Exercise 20.1

Question

\def\a{\alpha}
\def\b{\beta}
\def\d{\delta}
\def\e{\epsilon}
\def\r{\rho}

  \begin{enumerate}[label=(\alph*)]
  \item In ${\mathbb{R}}^n$, define
    \begin{gather*}
      d'(\mathbf{x} , \mathbf{y} ) = \left|x_1 - y_1\right| + \cdots + \left|x_n - y_n\right| \,.
    \end{gather*}
    Show that $d'$ is a metric that induces the usual topology of ${\mathbb{R}}^n$.
    Sketch the basis elements under $d'$ when $n = 2$.
  \item More generally, given $p \geq 1$, define
    \begin{gather*}
      d'(\mathbf{x} , \mathbf{y} ) = \left[\sum_{i=1}^n \left|x_i - y_i\right|^p\right]^{1/p}
    \end{gather*}
    for $\mathbf{x} , \mathbf{y}  \in {\mathbb{R}}^n$.
    Assume that $d'$ is a metric.
    Show that it induces the usual topology on ${\mathbb{R}}^n$.
  \end{enumerate}

Dan Whitman · Accepted Answer

\def\a{\alpha} \def\b{\beta} \def\d{\delta} \def\e{\epsilon} \def { ho} \begin{lemma}\label{lem:metric:pmon} If $x$ and $y$ are real and $x,y \geq 0$ then $x^p < y^p$ if and only if $x < y$, for all integers $p \geq 1$. \end{lemma} \begin{proof} First, if $x = 0$ then of course \begin{gather*} x < y \Leftrightarrow 0 < y \Leftrightarrow 0 < y^p \Leftrightarrow 0^p < y^p \Leftrightarrow x^p < y^p \end{gather*} for any $p \geq 1$, so assume it what follows that $x > 0$. We show this by induction on $p$. First, for $p=1$ we clearly have that $x^p = x$ and $y^p = y$ so that of course the biconditional holds. Now suppose that $x^p < y^p$ if and only if $x < y$. Suppose that $x < y$ so that $x^p < y^p$ follows by the induction hypothesis. We also have that $y > 0$ since $0 < x < y$ so that $y^p > 0$. Then \begin{align*} x^p &< y^p \ x \cdot x^p &< x \cdot y^p & ext{(since $x > 0$)} \ x \cdot x^p &< x \cdot y^p < y \cdot y^p & ext{(since $x < y$ and $y^p > 0$)} \ x^{p+1} &< y^{p+1} \,. \end{align*} Now suppose that it is not true that $x < y$ so that $x \geq y$. It then follows from the induction hypothesis that $x^p \geq y^p$. Then we have \begin{align*} x^p &\geq y^p \ x \cdot x^p &\geq x \cdot y^p & ext{(since $x > 0$)} \ x \cdot x^p &\geq x \cdot y^p \geq y \cdot y^p & ext{(since $x \geq y$ and $y^p \geq 0$ since $y \geq 0$)} \ x^{p+1} &\geq y^{p+1} \,. \end{align*} Hence by the contrapositive we have that $x^{p+1} < y^{p+1}$ implies that $x < y$. This completes the induction. \end{proof} \begin{corollary}\label{cor:metric:opmon} If $x$ and $y$ are real and $x,y \geq 0$ then $x^{1/p} < y^{1/p}$ if and only if $x < y$, for all integers $p \geq 1$. \end{corollary} \begin{proof} Consider any $p \geq 1$ and let $u = x^{1/p}$ and $v = y^{1/p}$. Then clearly we have $u,v \geq 0$ since $x,y \geq 0$. We then have by Lemma~ ef{lem:metric:pmon} that \begin{gather*} u^p < v^p \Leftrightarrow u < v \ (x^{1/p})^p < (y^{1/p})^p \Leftrightarrow x^{1/p} < y^{1/p} \ x < y \Leftrightarrow x^{1/p} < y^{1/p} \,, \end{gather*} which