TITLE: Definition 4.1 Principles of Mathematical Analysis
QUESTION [6 upvotes]: 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!

REPLY [4 votes]: 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 $E$.