$A'$ 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?
Best Answer
$\partial A = \bar A - \overset{\,\,\,\circ} A.$ Thus
$\overset{\,\,\,\circ} A \cup \partial A = \bar A$ because
$\overset{\,\,\,\circ} A \subseteq \bar A.$
In fact, because $\overset{\,\,\,\circ} A \subseteq A \subseteq \bar A,$
$A \cup \partial A = \bar A.$