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\title{hw3-solutions}
\begin{document}
\centerline{\Large Math 171 Homework 3}
\centerline{\small (due April 22)}
\vskip .2in
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 33.3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {33.3}
Let $f$ be defined on $[0, 1]$ by the formula
\[
f = \manylinedefinition{
x & \text{if $x$ is rational,} \\
0 & \text{if $x$ is irrational.}
}
\]
Prove that $f$ is continuous only at 0.
\sol
Firstly, we show that $f$ is \emph{not} continuous at any non-zero $x$ by
constructing a sequence \sequence x converging to $x$ such that its
image $\{f(x_n)\}$ does not converge to $f(x)$. We consider two cases:
\bi
\item
Case 1: $x$ is rational, so $f(x) = x \neq 0$.
In this case consider the sequence $x_n := x + \frac 1 {n\sqrt 2}$. Each
$x_n$ is irrational, so $f(x_n) = 0$. Then $\{f(x_n\}$ converges to $0$, and
not to $f(x) = x$.
\item
Case 2: $x$ is irrational, so $f(x) = 0$.
In this case consider a sequence of \emph{rational} numbers \sequence x
converging to $x$. We can find such a sequence by finding a rational
number $x_n$ in the interval $(x, x + 1/n)$ for each $n$ by Theorem 7.8.
Then $f(x_n) = x_n$ for each $n$, so the sequence $\{f(x_n)\}$ converges to
$x$ and not to $f(x) = 0$.
\ei
Secondly, we show that $f$ is continuous at 0 using Theorem 33.3. We claim
that given an $\e > 0$, for any $x \in \R$ such that $|x| < \e$
we have $|f(x)| < \e$ (we're using $f(0) = 0$ here). There are two cases
here again:
\bi
\item Case 1: $x$ is rational, so $f(x) = x$ and consequently $|f(x)| < \e$
by our assumption on $x$.
\item Case 2: $x$ is irrational, so $f(x) = 0$ and consequently
$|f(x)| = 0 < \e$.
\ei
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 35.5
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {35.5}
Let $(M, d)$ be a metric space and let $X$ be a subset of $M$. Prove that
$(X, d|_{X \times X})$ is a metric space.
\sol
We need to check that $d$ still satisfies the axioms of a metric
(Definition 35.1) after we restrict it to $X \times X$:
\bi
\item[(i)]
We know that $d(x, x) = 0$ for every $x \in M$, in particular, for every
$x \in X$.
Conversely, if $d(x, y) = 0$ with $x, y \in X$ then $x = y$ because
$x, y \in M$ and $d$ is a metric on $M$.
\item[(ii)]
We know that $d(x, y) = d(y, x)$ for any $x, y\in M$, in particular, for
$x, y \in X$.
\item[(iii)]
We know that the triangle inequality holds for $x, y, z \in M$, so in
particular it holds for $x, y, z \in X$.
\ei
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 35.6
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {35.6}
Let $l^\infty$ denote the set of all bounded real sequences, and let $c_0$
denote the set of all real sequences which converge to 0.
\bi
\itema
Prove that $l^1 \subset c_0 \subset l^\infty$.
\itemb
Prove that the containments in (a) are proper.
\itemc
Prove that
\[
d_{l^\infty}(\{a_n\}, \{b_n\}) := \sup \{|a_n - b_n| \| n \in \N\}
\]
defines a metric on $l^\infty$ (and thus by (a) together with Problem
35.5 also on $c_0$ and $l^1$)
\itemd
Let $d_{l^1}$ be the metric on $l^1$ defined in Example 35.5. Prove that
\[
d_{l^\infty}(x, y) \leq d_{l^1}(x,y)
\]
for $x, y\in l^1$.
\ei
\sol
\bi
\itema
By Theorem 22.3, $l^1 \subset c_0$ and by Theorem 13.2, $c_0 \subset l^\infty$.
\itemb
The harmonic series $a_n = 1/n$ is an element of $c_0$ but not of $l_1$.
The constant series $b_n = 1$ is an element of $l_\infty$ but not $c_0$.
\itemc
We verify the axioms of a metric:
\bi
\item[(i)]
Assume \sequence a = \sequence b. Then $a_n = b_n$ for every $n \in N$, so
\[
d_{l^\infty}(\{a_n\}, \{a_n\}) = \sup\{0\} = 0.
\]
Conversely, if \sequence a $\neq$ \sequence b, then there exists an $n \in \N$
such that $a_n - b_n \neq 0$, so $|a_n - b_n| > 0$ for that $n$. Hence
0 is not an upper bound of $\{|a_n - b_n| \| n \in \N\}$, so
$d_{l^\infty}(\{a_n\}, \{b_n\}) > 0$ in that case.
