$(X,\tau)$ is a Topological space where as $\tau$ is the topology on $X$, and
$A \subseteq X$ is an arbitrary subset, $\overline{A}$ is the closure $A^{\circ}$ the interior and $\partial A$ is the boundary of the set $A$.
Munkres topology defines the derived set of $A$, or a.k.a the set of all limit points of $A$ as:
$$A^{\prime}=\{x \in X \mid U \cap (A\backslash \{x\}) \neq \emptyset ,\ U \ \text{is an open subset containing} \ x \}.$$
Then in Munkres topology states that $\overline{A}=A \cup A^{\prime}$
which makes sense to me.
However if I compare that to the boudary of $A$ with
$$\partial A = \overline{A} \backslash A^{\circ}$$ which would lead me to
$$\overline{A}= A^{\circ} \cup \partial A = A \cup \partial A$$ since $A^{\circ} \subseteq A$. This brings me to the question of the relationship between $\partial A$ and $A^{\prime}$. I have read that they are not the same, but I don't understand how or why and if they are not the same then I think one must contain the other one right ?
I don't know how to approach this does someone have an idea? Also there a third set which I am confused about the set of all contact points but that totally lost me.
Best Answer
Let's try to understand through example :
Consider $(\Bbb{R}, \tau_{std}) $ and $A=(0, 1) $
Then $A'=[0, 1]$
And $\partial(A) =\{0,1\}$
So it's clear that not all limit points are in the boundary.
Is $\partial(A)\subset A'$ ? (N0)
Choose $x\in \partial(A) $ and a open set $x\in U$ then $U\cap A\neq \emptyset$ and $U\cap (X\setminus A) \neq \emptyset$
But to be a limit point of $A$ , $U$ have to contain a point of $A$ other than $x$.
$U$ may not contain such point.
For an example choose $A=\{0,1\}$ . Then both $0, 1$ are boundary points as any open interval containg $0$ intersect both $A$ and $\Bbb{R}\setminus A$ but none of them are limit points.
Task: A finite subset of $\Bbb{R}$ can't have any limit point.
Neither way set inclusion true.