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\noindent Solutions for Homework 3
\vskip .2in


\noindent{\bf Page 202/ \#1:} 
\bigskip

\noindent (i) Consider points $(x_1,y_1),(x_2,y_2)\in\mathbf{R}^2$.
Then for the metric $\rho_{max}$, we have 
$$
\rho_{max} ((x_1,y_1),(x_2,y_2)) = \max\{|x_1 - x_2|,|y_1-y_2|\} \geq 0,
$$ 
since we are taking the maximum of two non-negative numbers.  Moreover,
the maximum is zero if and only if both absolute values vanish, in 
which case $x_1 = x_2$ and $y_1 = y_2$.  For the metric $\rho_1$,
we have
$$
\rho_1((x_1,y_1),(x_2,y_2)) = |x_1 - x_2| + |y_1 - y_2|
                            \geq 0
$$
since we are adding non-negative numbers.  Moreover, the sum is zero if and
only if both absolute values are zero, in which case $x_1 = x_2$ and
$y_1 = y_2$ again.
\medskip

\noindent (ii) For the metric $\rho_{max}$ we have
\begin{eqnarray*}
\rho_{max}((x_1,y_1),(x_2,y_2)) & = & \max\{|x_1 - x_2|,|y_1-y_2|\} \\ 
                                & = & \max\{|x_2 - x_1|,|y_2-y_1|\} \\
				& = & \rho_{max}((x_2,y_2),(x_1,y_1)).
\end{eqnarray*}
The argument for the metric $\rho_1$ is similar.
\medskip

\noindent (iii) Let $(x_3,y_3)\in\mathbf{R}^2$ be a third point.  Then 
\begin{eqnarray*}
\rho_{max}((x_1,y_1),(x_3,y_3)) & = & \max\{|x_1 - x_3|,|y_1-y_3|\}  \\ 
                                & \leq & \max\{|x_1 - x_2|+|x_2 - x_3|,
                                           |y_1 - y_2|+|y_2 - y_3| \} \\
				& \leq & \max\{|x_1 - x_2|,|y_1 - y_2|\} 
				     + \max\{|x_2 - x_3|,|y_2-y_3|\} \\
			        & = & \rho_{max}((x_1,y_1),(x_2,y_2))
				     + \rho_{max}((x_2,y_2),(x_3,y_3))
\end{eqnarray*}       
\bigskip

\noindent{\bf Page 202/ \#6}
\bigskip

\noindent The facts that $\rho(x,y) = \rho(y,x)$ and that $\rho(x,y)\geq 0$
are pretty clear.  Moreover, we have $\rho(x,y) = 0$ if and only if
$$
\left|\frac{y-x}{xy}\right| = 0
$$
which only happens if $x = y = 0$ (remember that we're assuming $x,y\neq 0$.
Finally,
\begin{eqnarray*}
\rho(x,z) = \left|\frac1x - \frac1y\right| \\
	& = & \left|\frac1x - \frac1y + \frac1y -\frac1z\right| \\
        & \leq & \left|\frac1x - \frac1y\right| 
              + \left|\frac1y -\frac1z\right| \\
  	& = & \rho(x,y) + \rho(y,z), 
\end{eqnarray*}
for all $x,y,z\in (0,\infty)$, so the triangle inequality holds, too.
\qed
\newpage

\noindent{\bf Page 208/ \#4:}
\bigskip

\noindent The sets in (b) and (c) are complete, but the one in (a) is not.
For instance, the sequence $\{(1-1/n,0)\}_{n=1}^\infty$ converges to
the point $(1,0)\in\mathbf{R}^2$.  Therefore it is Cauchy.  But
$(1,0)$ does not belong to the original set, so the set is not complete.
\bigskip


\noindent{\bf Other Problem A.}

\bigskip
\noindent Many answers to this question are possible.  Here's mine.
Let $f_n:[0,1]\to\mathbf{R}$ be given by $f_n(x) = x^n$.  Then
$f_n(x)$ converges pointwise to the function $f:[0,1]\to\mathbf{R}$
satisfying $f(1) = 1$ and $f(x) = 0$ for $x\in[0,1)$.  If $f_n$
converged uniformly, then it would converge pointwise to the same
limiting function, and since $f_n$ is continuous for every $n$,
the limiting function would be continuous.  But the pointwise 
limit $f$ is not continuous, so we conclude that $f_n$ does not
converge uniformly.  On the other hand, if 
$\rho(f,g) = \int_0^1 |f(x) - g(x)|\,dx$, then
$$
\rho(f_n,f) = \int_0^1 |x^n - 0| = \frac{1}{n+1} 
$$
for every $n$ (note that the fact that $f(1) = 1$ won't affect the 
integral).  So $\lim_{n\to\infty} \rho(f_n,f) = 0$.
Therefore $f_n$ converges to $f$ with respect to the metric $\rho$.
\bigskip