\item[(ii)]
Symmetry: given \sequence a and \sequence b in $l^\infty$ we have
$|a_n - b_n| = |b_n - a_n|$ for every $n \in \N$, so
\[
\{|a_n - b_n| \| n \in \N\} = \{|b_n - a_n| \| n \in \N\}
\]
and, therefore,
\[
d_{l^\infty}(\{a_n\}, \{b_n\}) = d_{l^\infty}(\{b_n\}, \{a_n\}).
\]
\item[(iii)]
Given \sequence a, \sequence b and \sequence c in $l^\infty$, for every
$n \in \N$ we have that
\[
|a_n - b_n| \leq d_{l^\infty}(\{a_n\}, \{b_n\}) \and
|b_n - c_n| \leq d_{l^\infty}(\{b_n\}, \{c_n\}).
\]
By triangle inequality for the absolute value
\[
|a_n - c_n| \leq |a_n - b_n| + |b_n - c_n|,
\]
so
\[
|a_n - c_n| \leq d_{l^\infty}(\{a_n\}, \{b_n\}) +
d_{l^\infty}(\{b_n\}, \{c_n\})
\]
for every $n \in \N$. Thus, $d_{l^\infty}(\{a_n\}, \{b_n\}) +
d_{l^\infty}(\{b_n\}, \{c_n\})$ is an upper bound of
$\{|a_n - c_n| \| n \in \N\}$, so
\[d_{l^\infty}(\{a_n\}, \{b_n\}) \leq d_{l^\infty}(\{a_n\}, \{b_n\}) +
d_{l^\infty}(\{b_n\}, \{c_n\}),
\]
as desired.
\ei
\itemd
For every $n \in \N$,
\[
|x_n - y_n| \leq \sumint k |x_k - y_k| = d_{l^1}(x, y),
\]
so $d_{l^1}(x, y)$ is an upper bound on $\{|x_n - y_n| \| n \in \N\}$, and
hence its greater or equal to the least upper bound:
\[
d_{l^\infty}(x, y) \leq d_{l^1}(x, y).
\]
\ei
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 35.7
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {35.7}
Let $(M, d)$ be a metric space. Prove that
\[
d'(x, y) := \frac{d(x, y)}{1 + d(x, y)} \and
d''(x, y) := \min\{d(x, y), 1\}
\]
define metrics on $M$. Prove that $d'$ and $d''$ are bounded by 1.
\sol
We verify the axioms of a metric for $d'$ and $d''$:
\bi
\item[(i)]
\[
d'(x, x) = \frac {d(x, x)} {1 + d(x, x)} = \frac {0} {1 + 0} = 0.
\]
\[
d''(x, x) = \min\{d(x, x), 1\} =\min\{0, 1\} = 0.
\]
Conversely, if $x \neq y$ then $d(x, y) > 0$, so
\[
d'(x, y) = \frac{d(x, y)}{1 + d(x, y)} > 0
\]
because the numerator and the denominator are positive and
\[
d''(x, y) = \min\{d(x, y), 1\} > 0.
\]
because both $d(x, y)$ and 1 are positive.
\item[(ii)]
By assumption $d(x, y) = d(y, x)$, so
\[
d'(x, y) = \frac{d(x, y)}{1 + d(x, y)} = \frac{d(y, x)}{1 + d(y, x)}
= d'(y, x)
\]
and $\{d(x, y), 1\} = \{d(y, x), 1\}$, so
\[
d''(x, y) = \min\{d(x, y), 1\} = \min\{d(y, x), 1\} = d''(y, x).
\]
\item[(iii)]
The triangle inequality for $d'$. We have that
\[d'(x, y) = \frac{d(x, y)}{1 + d(x, y)} \geq
\frac{d(x, y)}{1 + d(x, y) + d(y, z)}
\]
and
\[
d'(y, z) = \frac{d(y, z)}{1 + d(y, z)} \geq
\frac{d(y, z)}{1 + d(x, y) + d(y, z)}. \\
\]
Thus,
\[d'(x, y) + d'(y, z) \geq \frac{d(x, y) + d(y, z)}{1 + d(x, y) + d(y, z)}
\geq \frac{d(x, z)} {1 + d(x, z)} = d'(x, z)
\]
where the latter inequality holds because
$d(x, y) + d(y, z) \geq d(x, z)$ and the function
$f(t) := \frac t {1 + t} = 1 - \frac 1 {1 + t}$ is increasing in $t$
for $t \in [0, \infty)$.
The triangle inequality for $d''$. We consider two cases:
\bi
\item[Case 1:] at least one of the two inequalities
$d(x, y) \geq 1$ or $d(y, z) \geq 1$
holds. In this case at least one of the two equalities $d''(x,y) = 1$
or $d''(y, z) = 1$ holds, so
\[
d''(x, y) + d''(y, z) \geq 1 \geq \min\{ d(x, z), 1\} = d''(x, z).
