Are singletons limit points? So I just started learning topology and I came across this problem.

Consider the topological space $(X,\tau)$where the set $X=\{a,b,c,d,e\}$ and $\tau=\{X,\emptyset,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e\}\}$
Determine the limit points of

i) $\{a\}$
Is $a$ a limit point for itself since the singleton a is an open set in the topology.
From the definition of limit point of a set  every neighbourhood of the point must contain a point of the set that id  different from a however when considering the singleton there are no different points of a so what is the correct answer im very confused.
Thanks in advance
 A: The limit points of a set $S$ are not an intrinsic property of $S$. They depend on how the set $S$ is immersed into an ambient topological space $X$.
To answer your question, you have to check if there are points $x\in X$ such that every punctured neighborhood of $x$ (i.e., the result of removing from an open neighborhood of $x$ the point $x$) passes through $a$.

*

*Now the punctured neighborhoods of ${a}$ cannot contain $a$.

*So let's examine the punctured neighborhoods of ${b}$. These are
$\{c,d,e\}$ and $X-\{b\}$. At least one does not contain $a$. So $b$
is not a limit point of $\{a\}$.

*The punctured neighborhoods of ${c}$ are $\{d\}$ and others. But you
already found one that does not contain $a$ so $c$ cannot be a limit
point of $a$.

And so on...
If you want some intuition about the concept of a limit point, then you can think that an element $x\in X$ is a limit point of a subset $S\subseteq X$ if $x$ can be well approximated by $S-\{x\}$.
A: The set of limit points of $\{a\}$ is empty.
