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Let $X$ be a topological Hausdorff space. Let $A\subset X$ some subset.

Let $L(A)$ denote the set of limit points of $A$.

Is it true that $L(A)$ is equal to the closure of $A$ $(=\bar{A})$ ?

I know that $L(A)\subset\bar{A}$ but I don't know how to prove or disprove the equlity.

I used this as a 'fact' in order to solve some question, for example with the retract of a Hausdorff space is closed and with the closed graph theorem, but now I am confused about wheter this assumption is true or not.

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    put $A=\{ 0 \} \subseteq \mathbb{R}$. $\phi =L(A)\subsetneq \bar{A}=\{ 0 \} $ – 2017-01-30
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    @Basti Why is $L(A)$ empty? – 2017-01-30
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    mmh check the definition of limit point: A point x in X is a limit point of A if every neighbourhood of x contains at least one point of A different from x itself. So, $L(A)=\phi$ because $A$ has only one element. – 2017-01-30
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    But sequence x, x, x, x, ... converges to x, so 0 belongs to L(A) – 2017-11-25
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    I think what the question author is asking is whether closure(A) equals A and its limit points. – 2017-11-25

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This is false for at least one definition of limit point.

If we define (as is very common) $x$ is a limit point of $A \subseteq X$ in the space $(X,\mathcal{T})$ iff for every $O \in \mathcal{T}$: if $x \in O$, then $O \cap (A \setminus \{x\}) \neq \emptyset$.

Another definition (which I learnt as adherent point): $x$ is an adherent point of $A \subseteq X$ in the space $(X,\mathcal{T})$ iff for every $O \in \mathcal{T}$: if $x \in O$, then $O \cap A \neq \emptyset$.

Then we have that $$\overline{A} = \{x: x \text{ adherent point of } A \}\text{.}$$ And $$\{x: x \text{ limit point of } A \} \subseteq \overline{A}\text{,}$$ without there being equality. The set $\{x: x \text{ limit point of } A \} $ is also denoted $A'$, the so-called derived set of $A$. $A$ could have isolated points, i.e. points $x \in A$ such that $\exists O \in \mathcal{T}: O \cap A = \{x\}$, and then for all $A$: $\overline{A} = A' \cup \{x: x \text{ isolated in } A\}$.

E.g. $A =\mathbb{Z}$ in the reals in the usual topology has $A' =\emptyset$, i.e. no limit points, but it equals its set of adherence points, so is closed.