\]
\item[Case 2:] $d(x, y) < 1$ and $d(y, z) < 1$. Hence,
$d''(x, y) = d(x, y)$ and $d''(y, z) = d(y, z)$. Therefore,
\[
d''(x, y) + d''(y, z) \geq d(x, z) \geq \min\{ d(x, z), 1\} = d''(x, z).
\]
\ei
\ei
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 36.8
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {36.8}
Let \sequence a $\in l^1$ and \sequence b $\in l^\infty$. Prove that
$\{a_nb_n\}\in l^1$.
\sol
This is exactly the statement of Theorem 26.4(i).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 36.11
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {36.11}
Let \sequence a be a sequence such that $\{a_nb_n\} \in l^1$ for every
sequence \sequence b $\in l^1$. Prove that \sequence a $\in l^\infty$. Show
(by example) that the above statement is false $l^\infty$ is replaced by $c_0$.
\sol
We prove the contrapositive: if \sequence a is not in $l^\infty$ then
there exists a sequence \sequence b in $l^1$ such that
$\{a_nb_n\} \notin l^1$.
If \sequence a is not in $l^\infty$, it is not bounded. Assume \sequence a
is not bounded from above (otherwise replace \sequence a with
$\{-a_n\}$). Then we can inductively
construct a subsequence \subsequence a such that $a_{n_k} > k$.
Let
\[
b_n := \left\{\begin{array}{ll}
\frac 1 {k^2} & \text{if $n = n_k$ for some $k$},\\
0 & \text{otherwise.}
\end{array}\right.
\]
Then $0 \leq b_n \leq 1/n^2$ for every $n \in \N$, so by the Comparison Test,
$b \in l^1$. However,
\[
a_nb_n = \left\{\begin{array}{ll}
\frac 1 {k} & \text{if $n = n_k$ for some $k$},\\
0 & \text{otherwise,}
\end{array}\right.
\]
so the subsequence
$\{\sum_{n = 1}^{n_k} a_nb_n\}_{k \in \N}
= \{\sum_{m = 1}^k \frac 1 k\}_{k \in \N},m $
of partial sums of $\{a_nb_n\}$ diverges. Hence, $\{a_nb_n\}$ is not
an element of $l^1$.
The constant sequence $a_n = 1$ (tautologically) satisfies the premise of
the problem and is not an element of $c_0$ because its limit is 1 and not 0.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 37.4
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {37.4}
Show (by examples) that Theorem 37.2 does not generalize to $l^2$, $c_0$ or
$l^\infty$.
\sol
Define the sequence $a^{(k)} = \{a_n^{(k)}\}$ by
\[
a^{(k)}_n := \left\{\begin{array}{ll}
1 & \text{if $n = k$,}\\
0 & \text{otherwise,}
\end{array}\right.
\]
i.e. $a^{(k)}$ has a 1 at index $k$ and zeros everywhere else.
Then $a^{(k)}$ is an element of $l^1$ (and hence of $c_0$ and $l^\infty$ too)
Then $\lim_{k \ra \infty} a_n^{(k)} = 0$ for every $n \in \N$ because
$a_n^{(k)} = 0$ for all $k > n$.
However, the sequence $\{a^{(k)}\}$ of sequences does not converge to
the sequence $\underline 0$ consisting of all zeros in $l^\infty$
(and, hence, by Problem 35.6(d) it does not converge to
$\underline 0$ in $l^1$ either) because
\[
d_{l^\infty}(a^{(k)}, \underline 0) = \sup\{0, 1\} = 1
\]
for every $k$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 37.10
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {37.10}
Let $(M, d)$ be a metric space and let $d'$ and $d''$ be defined as in
Exercise 35.7. Let \sequence a be a sequence in $M$ and let $a \in M$. Prove
that the following statements are equivalent:
\bi
\itema
\sequence a converges to $a$ in $(M, d)$.
\itemb
\sequence a converges to $a$ in $(M, d')$.
\itemc
\sequence a converges to $a$ in $(M, d'')$.
\ei
\sol
\bi
\item
(a) $\Rightarrow$ (b) and (c)
We have that $\limn d(a_n, a) = 0$ and also
\[
0 \leq d'(a_n, a) \leq d(a_n, a) \and 0 \leq d''(a_n, a) \leq d(a_n, a).
\]
Therefore, by Squeeze Theorem both $d'(a_n, a)$ and $d''(a_n, a)$ converge
to 0.
\item
(b) $\Rightarrow$ (a)
Given $\epsilon \in (0, 1)$, choose $N$ such that
\[d'(a_n, a) < \frac {\epsilon} {1 + \epsilon}
\]
for every $n \geq N$. Then
\[
d(a_n, a) = \frac 1 {1 - d'(a_n, a)} - 1 < \epsilon
\]
for every $n \geq N$. Hence, $\limn d(a_n, a) = 0$.