is of course the desired result. \end{proof} \begin{lemma}\label{lem:metric:polynom} For any $n,p \in {{\mathbb{Z}}_+}$ and a finite sequence $\left(x_i ight)_{i=1}^n$ where each $x_i \geq 0$, \begin{gather*} \sum_{i=1}^n x_i^p \leq \left(\sum_{i=1}^n x_i ight)^p \,. \end{gather*} \end{lemma} \begin{proof} For every $n \in {{\mathbb{Z}}_+}$, we show this by induction on $p$. For $p = 1$ we clearly have \begin{gather*} \sum_{i=1}^n x_i^p = \sum_{i=1}^n x_i \leq \sum_{i=1}^n x_i = \left(\sum_{i=1}^n x_i ight)^p \,. \end{gather*} Now suppose that the hypothesis is true for $p$. Then we have \begin{align*} \left(\sum_{i=1}^n x_i ight)^{p+1} &= \left(\sum_{i=1}^n x_i ight)\left(\sum_{i=1}^n x_i ight)^p \ &\geq \left(\sum_{i=1}^n x_i ight)\left(\sum_{i=1}^n x_i^p ight) & ext{(by the induction hypothesis since $\sum_{i=1}^n x_i \geq 0$)} \ &= \sum_{i=1}^n \sum_{j=1}^n x_i x_j^p \ &= \sum_{i=1}^n \left(x_i x_i^p + \sum_{j eq i} x_i x_j^p ight) \ &= \sum_{i=1}^n x_i^{p+1} + \sum_{i=1}^n \sum_{j eq i} x_i x_j^p \ &\geq \sum_{i=1}^n x_i^{p+1} \end{align*} since each $x_i x_j^p \geq 0$ so that the double sum is as well. This completes the induction. \end{proof} extbf{Main Problem.} (a) First, the basis elements of the metric topology induced by $d'$ are open intervals in ${\mathbb{R}}$, open diamonds in $n=2$, open octahedrons for $n = 3$, and the higher dimensional analogues for $n > 3$. A sketch of the ball $B_{d'}(0 imes 0, 1)$ in ${\mathbb{R}}^2$ is shown below: \def ikzgridclr{black!30!white} \def ikzgray{black!30!white} \def ikzgrayopac{0.3} \begin{center} \begin{tikzpicture}[scale=2] % Coordinate grid \draw[step=1cm, ikzgridclr,thin] (-1.5,-1.5) grid (1.5,1.5); % Set shape \filldraw[thick,dashed,fill=black,fill opacity= ikzgrayopac] (-1,0) -- (0,1) -- (1,0) -- (0,-1) -- cycle; % Axes \draw[thick,<->] (-1.5,0) -- (1.5,0); \draw[thick,<->] (0,-1.5) -- (0,1.5); % Point labels ode [above left] at (-1,0) {$(-1,0)$}; ode [above right] at (0,1) {$(0,1)$}; ode [above right] at (1,0) {$(1,0)$}; ode [below right] at (0,-1) {$(0,-1)$}; \end{tikzpicture} \end{center} Now we show that $d'$ is a metric and induces the usual topology of ${\mathbb{R}}^n$. \begin{proof} It is easy to see that $d'$ meets the properties required of a metric. Clearly $d'(\mathbf{x} , \mathbf{y} ) \geq 0$ since each $\left|x_i - y_i ight| \geq 0$, and $d'(\mathbf{x} , \mathbf{y} ) = 0$ if and only if each $x_i = y_i$ so that $\mathbf{x} = \mathbf{y} $. Also it is obvious that $d'(\mathbf{x} , \mathbf{y} ) = d'(\mathbf{y} , \mathbf{x} )$ since each $\left|x_i - y_i ight| = \left|y_i - x_i ight|$. For the triangle inequality, we simply have that \begin{align*} d'(\mathbf{x} , \mathbf{z} ) &= \sum_{i=1}^n \left|x_i - z_i ight| \ &\leq \sum_{i=1}^n \left(\left|x_i - y_i ight| + \left|y_i - z_i ight| ight) & ext{(since each $\left|x_i-z_i ight| \leq \left|x_i-y_i ight| + \left|y_i-z_i ight|$)} \ &= \sum_{i=1}^n \left|x_i - y_i ight| + \sum_{i=1}^n \left|y_i - z_i ight| \ &= d'(\mathbf{x} , \mathbf{y} ) + d'(\mathbf{y} , \mathbf{z} ) \,. \end{align*} We now show that the metric topology induced by $d'$ is the same as that induced by the square metric $ $, which shows the desired result since the square metric induces the standard product topology on ${\mathbb{R}}^n$ by Theorem~20.3. First consider any $\mathbf{x} \in {\mathbb{R}}^n$ and any $\e > 0$. Let $\d = \e$ and consider any $\mathbf{y} \in B_{d'}(\mathbf{x} , \d)$. Suppose also that $j$ is an index in $\left\{1, \ldots, n ight\}$ where \begin{gather*} (\mathbf{x} , \mathbf{y} ) = \max\left\{\left|x_1 - y_1 ight|, \ldots, \left|x_n - y_n ight| ight\} = \left|x_j - y_j ight| \,. \end{gather*} Since $\mathbf{y} \in B_{d'}(\mathbf{x} , \d)$, we have \begin{gather*} d'(\mathbf{x} ,\mathbf{y} ) = \sum_{i=1}^n \left|x_i - y_i ight| < \d = \e \ \left|x_j - y_j ight| + \sum_{i eq j} \left|x_i - y_i ight| < \e \ \left|x_j - y_j ight| < \e - \sum_{i eq j} \left|x_i - y_i ight| \leq \e \ (\mathbf{x} , \mathbf{y} ) < \e \end{gather*} since of course $\sum_{i eq j} \left|x_i - y_i ight| \geq 0$. Therefore $\mathbf{y} \in B_ (\mathbf{x} ,\e)$, which shows $B_{d'}(\mathbf{x} , \d) \subset B_ (\mathbf{x} , \e)$ so that the metric topology of $d'$ is finer the the metric topology of $ $ by Lemma~20.2. Now again consider and $\mathbf{x} \in {\mathbb{R}}^n$ and $\e > 0$, and this time let $\d = \e/n$. Consider any $\mathbf{y} \in B_ (\mathbf{x} , \d)$ and again suppose also that $j$ is an index in $\left\{1, \ldots, n ight\}$ where \begin{gather*} (\mathbf{x} , \mathbf{y} ) = \max\left\{\left|x_1 - y_1 ight|, \ldots, \left|x_n - y_n ight| ight\} = \left|x_j - y_j ight| \,. \end{gather*} We then have \begin{gather*} \left|x_j - y_j ight| = (\mathbf{x} , \mathbf{y} ) < \d = \e/n \ n \left|x_j - y_j ight| < \e \,. \end{gather*} We also have \begin{align*} d'(\mathbf{x} , \mathbf{y} ) &= \sum_{i=1}^n \left|x_i - y_i ight| \ &\leq \sum_{i=1}^n \left|x_j - y_i ight| & ext{(since each $\left|x_i - y_i ight| \leq \left|x_j - y_j ight|$)} \ &= n \left|x_j - y_j ight| \ &< \e \end{align*} so that $\mathbf{y} \in B_{d'}(\mathbf{x} , \e)$. Hence $B_ (\mathbf{x} , \d) \subset B_{d'}(\mathbf{x} , \e)$ so that the metric topology of $ $ is also finer than that of $d'$ again by Lemma~20.2. Therefore it must be that the two topologies are equal since each is finer than the other. \end{proof} (b) Let $d$ denote the metric defined in part~(a), that is \begin{gather*} d(\mathbf{x} , \mathbf{y} ) = \sum_{i=1}^n \left|x_i - y_i ight| \,. \end{gather*} First we show that the metric topology induced by $d'$ is finer than that induced by $ $. So consider any $\mathbf{x} \in {\mathbb{R}}^n$ and $\e > 0$. Let $\d = \e$ and suppose that $\mathbf{y} \in