Exercise 21.12

Question

\def\a{\alpha}
\def\b{\beta}
\def\d{\delta}
\def\e{\epsilon}
\def\r{\rho}

Prove continuity of the algebraic operations on ${\mathbb{R}}$, as follows: Use the metric $d(a,b) = \left|a-b\right|$ on ${\mathbb{R}}$ and the metric on ${\mathbb{R}}^2$ given by the equation
  \begin{gather*}
    \r((x,y),(x_0,y_0)) = \max\left\{\left|x-x_0\right|,\left|y-y_0\right|\right\} \,.
  \end{gather*}
  \begin{enumerate}[label=(\alph*)]
  \item Show that addition is continuous.
    [Hint: Given $\e$, let $\d=\e/2$ and note that
      \begin{gather*}
        d(x+y, x_0+y_0) \leq \left|x-x_0\right| + \left|y-y_0\right| \,.]
    \end{gather*}
  \item Show that multiplication is continuous.
    [Hint: Given $(x_0,y_0)$ and $0 < \e < 1$, let
      \begin{gather*}
        3\d = \e/(\left|x_0\right| + \left|y_0\right| + 1)
      \end{gather*}
      and note that
      \begin{gather*}
        d(xy, x_0 y_0) \leq \left|x_0\right|\left|y-y_0\right| + \left|y_0\right|\left|x-x_0\right| + \left|x-x_0\right|\left|y-y_0\right| \,.]
    \end{gather*}
  \item Show that the operation of taking reciprocals is a continuous map from ${\mathbb{R}}nz$ to ${\mathbb{R}}$.
    [Hint: Show the inverse image of the interval $(a,b)$ is open.
      Consider five cases, according as $a$ and $b$ are positive, negative, or zero.]
  \item Show that the subtraction and quotient operations are continuous.
  \end{enumerate}

