Show that x is an accumulation point of E iff ... Show that a point $x$ is an accumulation point of a set $E$ if and only if for every $\epsilon>0$ there are at least two points belonging to the set $E\cap (x-\epsilon, x+\epsilon)$.
My attempt:
Any point $x$ (not necessarily in $E$) is said to be an accumulation point of $E$ provided that for every $\epsilon>0$ the intersection $E\cap (x-\epsilon, x+\epsilon)$ contains infinitely many points.
One direction is easy: If $x$ is an accumulation point, there are infinitely many points in the intersection. So, there are at least two points. How do I prove the other direction?
 A: Select $\epsilon_1>0$ and an associated $x_1\in (x-\epsilon_1, x+\epsilon_1)$. Define $\delta_1 = |x-x_1|$ and let $\epsilon_2<\delta_1$. Define inductively $x_i$ to be some point not equal to $x$ in $(x-\delta_{i-1}, x+\delta_{i-1})$. By definition $x_i\ne x_j$ for any $j<i$. By hypothesis $x_n$ exists for all $n$, because there are always two points, and not both of them can be $x$, hence there are infinitely many points in such a neighborhood.
A: Let $\epsilon > 0$ be fixed. Consider those two points you know to exist, one of them is the point $x$ itself, the other one must be another point, say $y_1$. Then for $\epsilon_1 = |x-y_1|$, and $\epsilon_1 < \epsilon$ (why?), the corresponding neighborhood, $(x-\epsilon_1,x+\epsilon_1)$, you will contain at least two points, one, again could be $x$, but the other will be some other point, $y_2$. And $y_2 \neq y_1$ (why?). Then letting $\epsilon_2 = |x-y_2|$, you can construct infinitely many points, $y_1, y_2, \ldots$ such that for all $i\in\{1,2,\ldots,\}$, $y_i\in(x-\epsilon,x+\epsilon)$.
