Definition 4.1
Let X and Y be metric spaces; suppose E $\subset$ X, if $f$ maps E into Y and $p$ is a limit point of E. We write $f(x)$ $\to$ $q$ as $x$ $\to$ p if there is a point $q$ $\in$Y with the following property:
$\forall \epsilon>0, \exists\delta>0 $ s.t
$d(f(x),q)<\epsilon$ for all points $x\in E$ for which $0<d(x,q)<\delta$
I have a question about p being the limit point. Is it necessary for this definition? What if $p$ is an isolated point?
Thanks in advance!
It doesn't make much sense to define limits at isolated points. Suppose we leave the definition exactly how it is, except we drop the requirement that $p$ is a limit point of $E$. Let's see what happens when $p$ is not a limit point of $E$. In this case we can show that for every $q\in Y$, $f(x)\to q$ as $x\to p$. Here's why:
Suppose $p$ is not a limit point of $E$. So there is a $\delta>0$ such that for all $x\in E$, if $d(x,p)<\delta$, then $x=p$. In this case, we say that "$x$ is an isolated point of $E$". Now let $q\in Y$, and let $\epsilon>0$, and pick $\delta>0$ such that for all $x\in E$, if $d(x,p)<\delta$, then $x=p$. Since there is no $x\in E$ with $0<d(x,p)<\delta$, it is vacuously true that $d(f(x),q)<\epsilon$ for all $x\in E$ with $0<d(x,p)<\delta$.
So if $p$ is not a limit point, then for every $q\in Y$, $f(x)\to q$ as $x\to p$. This is why the definition requires that $p$ is a limit point of $E$.