\item
(c) $\Rightarrow$ (a)
Since $\limn d''(a_n, a) = 0$, we can choose $N$ such that
$d''(a_n, a) < 1$ for all $n \geq N$. Using the definition of $d''$ we get
that for all $n \geq N$, $d''(a_n, a) = d(a_n, a)$, so
$\limn d(a_n, a) = 0$.
\ei
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 40.10
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {40.10}
Let \sequence a $\in l^\infty$. Prove that $f$ defined by
\[
f(\{b_n\}) := \sumint n a_n b_n
\]
is a continuous real-valued function on $l^1$.
\sol
To show continuity of $f$ we need to show that given \sequence b $\in l^1$ and
$\e > 0$ we can find $\delta > 0$ such that for every
\sequence {\tilde b} $\in l^1$ such that
$d_{l^1}(\{b_n\}, \{\tilde b_n\}) < \delta$ we have that
\[
|f(\{b_n\}) - f(\{\tilde b_n\})| < \e.
\]
Assume \sequence a is bounded by $M$ with $M > 0$ in absolute value:
$|a_n| < M$ for every $n \in \N$. We show that $\delta := \e / M$ works.
By Theorem 23.1 applied to series $\sumint n a_nb_n$ and
$\sumint n a_n\tilde b_n$ we have that
\[
f(\{b_n\}) - f(\{\tilde b_n\}) = \sumint n a_n(b_n - \tilde b_n).
\]
Hence, by Theorem 26.2
\[
|f(\{b_n\}) - f(\{\tilde b_n\})| \leq \sumint n |a_n(b_n - \tilde b_n)|
\]
By Theorem 26.3(i) we have that
\[
\sumint n |a_n(b_n - \tilde b_n)| \leq \sumint n M |b_n - \tilde b_n| =
M d_{l^1}(\{b_n\}, \{\tilde b_n\}) < \epsilon.
\]
Therefore,
\[
|f(\{b_n\}) - f(\{\tilde b_n\})| < \epsilon,
\]
as desired.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {1. Product metrics, continuity and composition.}
If $(M_1, d_1)$ and $(M_2, d_2)$ are metric spaces then one can define a
distance function $d$ on the Cartesian product $M_1 \times M_2$ by
\beq
\label{eq:l1-metric}
d((m, n), (m', n')) := d_1(m, m') + d_2(n, n').
\eeq
\bi
\itema
Show that $d$ defines a metric on $M_1 \times M_2$ called the
\emph{(standard) product metric.}
\itemb
In fact, $d$ is not the only natural metric on $M_1 \times M_2$. For instance,
show that
\beq
\label{eq:l2-metric}
d_{\ltwo}((m, n), (m', n')) := \sqrt{ d_1(m, m')^2 + d_2(n, n')^2}
\eeq
defines another metric on $M_1 \times M_2$, called the
\emph{$\ltwo$ product metric}.
Then, also show that there is an isometry between $\R^n$ with its Euclidean
metric and $\R^k \times \R^{n-k}$ where each of $\R^k$ and $\R^{n - k}$ are
equipped with their Euclidean metrics and the product is equipped with the
$l^2$ product metric.
\itemc
Usually, unless otherwise specified, we will think of the product
$M_1 \times M_2$ as a metric space with respect to the standard product metric.
Show that if $N$ is a metric space and $f_i: N \ra M_i$ are continuous maps
for $i = 1, 2$, then the map $(f_1, f_2): N \ra M_1 \times M_2$ is also
continuous.
Now, show that the identity map
$(M_1 \times M_2, d) \ra (M_1 \times M_2, d_{\ltwo})$
is a continuous map from $M_1 \times M_2$ with its standard product metric to
$M_1 \times M_2$ with its $\ltwo$ product metric.
Conclude from Corollary 40.6 of the book (which says that the composition of
continuous functions is continuous) that if $f_1$ and $f_2$ are continuous,
then $(f_1, f_2): N \ra M_1 \times M_2$, where now $M_1 \times M_2$ is
equipped with the $\ltwo$ product metric, is continuous too.
\itemd
Prove that the following functions from $\R^2$ to $\R$ (where $\R^2$ is
equipped with its Euclidean metric) are continuous:
\bi
\item
$(x, y) \mapsto xy$,
\item
$(x, y) \mapsto x + y$.
\ei
Deduce a new proof of Theorem 40.4 parts (ii) and (v) from this, part (c),
and the fact that the compositions of continuous functions are continuous
(Corollary 40.6). Finally, indicated (but do not prove) how the remaining parts
of Theorem 40.4 are similarly consequences of the continuity of some basic
real-valued functions on $\R$ or $\R \times \R$ along with the part (c)
and Corollary 40.6.
