Exercise 20.3

Question

\def\a{\alpha}
\def\b{\beta}
\def\d{\delta}
\def\e{\epsilon}
\def{ho}

ewcommand{\vect}[1]{\mathbf{#1}}
\def\va{\vect{a}}
\def\vb{\vect{b}}

  Let $X$ be metric space with metric $d$.
  \begin{enumerate}[label=(\alph*)]
  \item Show that $d : X 	imes X 	o {\mathbb{R}}$ is continuous.
  \item Let $X'$ denote a space having the same underlying set as $X$.
    Show that if $d : X' 	imes X' 	o {\mathbb{R}}$ is continuous, then the topology of $X'$ is finer than the topology of $X$.
  \end{enumerate}
  One can summarize the result of this exercise as follows: If $X$ has a metric $d$, then the topology induced by $d$ is the coarsest topology relative to which the function $d$ is continuous.

Dan Whitman · Accepted Answer

\def\a{\alpha} \def\b{\beta} \def\d{\delta} \def\e{\epsilon} \def { ho} ewcommand{\vect}[1]{\mathbf{#1}} \def\va{\vect{a}} \def\vb{\vect{b}} (a) We use Theorem~18.1 part~(4) to show that $d$ is continuous. So consider any $x_1 imes x_2 \in X imes X$ and any neighborhood $V$ of $z = d(x_1,x_2)$, noting that $V \subset {\mathbb{R}}$ since ${\mathbb{R}}$ is the range of $d$. Since $V$ is open in ${\mathbb{R}}$, there is a basis element $B = (a,b)$ containing $z$ where $B \subset V$. Hence $a < z < b$. Now let $\e = \min\left\{(z-a)/2, (b-z)/2 ight\}$, noting that $\e > 0$ since $z > a$ and $b > z$. Next define $U_1 = B_d(x_1, \e)$ and $U_2 = B_d(x_2, \e)$ so that they are both basis elements and therefore open sets of the metric space $X$. It then follows that $U = U_1 imes U_2$ is a basis element and therefore an open set of the product space $X imes X$. Clearly we have that $x_1 \in B_d(x_1, \e) = U_1$ and $x_2 \in B_d(x_2, \e) = U_2$ so that $U$ contains $x_1 imes x_2$ and so is a neighborhood of $x_1 imes x_2$. We claim that $d(U) \subset B$. To see this, consider any $w \in d(U)$ so that there is a $y_1 imes y_2 \in U = U_1 imes U_2$ such that $w = d(y_1, y_2)$. Therefore $y_1 \in U_1 = B_d(x_1, \e)$ so that $d(y_1, x_1) < \e$, and similarly $d(y_2, x_2) < \e$ since $y_2 \in U_2 = B_d(x_2, \e)$. Then, since $d$ is a metric, we have \begin{align*} z = d(x_1, x_2) &\leq d(x_1, y_1) + d(y_1, x_2) \ &\leq d(x_1, y_1) + d(y_1, y_2) + d(y_2, x_2) \ &= d(y_1, x_1) + d(y_1, y_2) + d(y_2, x_2) \ &< \e + w + \e \ z &< w + 2\e \leq w + 2(z-a)/2 = w + z - a \ a &< w \,. \end{align*} Similarly, we have \begin{align*} w = d(y_1, y_2) &\leq d(y_1, x_1) + d(x_1, y_2) \ &\leq d(y_1, x_1) + d(x_1, x_2) + d(x_2, y_2) \ &= d(y_1, x_1) + d(x_1, x_2) + d(y_2, x_2) \ &< \e + z + \e \ w &< z + 2\e \leq z + 2(b-z)/2 = z + b - z \ w &< b \,. \end{align*} We, therefore, have that $a < w < b$ so that $w \in (a,b) = B$. This of course shows that $d(U) \subset B$ since $w$ was arbitrary. Moreover, we have that $B \subset V$ so that clearly $d(U) \subset V$, which completes the proof of Theorem~18.1 part~(4) so that $d$ is continuous. (b) Let $U$ be any open set of $X$ and consider any $x \in U$. Then clearly there is a basis element $B_d(y, \e)$, for some $\e > 0$ and $y \in U$, of the metric topology $X$ that contains $x$ and where $B_d(y, \e) \subset U$. Now, since $d$ is continuous with respect to $X' imes X'$, it follows from \href{./exercise_18-11}{Exercise~18.11} that the function $d_y(z) = d(y,z)$ is a continuous function from $X'$ to ${\mathbb{R}}$. Since clearly the interval $(-\infty, \e)$ is open in ${\mathbb{R}}$, it then follows that the set \begin{gather*} {d_y}^{-1}((-\infty, \e)) = \left\{z \in X' \mid d_y(z) < \e ight\} = \left\{z \in X \mid d(y, z) < \e ight\} = B_d(y, \e) \end{gather*} is also open in $X'$. Thus $B_d(y,\e)$ is an open set in $X'$ containing $x$ such that $B_d(y, \e) \subset U$. This shows that $U$ is also open in $X'$ by Exercise~13.1 since the point $x \in U$ was arbitrary. This suffices to show the desired result.

Exercise 20.3

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Exercise 20.3

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