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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 $d(p,q)<r$, then $q \in S$ (in other words: you can make a small enough ``ball" around $p$ that all points in the ball are also in $S$).
\begin{enumerate}
\item Let $(M,d)=(\mathbb{R}^2,d)$, where $d$ is the Euclidean metric
\[ d((x_1,y_1),(x_2,y_2))=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.\]
Is the set $\mathbb{R}=\{(x,0) \colon x \in \mathbb{R}\} \subset \mathbb{R}^2 \}$ a closed, an open, both closed and open, or neither a closed nor an open subset of $\mathbb{R}^2$? Why or why not? (\textit{hint: the set of points a certain distance away from a point in $\mathbb{R}^2$, in this metric, looks like a little disk (not including its boundary)})

\item Let $(M,d)$ be a metric space and let $S=\emptyset$ (i.e. the empty set). Is $\emptyset$ an open set, a closed set, both open and closed, or neither open nor closed? Why or why not?

\item Let $(M,d)$ be a metric space and let $x \in M$ be some arbitrary point. Prove that $S=\{x\}$ is a closed set.
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