TITLE: Proof: Boundary point not in set implies limit point
QUESTION [0 upvotes]: Problem:
Suppose that $\Omega \subseteq \mathbb{R}^n$. Prove that if $\vec{x}\notin \Omega$ and $\vec{x}$ is a boundary point of $\Omega$, then $\vec{x}$ is a limit point of $\Omega$.
Working:
Suppose that $\Omega \subseteq \mathbb{R}^n$. Let $\vec{x}$ be a boundary point of $\Omega$ such that $x\notin \Omega$. Since $\vec{x}$ is a boundary point of $\Omega$, we know that every ball around $\vec{x}$ contains points in $\Omega$ and points not in $\Omega$. 
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Therefore, there exists some sequence $\left\{\vec{x_i}\right\}$ in $\Omega$ with limit $\vec{x}$ and $\vec{x_i} \ne \vec{x}$. Hence, $\vec{x}$ is a limit point of $\Omega$.

REPLY [0 votes]: If $x \in \partial \Omega$, then for every $\epsilon >0$ you have $B(x,\epsilon)\cap (\Bbb R^n \setminus \Omega)\neq \varnothing$ and $B(x,\epsilon)\cap \Omega \neq \varnothing$.
We will use the $B(x,\epsilon)\cap \Omega \neq \varnothing$ part. Since $x \not\in \Omega$, this in fact says that for all $\epsilon > 0$, $(B(x,\epsilon)\setminus \{x\})\cap \Omega \neq \varnothing$, that is, $x \in \Omega'$.
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