\ei
\sol
\bi
\itema
We check the axioms of a metric for $d$:
\bi
\item[(i)]
We have $d_1(m, m) = 0$ and $d_2(n, n) = 0$, hence
\[
d((m, n), (m, n)) = d_1(m, m) + d_2(n, n) = 0.
\]
Conversely, if $(m, n) \neq (m', n')$ then at least one of the following
non-equalities hold: $m \neq m'$ or $n \neq n'$. Hence, at least one of
the following inequalities hold: $d_1(m, m') > 0$ or $d_2(n, n') > 0$. Thus,
\[
d((m, n), (m', n')) = d_1(m, m') + d_2(n, n') > 0.
\]
\item[(ii)] Symmetry:
\[
d((m, n), (m', n')) = d_1(m, m') + d_2(n, n') = d_1(m', m) + d_2(n', n)
= d((m', n'), (m, n)).
\]
\item[(iii)] Triangle inequality: we know that
\[
d_1(m, m'') \leq d_1(m, m') + d_1(m', m'') \and
d_2(n, n'') \leq d_2(n, n') + d_2(n', n'').
\]
Hence
\begin{align*}
d((m, n), (m'', n''))
& = d_1(m, m'') + d_2(n, n'') \\
& \leq (d_1(m, m') + d_1(m', m'')) + (d_2(n, n') + d_2(n', n'')) \\
& = (d_1(m, m') + d_2(n, n')) + (d_1(m', m'') + d_2(n', n'')) \\
& = d((m, n), (m', n')) + d((m', n'), (m'', n'')).
\end{align*}
\ei
\itemb
We check the axioms of a metric for $d_{\ltwo}$ in the same manner as
in the previous part:
\bi
\item[(i)]
We have $d_1(m, m) = 0$ and $d_2(n, n) = 0$, hence
\[
d_{\ltwo}((m, n), (m, n)) = \sqrt{d_1(m, m)^2 + d_2(n, n)^2} = 0.
\]
Conversely, if $(m, n) \neq (m', n')$ then at least one of the following
non-equalities hold: $m \neq m'$ or $n \neq n'$. Hence, at least one of
the following inequalities hold: $d_1(m, m') > 0$ or $d_2(n, n') > 0$. Thus,
\[
d_{\ltwo}((m, n), (m', n')) = \sqrt{ d_1(m, m')^2 + d_2(n, n')^2} > 0.
\]
\item[(ii)] Symmetry:
\[
d_\ltwo((m, n), (m', n')) = \sqrt{d_1(m, m')^2 + d_2(n, n')^2}
\]
\[
= \sqrt{d_1(m', m)^2 + d_2(n', n)^2}
= d_\ltwo((m', n'), (m, n)).
\]
\item[(iii)] Triangle inequality: as before, we know that
\[
d_1(m, m'') \leq d_1(m, m') + d_1(m', m'') \and
d_2(n, n'') \leq d_2(n, n') + d_2(n', n'').
\]
Hence,
\begin{align*}
d((m, n), (m'', n''))
& = \sqrt{d_1(m, m'')^2 + d_2(n, n'')^2} \\
& \leq \sqrt{(d_1(m, m') + d_1(m', m''))^2 + (d_2(n, n') + d_2(n', n''))^2} \\
& \leq \sqrt{d_1(m, m')^2 + d_2(n, n')^2}
+ \sqrt{d_1(m', m'')^2 + d_2(n', n'')^2} \\
& = d((m, n), (m', n')) + d((m', n'), (m'', n'')),
\end{align*}
where the latter inequality holds by Lemma \ref{lem:minkowski}.
\ei
\begin{lemma}
\label{lem:minkowski}
Any real numbers $a_1, a_2, b_1, b_2$ satisfy
\[
\sqrt{(a_1 + a_2)^2 + (b_1 + b_2)^2} \leq
\sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2}
\]
\end{lemma}
\begin{proof}
The desired inequality is equivalent to the following triangle inequality in
$\R^2$:
\[
d_{\R^2}((0, 0), (a_1 + a_2, b_1 + b_2)) \leq
d_{\R^2}((0, 0), (a_1, b_1)) + d_{\R^2}((a_1, b_1), (a_1 + a_2, b_1 + b_2)).
\]
\end{proof}
We can give an isometry $f: \R^k \times \R^{n-k} \ra \R^n$ via concatenation
of vectors:
\[
f((a_1, \ldots, a_k), (b_1, \ldots, b_{n - k}))
= (a_1, \ldots, a_k, b_1, \ldots, b_{n - k}).
\]
It is an isometry because, given $a, a'\in \R^k$ and $b, b'\in \R^{n-k}$ we
have that
\begin{align*}
d_{\R^k \times \R^{n - k}}((a, b), (a', b'))
& = \sqrt{d_{\R^k}(a, a')^2 + d_{\R^{n - k}}(b, b')^2} \\
& = \sqrt{\sum_{j = 1}^k|a_j - a'_j|^2
+ \sum_{j = 1}^{n - k} |b_j - b'_j|^2} \\
& = d_{\R^n}(f(a, b), f(a', b')).
