Show that $\overline{A} \subseteq \overset{\,\,\,\circ} A\cup \partial A$ $A'$ denotes the set of limit points of $A$ and $\partial A$ is the boundary of $A$.
$\overset{\,\,\,\circ} A$ denotes the set of interior points of $A$.

Show that $\overline{A} \subseteq \overset{\,\,\,\circ} A \cup \partial A$.

Suppose  $x \in \overline{A} \implies x \in A \cup A'$.
Suppose $x \not\in A$ (What if $x\in A$), then $x\in A'$.
$\implies$ for all open sets $U$ containing $x$, $U$ has another point in $A$ not equal to $x$.
$\implies U \cap A \neq \varnothing$ and $U \cap (X\setminus A) \neq \varnothing$.
$\implies x \in \partial A$.
This proof is not complete; what if $x\in A$? Can someone help me out?
 A: $\partial A = \bar A - \overset{\,\,\,\circ} A.$  Thus
$\overset{\,\,\,\circ} A \cup \partial A = \bar A$ because
$\overset{\,\,\,\circ} A \subseteq \bar A.$
In fact, because $\overset{\,\,\,\circ} A \subseteq A \subseteq \bar A,$
$A \cup \partial A = \bar A.$
A: If $x \in A \cup A'$ then suppose $x \notin\overset{\,\,\,\circ} A$, then for all open neighbourhoods $B$ (or balls around $x$ if you prefer) of $x$ we have that $B \nsubseteq A$, so for all such $B$: $B \cap (X\setminus A) \neq \emptyset$. Both $x \in A$ and $x \in A'$ imply that also for all such $B$: $B \cap A \neq \emptyset$.
So combined this means that $x \in \partial A$: all its neighbourhoods/balls intersect $A$ and its complement.
This shows $\overline{A} \subseteq\overset{\,\,\,\circ} A \cup \partial A$.
A: Let $A^c$ denote the complement of $A.$
The definition of $\partial A$ is $\bar A \cap \overline {A^c}.$
If $p\in \bar A$ then....
(i) If there exists an open set $U$ such that $p\in U$ and $U\cap A^c=\emptyset$ then $U\subset A$  so $p\in U\subset \overset{\,\,\,\circ} A$ so $p\in\overset{\,\,\,\circ} A$.
(ii) If every open $U$ such that $p\in U$  has non-empty intersection with $A^c$ then $p\in \overline {A^c}$ so $p\in \bar A\cap \bar {A^c}=\partial A.$
In both cases we have $p\in \overset{\,\,\,\circ} A\cup \partial A.$
Therefore $\bar A\subset \overset{\,\,\,\circ} A\cup \partial A.$
BTW, since $$\overset{\,\,\,\circ} A\cup \partial A\subset A\cup \partial A=A\cup (\bar A \cap \overline {A^c})\subset$$ $$\subset A\cup \bar A=\bar A$$ we also have $\overset{\,\,\,\circ} A\cup \partial A \subset \bar A.$ So we have $\bar A=\overset{\,\,\,\circ} A\cup \partial A.$
All of this  holds in every topological space.
