The equivalence of two definitions for accumulation point Usually, we define accumulation point as:
1) A point x is called accumulation point of a set X if for every ε>0, the ε-neighborhood of x contains an element of X \ {x}.
But today I also saw another one:
2) A point x is called accumulation point of a set X if for every ε>0, the ε-neighborhood of x contains infinitely many elements of X.
I think they are equivalent, but how can I prove it? Could you please help me? Thank you.
 A: Your question implies that the underlying topological space is a metric space. These two definitions are indeed equivalent under this assumption. It's worth noting, however, that these two definitions correspond to different concepts in the more general setting of topological spaces.
Let's show that the definitions are equivalent for metric spaces. it's easy to see that (2) implies (1). Now assume (1) holds. Let $p_1 \in B_\epsilon(x) - \{x\}$. Let $\epsilon_2 = d(x, p_1)$. Obtain $p_2 \in B_{\epsilon_2}(x) - \{x\}$. Repeat this process to get an infinite sequence of distinct elements $\{p_i\} \subset B_\epsilon(x) - \{x\}$.
A: $(2)\implies(1)$
If $U$ is an $\epsilon$-neighborhood of $x$ and contains infinitely many points of $X$ then evidently $U-\{x\}\neq\varnothing$.
$\neg(2)\implies\neg(1)$
Let it be that some $\epsilon>0$ exists such that $X\cap B(x,\epsilon)$ is finite.
Then consequently: $$\epsilon':=\min\{d(x,y)\mid y\in (X-\{x\})\cap B(x,\epsilon)\}>0$$
so that: $$(X-\{x\})\cap B(x,\epsilon')=\varnothing$$
A: 1 is usually called a limit point.
In a Hausdorff space, which metric spaces are, they are equivalent.
Clearly 2 implies 1.
Now assume x is an limit point of X.
Let U be an open ball about.
Thus exists a in X, different than a.
Find an r-ball about x that excludes a.
Whence exists b in X different than a and different than x.
Proceed thusly to find a sequence of distinct points, a, b, c,... in ever smaller balls.
Whereupon x is an accumulation point of X (definition 2). 
