# Statement(s) > [!NOTE] Theorem (Complement of Open Subset of $\mathbb{R}^n$ is Necessarily Closed) > A subset $X \subset \mathbb{R}^n$ is [[Open subsets of metric spaces|open]] if, and only if, its the [[Set Difference|set difference]] $\mathbb{R}^n\setminus X$ is [[Closed Sets|closed]]. **Notation**: We may write $X^c$ instead of $\mathbb{R}^n \setminus X$. # Proof(s) ###### Proof of Complement of Open Subset of $\mathbb{R}^n$ is Necessarily Closed ($\implies$) We take $X$ is open to mean that $\forall x\in X: \exists \varepsilon>0: \forall y\in \mathbb{R}^n:\lVert y-x \rVert<\varepsilon \implies y\in X.$Suppose $X^c$ is non-empty. Let $(x_{j})\subset X^c$ which converges to $x\in \mathbb{R}^n$. BWOC, suppose $x \not\in X^c$. Since $X$ is open, there exists $\varepsilon>0$ such that $\lVert y-x \rVert<\varepsilon \implies y\in X$. But $x_{j}\to j$ as $j\to \infty$ so there exists $N\in \mathbb{N}$ such that $j \geq N \implies \lVert x_{j}-x \rVert <\varepsilon \implies x_{j} \in X $which contradicts the assumption that $(x_{j})\subset X^c$, i.e. $x_{j}\not \in X$ for all $j\in \mathbb{N}$. ($\impliedby$) Conversely, let $X\subset \mathbb{R}^n$ be a closed which we take to mean that for all $(x_{j})\subset U$ such that $\lim_{ j \to \infty }x_{j}=x\in \mathbb{R}$ implies $x\in X$. To prove that $X^c$ is open, we have to show that, given $y\in X^c$, there exists $\varepsilon>0$ such that for all $x\in X$, $\lVert x-y \rVert \geqslant \varepsilon$. BWOC, suppose otherwise, then, we could find $y\in X^c$ and a sequence $x_{j}$ in $X$ such that $\lVert x_{j} - y \rVert < \frac{1}{j}$. But then $\lim_{ j \to \infty }x_{j}=y$ and, since $X$ is closed, $y\in X$ which contradicts the fact that $y\in X^c$.