continuity at isolated point

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 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.


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.


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.