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

Question

$A'$ 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?

Answer 1

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.

Answer 2

$\partial A = \bar A - \overset{\,\,\,\circ} A.$ Thus
$\overset{\,\,\,\circ} A \cup \partial A = \bar A$ because
$\overset{\,\,\,\circ} A \subseteq \bar A.$

In fact, because $\overset{\,\,\,\circ} A \subseteq A \subseteq \bar A,$
$A \cup \partial A = \bar A.$

Answer 3

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