=n$ is continuous. There is no limit in any point because $\Bbb N$ is composed of isolated points... but as I said you dont need any limit for the definition of continuity in isolated points. The isolated points are trivially continuous. For every $\varepsilon>0$ you can have any $0<\delta<1$ and then the defintion of continuity hold for any point.

• dear i understand your solution and explanation clearly. but if someone said function is continuous at 2 so limit is also 2 (as according to the definition a function is continuous at point a then limit exists at point a) .... u only say me it is true or wrong if we say limit is 2{ particularly talking about this case}

• i got it so it would be wrong if we say limit is 2

• Yes, exactly @girishkumarchandora because in $\Bbb N$ doesnt exist convergent sequence to some point. In a set exist two kind of points (on some classification): isolated points and limit points, one is the opposite to the other. I cant explain here, too long.

• Yes, correct. No need to define limit as the prior is not satisfied

• 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.

• Sir i didn't get the last part what do you mean when you said $|x'_n-l|>0$ ? Also for such ${x'_n} \rightarrow c$ as $c$ is isolated then in nbd of c, ${f(x'_n) }$ doesnt make sense so how do you so it goes to $f(c) $ ?

• Sir what is $\mathcal{D}^{\mathbb{N}}$ ?

Recents

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Source: https://9to5science.com/continuity-at-isolated-point