\end{align*}
\itemc
Since $f_1, f_2$ are continuous, then given $x \in N$ and $\e > 0$
we can choose $\delta_1, \delta_2 > 0$ such that
for every $y \in N$ such that $d_{N}(x, y) < \delta_i$ we have
$d_{M_i}(f_i(x), f_i(y)) < \e/2$ (with $i = 1, 2$).
Let $\delta := \min(\delta_1, \delta_2)$.
Then for every $y \in y$ such that $d_N(x, y) < \delta$ we have that
\begin{align*}
d_{M_1 \times M_2}((f_1(x), f_2(x)), (f_1(y), f_2(y))
& = d_{M_1}(f_1(x), f_1(y)) + d_{M_2}(f_2(x), f_2(y)) \\
& < \frac \e 2 + \frac \e 2 = \e.
\end{align*}
Thus, $(f_1, f_2)$ is continuous.
To show that the identity $(M_1 \times M_2, d) \ra (M_1 \times M_2, d_{\ltwo})$
is continuous we show
\[
d_{\ltwo}(x, y) \leq d(x, y)
\]
for every $x, y \in M_1 \times M_2$. Indeed, assuming the inequality above,
given $\e > 0$ and $x \in M_1 \times M_2$ for every
$y \in M_1 \times M_2$ such that $d(x, y) < \e$ we also have
$d_{\ltwo}(x, y) < \e$.
The desired inequality is equivalent to the following two inequalities
\begin{align*}
& \sqrt{d_1(x_1, y_1)^2 + d_2(x_2, y_2)^2}
\leq d_1(x_1, y_1)^2 + d_2(x_2, y_2)^2 \\
\iff\> & d_1(x_1, y_1)^2 + d_2(x_2, y_2)^2
\leq d_1(x_1, y_1)^2 + 2 d_1(x_1, y_1) \cdot d_2(x_2, y_2)
+ d_2(x_2, y_2)^2,
\end{align*}
with the latter one is being obviously true.
We have that $(f_1, f_2): (N, d_{N}) \ra (M_1 \times M_2, d_{\ltwo})$ is a
composition of two continuous continuous functions:
$(f_1, f_2): (N, d_{N}) \ra (M_1 \times M_2, d)$ and
identity $(M_1 \times M_2, d) \ra (M_1 \times M_2, d_{\ltwo})$, hence it
is continuous by Corollary 40.6.
\itemd
By Theorem 40.2, to show continuity of $(x, y) \mapsto xy$ and
$(x, y) \mapsto x + y$ it suffices to show that for any sequence
$\{(x_n, y_n)\}$ points in $\R^2$ converging to some $(x, y) \in \R^2$,
the sequences $\{x_ny_n\}$ and $\{x_n + y_n\}$ converge to $xy$ and $x + y$,
respectively.
By Theorem 37.2, \sequence x and \sequence y converge to $x$ and $y$,
respectively. Hence, the desired conclusion follows by Theorem 12.6 and
12.2, respectively.
The remaining parts of Theorem 40.4 follow from the continuity of
\bi
\item[(i)]
$x \mapsto |x|: \R \ra \R$,
\item[(iii)]
$x \mapsto cx: \R \ra \R$,
\item[(iv)]
$(x, y) \mapsto x - y: \R^2 \ra \R$,
\item[(vi)]
$(x, y) \mapsto x/y: \R \times \R \backslash \{0\} \ra \R$.
\ei
\ei
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%problem 2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pb {2. Pseudometrics and equivalence relations.}
Recall that an \emph{equivalence relation} on a set $X$ is any binary relation
on $X$, denoted $\sim$, which satisfies the following properties:
\bi
\item
\emph{(reflexive property)}
$x \sim x$, $\forall x \in X$;
\item
\emph{(symmetric property)}
$x \sim y$, implies $y \sim x$, $\forall x, y \in X$;
\item
\emph{(transitive property)}
If $x \sim y$ and $y \sim z$ then $x \sim z$, $\forall x,y,z \in X$.
\ei
One simple example of an equivalence relation is \emph{equality}: it is clear
that $x = x$, $x = y$ if $y = x$ and if $x = y$ and $y = z$ then $x = z$.
Now, let $M$ be a set. A function $d: M \times M \ra [0, \infty)$ is called
a \emph{pseudometric} if $d$ satisfies
\bi
\item[(i)]
$d(x, x) = 0$, $\forall x \in M$;
\item[(ii)]
$d(x, y) = d(y, x)$, $\forall x,y \in M$ and
\item[(iii)]
$d(x, z) \leq d(x, y) + d(y, z)$, $\forall x, y, z \in M$.
