Def'n: Let $(X, T )$ be a topological space and let A be a subset of X. A point x in X is a limit point of A (or cluster point or accumulation point) if every open set U with $x \in U $ also contains a point of A other than x itself. We let $A'$ denote the set of limit points of A.
Theorem: Let $(X, T )$ be a topological space. If A is any subset of X, then $Cl(A) = A ∪ A' $
This is somewhat confusing to me because looking at the definition of $A'$ its hard to understand when $CL(A) \neq A' $ except when The topology contains singletons. Is my understanding correct or is there anther instance where this happens?
Can it be listed by class $T_0, T_1 , T_2 $ (ofc other than the discreet topology) for example i believe $Cl(A) = A' $ if the topology is metric.
The closure of $A$ can be alternatively defined as all adherent points of $A$: all $x$ such that for any open set $O$ that contains $x$, $O \cap A \neq \emptyset$. There are two cases for being an adherent point $x$: $x \in A$ and in that case $O \cap A \ni x$ so it's clearly non-empty, and so trivially all points of $A$ are adherent points of $A$, and $x \notin A$ and in that case any open set $O$ that contains $x$ must intersect $A$ in a point not equal to $x$ obviously (as $x \notin A$, $x \notin O \cap A$), so those adherent points are limit points, so are in $A'$.
This immediately shows (using this definition of closure) that $\operatorname{Cl}(A) = A \cup A'$.
Points of $A$ need not be limit points, as they can be isolated points of $A$, which means that there is an open set of $X$ such that $O \cap A = \{x\}$. This happens for all finite subsets of $T_1$ spaces, and sets like $\mathbb{Z}$ in the reals (usual topology). Points of $A$ are either in $A'$ or an isolated point of $A$. So we either use the adherent point definition or the $A \cup A'$ definition.
The adherent point definition makes it relatively easy to see that $\operatorname{Cl}(A)$ is the smallest closed subset that contains $A$. To force $A \subseteq \operatorname{Cl}(A)$ we need to include $A$ explicitly, only taking the limit points could make $A'$ smaller than $A$ (as in the case of $A = \mathbb{Z}$ which has $A' = \emptyset$).