B_{d'}(\mathbf{x} , \d)$ so that \begin{gather*} d'(\mathbf{x} , \mathbf{y} ) = \left(\sum_{i=1}^n \left|x_i - y_i ight|^p ight)^{1/p} < \d = \e \,. \end{gather*} Suppose that $j$ is an index in $\left\{1, \ldots, n ight\}$ where \begin{gather*} (\mathbf{x} , \mathbf{y} ) = \max\left\{\left|x_1 - y_1 ight|, \ldots, \left|x_n - y_n ight| ight\} = \left|x_j - y_j ight| \,. \end{gather*} Then \begin{gather*} \left|x_j - y_j ight|^p \leq \left|x_j - y_j ight|^p + \sum_{i eq j} \left|x_i - y_i ight|^p = \sum_{i=1}^n \left|x_i - y_i ight|^p \end{gather*} so that, by Corollary~ ef{cor:metric:opmon}, we have \begin{gather*} \left(\left|x_j - y_j ight|^p ight)^{1/p} \leq \left(\sum_{i=1}^n \left|x_i - y_i ight|^p ight)^{1/p} < \e \ \left|x_j - y_j ight| < \e \ (\mathbf{x} , \mathbf{y} ) < \e \,. \end{gather*} Therefore $\mathbf{y} \in B_ (\mathbf{x} , \e)$ so that $B_{d'}(\mathbf{x} , \d) \subset B_ (\mathbf{x} , \e)$. This suffices to show that the metric topology induced by $d'$ is finer than that induced by $ $ by Lemma~20.2. Now we show that the metric topology induced by $d$ is finer than that induced by $d'$. So again consider any $\mathbf{x} \in {\mathbb{R}}^n$ and $\e > 0$. Again let $\d = \e$ and suppose that $\mathbf{y} \in B_d(\mathbf{x} , \d)$ so that \begin{gather*} d(\mathbf{x} , \mathbf{y} ) = \sum_{i=1}^n \left|x_i - y_i ight| < \d = \e \,. \end{gather*} Then, since each $\left|x_i - y_i ight| \geq 0$, we have by Lemma~ ef{lem:metric:polynom} that \begin{align*} \sum_{i=1}^n \left|x_i - y_i ight|^p &\leq \left(\sum_{i=1}^n \left|x_i - y_i ight| ight)^p \ \left(\sum_{i=1}^n \left|x_i - y_i ight|^p ight)^{1/p} &\leq \left[\left(\sum_{i=1}^n \left|x_i - y_i ight| ight)^p ight]^{1/p} \ d'(\mathbf{x} , \mathbf{y} ) &\leq \sum_{i=1}^n \left|x_i - y_i ight| < \e \,, \end{align*} where we have used Corollary~ ef{cor:metric:opmon} in the second step. Thus $\mathbf{y} \in B_{d'}(\mathbf{x} , \e)$ so that $B_d(\mathbf{x} , \d) \subset B_{d'}(\mathbf{x} , \e)$. This of course shows that the metric topology induced by $d$ is finer than that induced by $d'$ by Lemma~20.2 again. Thus we have shown that the metric topology induced by $d'$ is finer than that induced by $ $, and also that that induced by $d$ is finer than that induced by $d'$. But it was shown in part~(a) and Theorem~20.3 that those induced by $d$ and $ $ are the same topology, which is is the usual product topology on ${\mathbb{R}}^n$. Hence if $\mathcal{T}_p$ denotes this usual product topology, we have \begin{gather*} \mathcal{T}_p = \mathcal{T}_ \subset \mathcal{T}_{d'} \subset \mathcal{T}_d = \mathcal{T}_ = \mathcal{T}_p \,. \end{gather*} So it must be that the metric topology induced by $d'$ is this topology as well as desired.

Exercise 20.1

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Exercise 20.1

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