\ei
Namely, $d$ satisfies all of the conditions of a metric except that
$d(x, y) = 0$ need not imply $x = y$ (note that condition (i) is weaker than
the respective condition for a metric). A \emph{pseudometric space} is a pair
$(M, d)$ of a set and a pseudometric on it.
\bi
\itema
Show that $\R^2$ with the function $d(x, y) = |y_2 - x_2|$ is a pseudometric
space but not a metric space (that is, $d(x, y) = 0$ does not imply $x = y$).
\itemb
Given a pseudometric space $(M, d)$, consider the following binary relation
on $M$: say $x \sim y$ if $d(x, y) = 0$. Show that $\sim$ is an equivalence
relation.
\itemc
Recall that given a set $X$ and an equivalence relation $\sim$ on $X$ one
can partition $X$ into a collection of \emph{equivalence classes}. An
equivalence class is a subset of $X$ consisting of all elements that are
similar to a given element. Any element $a \in X$ belongs to a single
equivalence class called $[a]$:
\[
[a] := \{ x \in X \| x \sim a \}.
\]
Two elements $a$ and $b$ have the same equivalence class (i.e. $[a] = [b]$)
if and only if $a \sim b$. In that case $a$ and $b$ are different
\emph{representatives} of the same equivalence class.
Show that given a pseudometric space $(M, d)$, the function $d(x, y)$ only
depends on the equivalence classes $[x], [y]$ with respect to the
equivalence relation $\sim$.
\itemd
Given a set with an equivalence relation $(X, \sim)$ we can define the
\emph{quotient} of $X$ by $\sim$ as the set of distinct equivalence classes:
\[
X / \sim := \{ [a] \| a \in X \}.
\]
Given a pseudometric space $(M, d)$, define $M^* := M / \sim$ where $\sim$
is the equivalence relation defined above. Define
$d^*: M^* \times M^* \ra [0, \infty)$ by $d^*(\alpha, \beta) := d(x, y)$
for any representatives $x \in \alpha, y \in \beta$. By the previous section,
$d(x, y)$ only depends on the equivalence classes $[x], [y]$, do $d^*$ is
well-defined.
Show that $d^*$ gives a metric on $M^*$.
\iteme
Let's return to the example $X = (\R^2, d(x, y) = |y_2 - x_2|)$. Show that
the induced metric space $(X^*, d^*)$ is \emph{isometric} to $\R$ with its
Euclidean metric.
\itemf
Let $\mc F(\R)$ be the set of functions $f: \R \ra \R$. Fixing a pair of
distinct $x_0, x_1 \in \R$, define a function
\[
d: \mc F(\R) \times \mc F(\R) \ra [0, \infty)
\]
by
\[
d(f, g) = |f(x_0) - g(x_0)| + |f(x_1) - g(x_1)|.
\]
Show that $d$ defines a pseudometric on $\mc F(\R)$ but that $d(x, y) = 0$
does not imply $x = y$. Show that the resulting quotient metric space
$\mc F(\R)^*$ is isometric to $\R^2$ with its $\ell^1$ metric.
\ei
\sol
\bi
\itema
We check the axioms of a pseudometric space for $d$:
\bi
\item[(i)] $d(x, x) = |x_2 - x_2| = 0$.
\item[(ii)]
$d(x, y) = |y_2 - x_2| = |x_2 - y_2| = d(y, x)$
\item[(iii)]
$d(x, z) = |z_2 - x_2| \leq |y_2 - x_2| + |z_2 - y_2| = d(x, y) + d(y, z)$.
\ei
However, $d$ is not a metric because $d((0, 0), (1, 0)) = 0$.
\itemb
We have that $x \sim y$ if and only if $x_2 = y_2$. We check the axioms of
an equivalence for this relation:
\bi
\item Reflexivity: $x \sim x$ because $d(x, x) = 0$.
\item Symmetry: if $x \sim y$ then $d(x, y) = 0$. By symmetry of the
metric $d(y, x) = 0$, hence $y \sim x$.
\item Transitivity: if $x \sim y$ and $y \sim z$ then $d(x, y) = 0$ and
$d(y, z) = 0$, so by triangle inequality
\[
d(x, z) \leq d(x, y) + d(y, z) = 0.
\]
Since $d(x, z)$ cannot be negative, $d(x, z) = 0$, so $x \sim z$
\ei
\itemc
It suffices to show that $d(x, y) = d(x, y')$ whenever $y \sim y'$. Indeed,
in that case whenever $x \sim x'$ and $y \sim y'$ we have the following
chain of equalities
\[
d(x, y) = d(x, y') = d(y', x) = d(y', x') = d(x', y'),
\]
showing the independence of $d$ on the choice of representatives $x$ and $y$.
