Convergence and Closed Sequences Suppose $C$ is a subset of $\Bbb R^n$ with the following property: whenever $\{x_i\}_{i=1}^\infty$ is a sequence in $C$ that converges to a point $p\in R^n$, it follows that $p\in C$. Prove that $C$ is closed.
 A: Let $x\in A = \mathbb{R}^n\setminus C$ be given. We will show that there exists a neighborhood $U$ of $x$ with $U\subseteq A$. Suppose toward a contradiction that no such neighborhood exists. Then for all $n\in\mathbb{N}$ the open ball $B(x, \frac{1}{n})$ must have nonempty intersection with $C$. So for all $n\in\mathbb{N}$ we can choose $x_n\in C\cap B(x, \frac{1}{n})$. But then $x_n$ is a sequence of points in $C$ converging to $x$ so by assumption $x\in C$. But this contradicts our assumption that $x\in \mathbb{R}^n\setminus C$. So for all $x\in A$ there exists some neighborhood $U$ of $x$ contained in $A$ and in particular $A$ is open. Thus $C$ is closed since it is the complement of an open set.
A: In general,$S$$\subseteq$ $X$ is closed in a metric space $(X,d)$ iff a sequence {$x_n$} in $S$ and $x_n$ $\rightarrow$ $x$ (in $X$) implies that $x$ $\in$ $S$ .
Assume every convergent sequence in $C$ converges to a point in $C$.Let $p$ $\in$ $\overline{C}$.
By definition, for any $\epsilon$ $>$0, there exists an open ball $B(p,\epsilon) $ s.t $B(p,\epsilon)\cap$ $C$ $\neq$ $\emptyset$. Then, it's possible to construct sequences {$x_n$} & {$B(p,$1/$n$)} s.t $x_n$ $\in$ $B(p,$1/$n$) for all $n$. Now, $x_n$ $\rightarrow$ $p$. More importantly, $p$ $\in$ $C$. So, $C$ contains all the points in its closure. Hence, $C$ is closed.
