Boundary Criterion for Open/Closed Sets in a Topological Space

# Boundary Criterion for Open/Closed Sets in a Topological Space

Proposition 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is open if and only if $A \cap \partial A = \emptyset$, that is, $A$ and its boundary $\partial A$ are disjoint. |

**Proof**$\Rightarrow$ Let $A$ be open. Then $A = \mathrm{int} (A)$. Observe that if $x \in \partial A$ then $x \in \overline{A} \setminus \mathrm{int}(A)$, so $x \not \in \mathrm{int}(A) = A$. Thus:

\begin{align} \quad A \cap \partial A = \emptyset \end{align}

- $\Leftarrow$ Suppose that $A \cap \partial A = \emptyset$. Let $x \in A$. Then $x \not \in \partial A$. So $x \not \in \overline{A} \setminus \mathrm{int}(A)$. So $x \in \mathrm{int}(A)$ and thus, $A = \mathrm{int}(A)$, i.e., $A$ is closed.

Proposition 2: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is closed if and only if $\partial A \subseteq A$, that is, the boundary $\partial A$ is a subset of $A$. |

**Proof:**$\Rightarrow$ Let $A$ be closed. Then $X \setminus A$ is open, and by proposition 1 we have that $X \setminus A$ and its boundary, $\partial (X \setminus A)$ are disjoint. But $\partial A = \partial (X \setminus A)$. Therefore, $X \setminus A$ and $\partial A$ are disjoint. Therefore $\partial A \subseteq A$.

- $\Leftarrow$ Suppose that $\partial A \subseteq A$. Then $(X \setminus A) \cap \partial A = \emptyset$. Equivalently, $(X \setminus A) \cap \partial (X \setminus A) = \emptyset$. So by proposition 1, $X \setminus A$ is open. Thus $A$ is closed. $\blacksquare$