By triangle inequality we have that
\[
d(x, y) \leq d(x, y') + d(y', y) = d(x, y')
\]
and
\[
d(x, y') \leq d(x, y) + d(y, y') = d(x, y).
\]
Thus, $d(x, y) = d(x, y')$, as desired.
\itemd
All of the axioms of a metric follow by the axioms of a pseudo-metric
and a coherent choice of representatives: given
$\alpha, \beta, \gamma \in X / \sim$ choose $x \in \alpha$, $y \in \beta$
and $z \in \gamma$. Then
\bi
\item[(i)] $d^*(\alpha, \alpha) = d(x, x) = 0$. Conversely, if
$d^*(\alpha, \beta) = 0$, then $d(x, y) = 0$, so $x \sim y$, so
$\alpha = [x] = [y] = \beta$.
\item[(ii)]
$d^*(\alpha, \beta) = d(x, y) = d(y, x) = d^*(\beta, \gamma)$.
\item[(iii)]
$d^*(\alpha, \gamma) = d(x, z) \leq d(x, y) + d(y, z)
= d^*(\alpha, \beta) + d^*(\beta, \gamma).$
\ei
\iteme
We show that a map $f: (X^*, d^*) \ra \R$ given by
\[
f([x]) = x_2
\]
is an isometry. We start by noting that $f$ is well-defined,
i.e. if $x$ and $y$ are two different representatives of $\alpha$, then
$x \sim y$, so $x_2 = y_2$, hence they given the same $f(\alpha)$ when
chosen as representatives for $\alpha$.
Next, we show that the function $f$ is bijective. The function $f$ is
injective because if $f([x]) = f([y])$ then $x_2 = y_2$, so $x \sim y$ and
consequently $[x] = [y]$. The function $f$ is surjective because given
any $t \in \R$ we have that $f([(0, t)]) = t$.
Finally, the function $f$ preserves distances:
\[
|f(x) - f(y)| = |x_2 - y_2| = d^*(x, y).
\]
\itemf
We verify the axioms of a pseudometric for $d$:
\bi
\item[(i)]
$d(f, f) = |f(x_0) - f(x_0)| + |f(x_1) - f(x_1)| = 0.$
\item[(ii)]
\begin{align*}
d(f, g)
& = |f(x_0) - g(x_0)| + |f(x_1) - g(x_1)| \\
& = |g(x_0) - f(x_0)| + |g(x_1) - f(x_1)| \\
& = d(g, f)
\end{align*}
\item[(iii)]
\begin{align*}
d(f, h)
& = |f(x_0) - h(x_0)| + |f(x_1) - h(x_1)| \\
& \leq (|f(x_0) - g(x_0)| + |g(x_0) - h(x_0)|)
+ (|f(x_1) - g(x_1)| + |g(x_1) - h(x_1)|) \\
& = (|f(x_0) - g(x_0)| + |f(x_1) - g(x_1)|)
+ (|g(x_0) - h(x_0)| + |g(x_1) - h(x_1)|) \\
& = d(f, g) + d(g, h).
\end{align*}
\ei
Also, $d$ is not a metric because
\[
d(x \mapsto 0, x \mapsto (x - x_0)(x - x_1) ) = 0.
\]
with $x \mapsto 0$ and $x \mapsto (x - x_0)(x - x_1)$ being two distinct
functions $\R \ra \R$.
We show that the evaluation function
$\ev: \mc F(\R)^* \ra (\R^2, d_{\ell^1})$ given
by $\ev([f]) := (f(x_0), f(x_1))$ is an isometry. As in the previous
part we start by showing that the function $\ev$ is well-defined:
i.e. if $f$ and $g$ are the representative of the same equivalence class
than they give the same element of $\R^2$. If $[f] = [g]$ then $f \sim g$,
so
\[
0 = d(f, g) = |f(x_0) - g(x_0)| + |f(x_1) - g(x_1)|.
\]
Therefore, $(f(x_0), f(x_1)) = (g(x_0), g(x_1))$, so $\ev$ is well-defined.
Next we show that the evaluation function $\ev$ is bijective. The
evaluation function is injective because if $\ev([f]) = \ev([g])$ then
$f(x_0) = g(x_0)$ and $f(x_1) = g(x_1)$, so $f \sim g$ and, consequently,
$[f] = [g]$. The evaluation function is surjective, because given
$(a, b) \in \R$ we have $\ev([f]) = (a, b)$ with
\[
f(x) := \left\{\begin{array}{ll}
a & \text{if $x = x_0$,} \\
b & \text{if $x = x_1$,} \\
0 & \text{otherwise}.
\end{array}
\right.
\]
Finally, $\ev$ preserves distances:
\[
d_{\ell^1}(\ev([f]), \ev([g])) = |f(x_0) - g(x_0)| + |f(x_1) - g(x_1)| =
d^*([f], [g]).
\]
\ei
\end{document}