Show that $x\in A$ is an isolated point of $A$ iff $\big(B_\epsilon(x)\backslash \{x\}\big)\cap A = \emptyset$.

Question

Exercise: If $x\in A$ is not a limit point of $A$, then $x$ is called an isolated point of $A$. Show that $x\in A$ is an isolated point of $A$ if and only if $\big(B_\epsilon(x)\backslash \{x\}\big)\cap A = \emptyset$ for some $\epsilon>0$.

Question: How do I solve this exercise? Isolated points have not been mentioned thus far in my book. The $\Leftarrow$ direction of the proof is quite easy, but I don't see how from this given definition of isolated points the $\Rightarrow$ direction is proven. $$x\in A \text{ not a limit point }\Rightarrow x\in A \text{ is an isolated point}$$ doesn't imply $$x\in A \text{ is an isolated point } \Rightarrow x\in A \text{ is not a limit point}$$ Right? (based on the above definition)

Answer 1

We have:

$x$ is a limit point of $A$ iff for every $B_{\epsilon}$ there is an $y \in A$ such that: $x \ne y$ and $y \in B_{\epsilon}$.

Thus, negating it, we get:

$x$ is not a limit point of $A$ iff there is a $B_{\epsilon}$ such that, for every $y \in A$: if $x \ne y$, then $y \notin B_{\epsilon}$.

This amounts to saying that there exists a neighborhood of $x$ which does not contain any other points of $A$, and this is the definition of isolated point.