Boundary point sequence proof > Let $A\subset \mathbb{R}^p$ and $x\in \mathbb{R}^p$. Then $x$ is a boundary point of $A$ if and only if there is a sequence $(a_n)$ of elements in $A$ and a sequence $(b_n)$ of elements in $ \cal{C} (A)$ (where $\cal{C} (A)$ is the complement of $A$) such that $$lim(a_n)=x=lim(b_n)$$ Let $x$ be a boundary point of $A$, then $x$ is not an interior point to $A$. Since $x$ is a boundary point, does it necessary mean that it intersects $ \cal{C} (A)$? If not, how can I continue proceeding proving this?

A boundary point is either in $A$ or not. (For example, the boundary of $(-\infty, 0)$ is $\\{0\\}$, the boundary of $(-\infty, 0]$ is also $\\{0\\}$.)

The key point is that if $x$ is in the boundary of $A$, then any **open** set containing $x$ must intersect both $A$ **and** $A^c$.

To wit, suppose $x \in \partial A = \overline{A} \cap \overline{A^c}$. Let $x \in U$, where $U$ is open. Then $U$ must intersect $A$ and $A^c$. To see the latter, suppose $U$ does not intersect $A$, then $A \subset \overline{A} \subset U^c$ (since $U^c$ is closed), which contradicts $x \in U$. A similar argument applies to $A^c$.

Now choose $U_n = B(x,\frac{1}{n})$, and pick $a_n \in U_n \cap A$, $b_n \in U_n \cap A^c$. Clearly $a_n \in A$, and $a_n \to x$. Similarly $b_n \in A^c$ and $b_n \to x$.