Answer to Question #114844 in Differential Geometry | Topology for Sheela John

Question

Prove that the boundary of a subset A of a metric space X is always a closed set

Answer 1

The boundary $\partial A$ of $A$ is the closure of $A$ minus the interior of $A$. We need to prove that every $x\in X$ such that every ball around $x$ intersects $\partial A$ belongs to $\partial A$. Let $x\in X$ such that every ball around $x$ intersects $\partial A$.

• Since $\partial A$ is included in the closure of $A$, every ball around $x$ intersects the closure of $A$. Since every closure is a closed set, $x$ belongs to the closure of $A$.
• Assume that $x$ is in the interior of $A$. Since the interior of $A$ is an open set, there is a ball $B$ around $x$ that is included in the interior of $A$. Hence $B$ does not intersect $\partial A$. This contradicts the condition that every ball around $x$ intersects $\partial A$. Therefore, $x$ is not in the interior of $A$.

Since $x$ belongs to the closure of $A$ and is not in the interior of $A$, $x\in \partial A$ .