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\flushleft\underline{Homework 3 --- MATH 4590 Spring 2018} \\
\begin{enumerate}
\item Prove that if $M=\mathbb{R}$ and $d(x,y)=|x-y|$, then $(M,d)$ is a metric space.  \\
\textit{Solution}: We must show three things: 
\begin{enumerate}[a.)]
\item (Positive definite) $d(x,y) \geq 0$ and $d(x,y)=0$ if and only if $x=y$
\item (Symmetric) $d(x,y)=d(y,x)$. and
\item (Triangle inequality) $d(x,y) \leq d(x,z)+d(z,y)$
\end{enumerate}
To prove $a.)$, note that the absolute value of any real number is always $\geq 0$. Furthermore, if $|x-y|=0$, then either $x-y=0$ or $-(x-y)=0$; in either case we have $x=y$. On the other hand, if $x=y$, then it is clear that $|x-y|=|0|=0$. This proves $a.)$. \\

To prove $b.)$, it is sufficient to note that the absolute value is symmetric, i.e. for any $x,y \in \mathbb{R}$, $|x-y|=|y-x|$. 

To prove $c.)$, follow the proof we did in class (or in the text). $\blacksquare$  
\item Prove the following ``$\epsilon$-principle": if for all $\epsilon > 0$ it holds that $|x-y| < \epsilon$, then $x=y$. \\
\textit{Proof}: Suppose that $x \neq y$. Then $|x-y| > 0$. Choose $\epsilon$ so that $0 < \epsilon < |x-y|$ (for instance, $\epsilon = \dfrac{|x-y|}{2}$ would work). But since for all $\epsilon$, $|x-y| \leq \epsilon$ we get
$$0 < \epsilon < |x-y| \leq \epsilon.$$
From this we must conclude $\epsilon < \epsilon$, which is absurd. Hence we have a contradiction. Therefore it must be the case that $x=y$. $\blacksquare$
\item Prove that if $M=\mathbb{R}^2$ and $d((x_1,y_1),(x_2,y_2))= \mathrm{max}\{|x_1-x_2|,|y_1-y_2|\}$, then $(M,d)$ is a metric space. Draw the ``unit circle" in this metric space, i.e. the set $\{x \in \mathbb{R}^2 \colon d(x,0)=1\}$. \\
\textit{Proof}: To show $a.))$, note that just as before, the absolute value function is non-negative by definition. Furthermore, if $d(x,y)=0$, then both $|x_1-x_2|=0$ and $|y_1-y_2|=0$, from which we may conclude that $x_1=x_2$ and $y_1=y_2$, in other words, $(x_1,y_1)=(x_2,y_2)$. Conversely, assume that $(x_1,y_1)=(x_2,y_2)$. Then 
\[d((x_1,y_1),(x_2,y_2))=\mathrm{max}\{|x_1-x_2|,|y_1-y_2|\}=\mathrm{max}\{0,0\}=0, \]
completing the proof of $a.)$.

To prove $b.)$, simply note that
$$d((x_1,y_1),(x_2,y_2))=\max\{|x_1-x_2|,|y_1-y_2|\}=\max\{|x_2-x_2|,|y_2-y_1|\}=d((x_2,y_2),(x_1,y_1)).$$

To prove $c.)$, we apply the triangle inequality for the absolute value in each piece of the maximum function:
$$|x_1-x_2| \leq |x_1-x_3|+|x_3-x_2|,$$
and
$$|y_1-y_2| \leq |y_1-y_3|+|y_3-y_2|.$$
Thus,
$$\begin{array}{ll}
d((x_1,y_1),(x_2,y_2)) &= \max \{|x_1-x_2|,|y_1-y_2|\} \\
&\leq \max \{|x_1-x_3|+|x_3-x_2|, |y_1-y_3|+|y_3-y_2|\} \\
&\leq \max \{|x_1-x_3|,|y_1-y_3|\} + \max\{|x_3-x_2|,|y_3-y_2|\} \\
&=d((x_1,y_1),(x_3,y_3))+d((x_2,y_2),(x_3,y_3)),
\end{array}$$
completing the proof. $\blacksquare$
\item Let $(M,d)$ be any metric space and let $(x_n)$ and $(y_n)$ be convergent sequences in $M$ with $x_n \rightarrow x$ and $y_n \rightarrow y$. Prove that the sequence $(d(x_n,y_n))$ is a convergent sequence in $\mathbb{R}$ with $d(x_n,y_n) \rightarrow d(x,y)$. \\
(\textit{note: you don't have a formula for the metric $d$ in $M$ nor do you know much about $M$ -- it could be a lot of things -- but you should be thinking of the metric in $\mathbb{R}$ as $d_{\mathbb{R}}(x,y)=|x-y|$}) \\
\textit{Proof}: Recall the reverse triangle inequality:
$$|d(x,z)-d(y,z)| \leq d(x,y).$$
Since $x_n \rightarrow x$, we know there exists an $N_x$ such that for all $n \geq N_x$, $d(x_n,x) < \dfrac{\epsilon}{2}$. Similarly, since $y_n \rightarrow y$, there is an $N_y$ such that for all $n \geq N_y$, $d(y_n,y) < \dfrac{\epsilon}{2}$. 

Now we shall prove that $d(x_n,y_n)\rightarrow d(x,y)$. To do this, let $\epsilon > 0$ and choose $N \geq \max \{ N_x, N_y\}$ and let $n \geq N$. It remains to show that $|d(x_n,y_n)-d(x,y)|<\epsilon$. To see this, compute
\[ \begin{array}{ll}
|d(x_n,y_n)-d(x,y)| &= |d(x_n,y_n)-d(x,y)+d(x_n,y)-d(x_n,y)| \\
&\leq |d(x_n,y_n)-d(x_n,y)| + |d(x_n,y)-d(x,y)| \\
&\stackrel{\text{reverse }\Delta \text{ ineq}}{\leq} d(y_n,y) + d(x_n,x) \\
&< \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} \\
&=\epsilon,
\end{array} \]
completing the proof. $\blacksquare$

\item Consider the metric space $(\mathbb{R},d)$ where $d(x,y)=|x-y|$. Prove that the function $f \colon \mathbb{R} \rightarrow \mathbb{R}$ with $f(x)=x$ is continuous. \\
\textit{Proof}: Let $\epsilon > 0$ and let $p \in \mathbb{R}$. Choose $\delta=\epsilon$. Then compute for $q \in \mathbb{R}$ with $|q-p|<\delta$:
$$|f(p)-f(q)|=|p-q|<\delta=\epsilon,$$
completing the proof. $\blacksquare$
\end{enumerate}
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