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\flushleft\underline{Homework 4 --- MATH 4590 Spring 2018} \\
Let $(M,d)$ be a metric space and let $S \subset M$. A point $p \in M$ is called a limit of $S$ provided that there exists a sequence $(p_n)$ \underline{in $S$} such that $p_n \rightarrow p$. We say that $S$ is closed if it contains all of its limits. We say that $S$ is open if $\forall p \in S \exists r > 0$ such that if $d(p,q) 0$ and choose $N=1$, then compute
$$d(p_n,p)=d(p_n,p)=d(x,x)=0 < \epsilon.$$
But since $(p_n)$ is the only sequence in $S$ and its limit lies in $S$, it follows that the limit (i.e. $x$) of all convergent sequences (there's only one) in $S$ lies in $S$. This means $S$ contains all of its limits. This means that $S$ is closed.
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