Why is the closure of a set not the same as its limit points? I'm getting confused about the notion of a limit point in a topological space.
In my notes I have:

If $L(A)$ is the set of all limit points of $A$, then $L(A)\subseteq
 \text{closure}(A)$, but the sets are not necessarily equal.

But the definition given is:


*

*$x\in X$ is a limit point of $A$ if there is a sequence $x_n\in A$ the
converges to $x$


Doesn't this imply $A\subseteq L(A)$, so $L(A)=\text{closure}(A)?$
Another set of notes I have has:


*

*$x\in X$ is a limit point of $A$ if any open neighbourhood $U$ containing $x$ has $U\bigcap A\neq\emptyset$


Again don't we have $A\subseteq L(A)?$
I can't see where my logic is going wrong...
 A: Your logic is fine: both of those definitions do imply that $A\subseteq L(A)$. The problem is that those definitions are non-standard. The usual definitions are as follows. First, for metric spaces:

$x\in X$ is a limit point of $A$ if there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A\color{red}{\setminus\{x\}}$ that converges to $x$.

In other words, the constant sequence at $x$ isn’t allowed. And for topological spaces in general:

$x\in X$ is a limit point of $A$ if any open nbhd $U$ containing $x$ has $U\cap\big(A\color{red}{\setminus\{x\}}\big)\ne\varnothing$.

That is, each open nbhd of $x$ must contain some point of $A$ other than $x$.
For example if $X=\Bbb R$, and $A=\{0\}\cup(1,2)$, $L(A)=[1,2]$, which does not contain $0$: $0$ is not a limit point of $A$ because no sequence in $(1,2)$ converges to it (first definition), or because $(-1,1)$ is an open nbhd of $0$ disjoint from the rest of $A$ (second definition).
A: Check the definition of limit point again. An important part of the definition is that $x_n\neq x$, and the $U$ is a neighborhood of $x$, but without $x$ itself (rather, a deleted neighborhood of $x$). The reason this is important is that we do not want isolated points to be limit points. For example, in the set $[0,1]\cup\{2\}$, the point $2$ should not be a limit point.
