TITLE: Limit points in topological space $X$
QUESTION [8 upvotes]: Let $X$ be a topological space. Let $A \subset X$. The point $x$ is a limit point of $A$ iff every neighborhood of $x$ contains a point $a$ of $A$ not equal to $x$.
I am thinking about the following question: If $x$ is a limit point of $A$ does it hold that there exists a sequence $x_n $ in $A$ converging to $x$?
Please can you tell me if my thoughts are correct: If $x$ has infinitely many  neighborhoods then there are infinitely many different points $x_n \in A$ and for larger $n$ they are nearer $x$ therefore $x_n \to x$ and it is true. If $x$ only has finitely many neighborhoods then it is also true: take the smallest neighborhood and the point $x_n = a \in A$ in that neighborhood and then the constant sequence $a$ converges to $x$ (even though $x\neq a$).

REPLY [3 votes]: It's not true. 
A few examples:

Take the set $\Omega$ of ordinals $\le\omega_1$ in the order topology. Then $\omega_1$ is a limit point but not a sequential limit point of $\Omega\setminus\{\omega_1\}$. 
Take $=\Bbb R^{\Bbb R}$ with the product topology and set
$$
E=\{f\in \Bbb R^{\Bbb R} \mid f(x)=0\ \text{or}\ 1\ \text{and}\ f(x)=0\ \text{only finitely often}\ \}.
$$ 
Then the zero function is a limit point of $E$ but not a sequential limit point of $E$.
In $\ell_2$ with the weak topology, take $E=\{e_n+ne_m\mid m>n\}$.  Then the zero vector is a limit point of $E$ but not a sequential limit point of $E$.