Dan Whitman · Accepted Answer

\def\a{\alpha} \def\b{\beta} \def\d{\delta} \def\e{\epsilon} \def { ho} First, we note that this exercise proves Lemma~21.4 in the text. We also note that the square metric $ $ as defined above induces the usual product topology on ${\mathbb{R}}^2$, which was shown in Theorem~20.3. (a) \begin{proof} We show the continuity of $+: {\mathbb{R}}^2 o {\mathbb{R}}$ using Theorem~21.1. So consider any $x_0 imes y_0 \in {\mathbb{R}}^2$ and any $\e>0$. Following the hint, let $\d=\e/2$ and consider any $x imes y \in {\mathbb{R}}^2$ where $ (x imes y,x_0 imes y_0) < \d$. Then we have that \begin{gather*} (x imes y,x_0 imes y_0) = \max\left\{\left|x-x_0 ight|,\left|y-y_0 ight| ight\} < \d = \e/2 \,, \end{gather*} from which it clearly follows that both $\left|x-x_0 ight| < \e/2$ and $\left|y-y_0 ight| < \e/2$. Then we have \begin{align*} d(+(x imes y), +(x_0 imes y_0)) &= d(x+y,x_0+y_0) = \left|x+y - (x_0+y_0) ight| \ &= \left|(x-x_0) + (y-y_0) ight| \leq \left|x-x_0 ight| + \left|y-y_0 ight| \ &< \frac{\e}{2} + \frac{\e}{2} = \e \,. \end{align*} This suffices to show that $+$ is continuous by Theorem~21.1 since $\e$ was arbitrary. \end{proof} (b) \begin{proof} We again use Theorem~21.1 so show that the multiplication function $ imes : {\mathbb{R}}^2 o {\mathbb{R}}$ is continuous. So consider any $x imes y \in {\mathbb{R}}^2$ and any $\e > 0$. Let \begin{gather*} \d = \min\left\{\frac{\e}{\left|x_0 ight| + \left|y_0 ight| + 1}, 1 ight\} \end{gather*} so that of course $\d > 0$. Now consider any $x imes y \in {\mathbb{R}}^2$ where $ (x imes y,x_0 imes y_0) < \d$. Then of course \begin{gather*} (x imes y,x_0 imes y_0) = \max\left\{\left|x-x_0 ight|,\left|y-y_0 ight| ight\} < \d \end{gather*} so that both $\left|x-x_0 ight| < \d$ and $\left|y-y_0 ight| < \d$. We then have \begin{align*} d( imes(x imes y), imes(x_0 imes y_0)) &= d(xy, x_0 y_0) = \left|xy - x_0 y_0 ight| \ &= \left|xy - x_0 y_0 + (x_0 y - x_0 y) + (x y_0 - x y_0) + (x_0 y_0 - x_0 y_0) ight| \ &= \left|(x_0 y - x_0 y_0) + (x y_0 - x_0 y_0) + (xy - x y_0 - x_0 y + x_0 y_0) ight| \ &= \left|x_0(y-y_0) + y_0(x-x_0) + (x-x_0)(y-y_0) ight| \ &\leq \left|x_0(y-y_0) ight| + \left|y_0(x-x_0) ight| + \left|(x-x_0)(y-y_0) ight| \ &= \left|x_0 ight|\left|y-y_0 ight| + \left|y_0 ight|\left|x-x_0 ight| + \left|x-x_0 ight|\left|y-y_0 ight| \ &< \left|x_0 ight| \d + \left|y_0 ight| \d + \d^2 \ &\leq \left|x_0 ight| \d + \left|y_0 ight| \d + \d \ &= \d (\left|x_0 ight| + \left|y_0 ight| + 1) \ &\leq \frac{\e}{\left|x_0 ight| + \left|y_0 ight| + 1} (\left|x_0 ight| + \left|y_0 ight| + 1) \ &= \e \,, \end{align*} where we have used the fact that $0 < \d \leq 1$, from which it follows that $\d^2 \leq d$. This suffices to show that $ imes$ is continuous by Theorem~21.1 since $\e$ was arbitrary. \end{proof} \def a{ frac{1}{a}} \def b{ frac{1}{b}} \def x{ frac{1}{x}} (c) \begin{proof} Define the reciprocal function $f: {\mathbb{R}}nz o {\mathbb{R}}$ by $f(x) = x$. To show that $f$ is continuous, it suffices to show that ${f}^{-1}(B)$ is open in ${\mathbb{R}}nz$ for every basis element $B$ of ${\mathbb{R}}$, which was shown back in \S 18. So let $B = (a,b)$ be any basis element of ${\mathbb{R}}$. Case: $a > 0$. Then it has to be that $0 < a < b$. We claim that ${f}^{-1}(B) = ( b, a)$, noting that $0 < b < a$ follows readily from the fact that $0 < a < b$. We have the following \begin{align*} x \in ( b, a) &\Leftrightarrow b < x < a \Leftrightarrow b < x \land x < a \ &\Leftrightarrow 1 < bx \land ax < 1 \Leftrightarrow x < b \land a < x \ &\Leftrightarrow a < x < b \Leftrightarrow f(x) = x \in (a,b) = B \ &\Leftrightarrow x \in {f}^{-1}(B) \,, \end{align*} noting that $x > a > 0$ and $ x > b > 0$. This of course shows that ${f}^{-1}(B) = ( b, a)$ as desired. Case: $a = 0$. Then we have $0 = a < b$ and hence also $0 < b$. We claim that ${f}^{-1}(B) = ( b,\infty)$. So first suppose that $x \in {f}^{-1}(B)$ so that $f(x) \in B = (a,b) = (0,b)$, and hence $0 < x < b$. Then $0 < 1 < b/x$, from which follows $0 < b < x$ so that $x \in ( b,\infty)$. Now consider any $x \in ( b,\infty)$ so that $0 < b < x$. Then $0 < 1 < bx$ so that $0 < x < b$, and hence $f(x) = x \in (0,b) = (a,b) = B$. Therefore $x \in {f}^{-1}(B)$, from which we conclude that ${f}^{-1}(B) = ( b,\infty)$ as desired. Case: $a < 0$. Then also $ a < 0$ and we have the following sub-cases: \begin{adjustwidth}{1cm}{} Case: $b > 0$. Then also $ b > 0$. Here we claim that ${f}^{-1}(B) = (-\infty, a) \cup ( b,\infty)$. So suppose that $x \in {f}^{-1}(B)$ so that $f(x) = x \in B = (a,b)$, and hence $a < x < b$. We know that $x eq 0$ since the domain of $f$ is ${\mathbb{R}}nz$. If $x < 0$ then we have $a < x$ so that $ax > 1$, and hence $x < a$ since $a < 0$. Therefore $x \in (\infty, a)$. On the other hand, if $x > 0$ then we have $ x < b$ so that $1 < bx$, and hence $ b < x$ since $b>0$. Therefore $x \in ( b,\infty)$. So either way $x \in (-\infty, a) \cup ( b, \infty)$ so that ${f}^{-1}(B) \subset (-\infty, a) \cup ( b, \infty)$. Now consider $x \in (-\infty, a) \cup ( b, \infty)$. If $x \in (-\infty. a)$ then $x < a < 0$ so that $ax > 1$, and hence $a < x < 0 < b$. If $x \in ( b, \infty)$ then $0 < b < x$ so that $1 < bx$, and hence $a < 0 < x < b$. Thus either way we have $a < x = f(x) < b$ so that $f(x) \in (a,b) = B$ and $x \in {f}^{-1}(B)$. Hence $(-\infty, a) \cup ( b, \infty) \subset {f}^{-1}(B)$, which completes the proof that ${f}^{-1}(B) = (-\infty, a) \cup ( b, \infty)$ as desired. Case: $b = 0$. Then we have $a < b = 0$. An argument analogous to the previous case when $a = 0 < b$ shows that ${f}^{-1}(B) = (-\infty, a)$. Case: $b < 0$. Then we have $a < b < 0$. An argument analogous to the previous case when $0 < a < b$ again shows that ${f}^{-1}(B) = ( b, a)$. \end{adjustwidth} Thus in all cases and sub-cases clearly ${f}^{-1}(B)$ is an open set of ${\mathbb{R}}nz$, which shows that $f$ is continuous as previously discussed since $B = (a,b)$ was an arbitrary basis element. \end{proof} (d) \begin{proof} Regarding the subtraction function $-: {\mathbb{R}}^2 o {\mathbb{R}}$, first define the function $f: {\mathbb{R}} o {\mathbb{R}}$ by $f(x) = -x$, which is clearly continuous by elementary calculus. We also know that the coordinate functions $\pi_1 : {\mathbb{R}}^2 o {\mathbb{R}}$ and $\pi_2: {\mathbb{R}}^2 o {\mathbb{R}}$ are also continuous. Therefore the composition $f \circ \pi_2 : {\mathbb{R}}^2 o {\mathbb{R}}$ is also continuous by Theorem~18.2 part~(c). We then have that the function $g: {\mathbb{R}}^2 o {\mathbb{R}}^2$ defined by $g(x imes y) = \pi_1(x imes y) imes (f \circ \pi_2)(x imes y)$ is continuous by Theorem~18.4. Then the composition $+ \circ g : {\mathbb{R}}^2 o {\mathbb{R}}$ is continuous, again by Theorem~18.2 part~(c), since it was shown in part~(a) that the addition function $+: {\mathbb{R}}^2 o {\mathbb{R}}$ is continuous. However, for any $x imes y \in {\mathbb{R}}^2$, we have \begin{align*} (+ \circ g)(x imes y) &= +(g(x imes y)) = +(\pi_1(x imes y) imes (f \circ \pi_2)(x imes y)) \ &= \pi_1(x imes y) + (f \circ \pi_2)(x imes y) = x + f(\pi_1(x imes y)) \ &= x + f(y) = x - y \ &= -(x imes y) \end{align*} so that $- = + \circ g$ is continuous just as we would like. Regarding the quotient function $/: {\mathbb{R}} imes ({\mathbb{R}}nz) o {\mathbb{R}}$, let $f: {\mathbb{R}}nz o {\mathbb{R}}$ be the reciprocal function, which we know is continuous by part~(c). Again the coordinate functions $\pi_1: {\mathbb{R}} imes ({\mathbb{R}}nz) o {\mathbb{R}}$ and $\pi_2: {\mathbb{R}} imes ({\mathbb{R}}nz) o {\mathbb{R}}nz$ are continuous so that the composition $f \circ \pi_2: {\mathbb{R}} imes ({\mathbb{R}}nz) o {\mathbb{R}}$ is also continuous by Theorem~18.2 part~(c). Then we have that the function $g: {\mathbb{R}} imes ({\mathbb{R}}nz) o {\mathbb{R}}^2$ defined by $g(x imes y) = \pi_1(x imes y) imes (f \circ \pi_2)(x imes y)$ is continuous by Theorem~18.4. Thus the composition $ imes \circ g : {\mathbb{R}} imes ({\mathbb{R}}nz) o {\mathbb{R}}$ is continuous again by Theorem~18.2 part~(c) since it was shown in part~(b) that the multiplication function $ imes: {\mathbb{R}}^2 o {\mathbb{R}}$ is continuous. Now, for any $x imes y \in {\mathbb{R}} imes ({\mathbb{R}}nz)$, we have that \begin{align*} ( imes \circ g)(x imes y) &= imes(g(x imes y)) = imes(\pi_1(x imes y) imes (f \circ \pi_2)(x imes y)) \ &= \pi_1(x imes y) \cdot (f \circ \pi_2)(x imes y) = x \cdot f(\pi_2(x imes y)) \ &= x \cdot f(y) = x \cdot frac{1}{y} = frac{x}{y} \ &= /(x imes y) \end{align*} so that $/ = imes \circ g$ is continuous as desired. \end{proof}

Exercise 21.12

Answers

Comments

Exercise 21.12

Answers

Comments

Add answer