\noindent {\bf Other Problem B.}
\bigskip 

\noindent (i) Suppose that $\{a_n\}_{n=1}^\infty$ converges uniformly
to an element $a\in\mathcal{M}$ with respect to the metric $\rho_1$.
Then given $\epsilon>0$, I set $\epsilon' = \epsilon/c_2$ and choose
$N\in\mathbf{N}$ large enough that $n\geq N$ implies that
$$
\rho_1(a_n,a) < \epsilon'.
$$
Then for $n\geq N$, we also have
$$
\rho_2(a_n,a) \leq c_2\rho_1(a_n,a) < c_2\epsilon' = \epsilon.
$$
Hence, $a_n\to a$ with respect to the metric $\rho_2$.  Repeating
the argument with $\rho_1$ and $\rho_2$ switched and with 
$c_2$ replaced by $1/c_1$ takes care of the other direction.
\qed
              
\medskip
\noindent (ii) Let 
$\mathbf{x} = (x_1,\dots,x_n), \mathbf{y} = (y_1,\dots,y_n) \in\mathbf{R}^n$
be points.  Let $j$ between $1$ and $n$ be chosen so that $|x_j - y_j|$
is as large as possible.  Then we have
\begin{eqnarray*}
\rho_{max}(\mathbf{x},\mathbf{y})^2 & = &   |x_j - y_j|^2 \\
	& \leq & \sum_{k=1}^n |x_k - y_k|^2 
        	= \rho_2(\mathbf{x},\mathbf{y})^2. \\
        & \leq & \sum_{k,\ell=1}^n |x_k-y_k||x_\ell - y_\ell| \\
	& = & \left(\sum_{k=1}^n |x_k - y_k|\right)^2 
   		= \rho_1(\mathbf{x},\mathbf{y})^2 \\
	& \leq & (n|x_j - y_j|)^2 
 		= n^2 \rho_{max}(\mathbf{x},\mathbf{y})^2.
\end{eqnarray*}
Taking square roots (all distances are non-negative so absolute value
signs are unnecessary) and summarizing gives:
$$
\rho_{max}(\mathbf{x},\mathbf{y})
	\leq \rho_2(\mathbf{x},\mathbf{y})
	\leq \rho_1(\mathbf{x},\mathbf{y})
	\leq n\rho_{max}(\mathbf{x},\mathbf{y}).
$$
So $\rho_1$ and $\rho_2$ are each comparable to $\rho_{max}$ (and
hence comparable to each other as well).  Therefore by part (i) a sequence
converges with respect to one of these three metrics if and only if it 
converges with respect to the other two.
\qed

\medskip 
\noindent (iii) By part (ii) of this problem, we know that the Euclidean
metric $\rho_2$ is comparable to the box metric $\rho_{max}$.
So if
$\mathbf{x}_j \to \mathbf{y} = (y_1,\dots,y_n) \in\mathbf{R}^n$,
we then have for each $1\leq k\leq n$ and all $j\in\mathbf{N}$ that
$$
0 \leq |x_{j,k} - y_k| \leq \rho_{max}(\mathbf{x}_j,\mathbf{y}) \leq
c_1\rho_2(\mathbf{x}_j,\mathbf{y}),
$$
for some $c_1>0$.
Since the right side tends to zero as $j\to\infty$, the Squeeze Theorem
tells us that $\lim_{j\to\infty}|x_{j,k} - y_k| = 0$---in other words,
for each $k$ between $1$ and $n$, we have $\lim_{j\to\infty} x_{j,k} = y_k$.

Now suppose that $\mathbf{x}_n\to\mathbf{y}$ coordinate-wise.  Then
we can use comparability of the metrics $\rho_1$ and $\rho_2$ to obtain
$$
0 \leq \rho_2(\mathbf{x}_j,\mathbf{y}) \leq c_2\rho_1(\mathbf{x}_j,\mathbf{y}) 
  = c_2(|x_{j,1} - y_1| + \dots + |x_{j,n} - y_n|).
$$
By assumption, all $n$ terms on the right side go to zero as $j\to\infty$,
so the Squeeze Theorem allows us to conclude that 
$\rho_2(\mathbf{x_j},\mathbf{y})$ goes to zero as well.  In other words,
$\mathbf{x_j}\to\mathbf{y}$ in the Euclidean metric.
\qed




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