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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)}) \\
\textit{Solution}: It is a closed set and it is not an open set. It is closed because given any converging sequence $p_n$ in $\mathbb{R}$ where $p_n \rightarrow p$, the point $p$ lives in $\mathbb{R}$: \\
\begin{figure}[h]
\begin{center}\begin{tikzpicture}[scale=3]
\draw (0,0) -- (2,0);
\draw[fill=black] (0.5,0) circle (0.05em);
\draw (0.5,-0.12) node {$\frac{1}{2}$};
\draw[fill=black] (0.6666,0) circle (0.05em);
\draw (0.6666,-0.12) node {$\frac{2}{3}$};
\draw[fill=black] (0.75,0) circle (0.05em);
\draw (0.75,-0.12) node {$\frac{3}{4}$};
\draw[fill=black] (0.8,0) circle (0.05em);
\draw[fill=black] (0.83333333,0) circle (0.05em);
\draw[fill=black] (0.85714,0) circle (0.05em);
\draw[fill=black] (0.875,0) circle (0.05em);
\draw[fill=black] (0.88888888,0) circle (0.05em);
\draw[fill=black] (0.9,0) circle (0.05em);
\draw[fill=black] (0.90909090909,0) circle (0.05em);
\draw[fill=black] (0.916666666,0) circle (0.05em);
\draw[fill=black] (0.9230769,0) circle (0.05em);
\draw[fill=black] (0.9285714,0) circle (0.05em);
\draw[fill=black] (0.944444444,0) circle (0.05em);
\draw[fill=black] (0.9523809,0) circle (0.05em);
\draw[fill=black] (0.9666666666,0) circle (0.05em);
\draw[fill=black] (0.971428,0) circle (0.05em);
\draw[fill=black] (0.9777777777,0) circle (0.05em);
\draw[fill=black] (0.9888888888,0) circle (0.05em);
\draw[fill=red] (1,0) circle (0.05em);
\draw (1,-0.12) node {$1$};
\end{tikzpicture}\end{center} 
\caption{The sequence $1-\dfrac{1}{n}$ in $\mathbb{R}$, which converges to $1$.}
\end{figure}
That fact can be proven: \\
\textbf{Theorem}: If $p_n=(x_n,0)$ is a convergent sequence in $(\mathbb{R}^2,d)$, where $d$ is the Euclidean metric, then  its limit is of the form $(q,0)$ (i.e. it lies in $\mathbb{R} \subset \mathbb{R}^2$). \\
\textit{Proof}: Suppose that the limit was not in $\mathbb{R} \subset \mathbb{R}^2$, then the limit would be of the form $(q,w)$ with $w \neq 0$. Note that the ball of radius $\dfrac{|w|}{2}$ does not touch the $x$-axis, i.e. the points 
$$M_{\frac{w}{2}}((q,w)) \cap \mathbb{R}=\left\{ (a,b) \in \mathbb{R}^2 \colon d((q,w),(a,b)) < \dfrac{w}{2} \right\} \cap \mathbb{R} = \emptyset.$$

\begin{figure}[h]
\begin{center}\begin{tikzpicture}
\draw (-1,0) -- (3,0); 
\draw (0,-1) -- (0,3);
\draw[dashed,fill=gray,opacity=0.5] (1,1) circle (0.5);
\draw[fill=blue] (1,1) circle (0.2em);
\draw (1.3,1.3) node {$(q,w)$};
\end{tikzpicture}\end{center}
\end{figure}

\newpage

Choose $\epsilon = \dfrac{|w|}{2}$. If the sequence $p_n$ was really converging to $(q,w)$, then there would be some $N \in \mathbb{N}$ such that for all $n \geq N$, $d((x_n,0),(q,w))<\epsilon$. However, this cannot happen since the points of the form $(x_n,0)$ are on the $x$-axis but all points in $M_{\frac{w}{2}}((q,w))$ are strictly above the $x$-axis. This is a contradiction, and therefore we must have the limit of $p_n$ lying on the $x$-axis. $\blacksquare$


\hrule

The set is not open because if you pick any point $q \in \mathbb{R}$, any ``ball" around $p$ does not contain only points of $\mathbb{R}$ -- there is always a $q \in \mathbb{R}^2$ such that $q$ is in the ball but not in $\mathbb{R}$: 

\begin{figure}[h]
\begin{center}\begin{tikzpicture}
\draw (-1,0) -- (3,0); 
\draw (0,-1) -- (0,3);
\draw[fill=red] (1,0) circle (0.2em);
\draw (1,-0.3) node {$(1,0)$};
\draw[dashed,fill=gray,opacity=0.5] (1,0) circle (2em);
\draw[fill=blue] (1,0.3) circle (0.2em);
\draw (1.3,0.3) node {$q$};
\end{tikzpicture}\end{center}
\end{figure}

\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? \\
\textit{Solution}: $\emptyset$ is both open and closed. It is closed because for all convergent sequences of point in $\emptyset$ (there are none...), the limit (which doesn't exist since no sequence exist) is also in $\emptyset$. It is open because given any point in $\emptyset$ (there isn't one), there is a ball small enough so that all points in the ball are also points of $\emptyset$ (but any ball is precisely $\emptyset$ here and $\emptyset \subset \emptyset$).

\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. \\
\textit{Solution}: Let $p_n$ be a sequence in $S$. Since $S$ contains only one point, we know this sequence \textit{must} be $p_1=x, p_2=x, p_3=x, \ldots$. This sequence converges to $x$: let $\epsilon > 0$ 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.
\end{enumerate}

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