Let X be a topological space and let $A \subset X$.
Let $\overline{A}$ be the closure of A and let $\partial A = \{p \in X \mid \forall U \ni p, U \text{ open}:U \cap A \ne \emptyset \text{ and } U \cap A^c \ne \emptyset\}$ be the boundary of A.
I wanna prove that $\partial A \subset \overline{A}$.
I've sketched the following proof, but I don't know if it's right.
Proof: If $a \in \partial A$, then $\forall U \ni a, U \text{ open}$ we have also that $U \cap A \ne \emptyset$. The are two cases:
- $a \in A$ (in this case we're not sure that $(U \backslash \{a\}) \cap A \ne \emptyset$).
- $a \notin A$. So there exists $a' \ne a$ such that $a' \in (U \backslash \{a\}) \cap A$, then $a \in A'$, where $A'$ is the derived set of $A$
In either case, $a \in A \cup A' = \overline{A}$. Finally $\partial A \subset \overline{A}$
Best Answer
No need to consider two cases. By definition of $\partial A$, every open set to which $a$ belongs contains elemeents of $A$ and therefore $\in\overline A$.