Exercise 21.3

Proof of $(a)$. $\rho$ satisfies properties $(1)$ and $(2)$ on p. 119 since the components do; it then suffices to show the triangle inequality. We first have $d_i(x_i,z_i)\le d(x_i,y_i)+d(y_i,z_i)$ for all $i$. Then, by definition of $\rho$, $d_i(x_i,z_i)\le \rho (x,y)+\rho (y,z)$ for all $i$. But since this is true for all $i$, we have that $\rho (x,z)\le \rho (x,y)+\rho (y,z)$.

We now show that this defines a metric for the product space. First let $B=\prod <!--\; FUNCTION\; APPLICATION\; -->U_i$ be a basis element of $\prod <!--\; FUNCTION\; APPLICATION\; -->X_i$, and let $x\in B$. For each $i$, there is an ${𝜖}_i$ such that $B_{d_i}(x_i,{𝜖}_i)\subseteq U_i$. Choosing $𝜖=\min <!--\; FUNCTION\; APPLICATION\; -->\{{𝜖}_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{𝜖}_n\}$, we see that $B_{\rho }(x,𝜖)\subseteq B$, since if $y\in B_{\rho }(x,𝜖)$, $d_i(x_i,y_i)\le \rho (x,y)<𝜖\le {𝜖}_i$, and so $y\in \prod <!--\; FUNCTION\; APPLICATION\; -->U_i$ as desired. Thus the metric topology is finer than the product topology.

Conversely, let $B_{\rho }(x,𝜖)$ be a basis element in the metric topology; since it is the product $B_{d_i}(x_i,𝜖)$, we see that the product topology is finer than the metric topology. These two facts imply the two topologies are equal. □

Proof of $(b)$. $D$ satisfies properties $(1)$ and $(2)$ on p. 119 since the components do; it then suffices to show the triangle inequality. We first have

for all $i$. But since this is true for all $i$, we have that

We now show that this defines a metric for the product space. Let $U$ be open in the metric topology and let $x\in U$; choose an open ball $B_D(x,𝜖)\subseteq U$. Choose $N$ such that $1∕N<𝜖$, and let

We claim $V\subseteq B_D(x,𝜖)\subseteq U$. Given $y\in \prod <!--\; FUNCTION\; APPLICATION\; -->X_i$, ${\overline{d}}_i(x_i,y_i)∕i\le 1∕N$ for $i\ge N$. Therefore,

If $y\in V$, this expression is less than $𝜖$, so $V\subseteq B_D(x,𝜖)$ as desired, and the product topology is finer than the metric topology.

Conversely, let $U=\prod <!--\; FUNCTION\; APPLICATION\; -->U_i$ where $U_i$ is open in $X_i$ for ${\alpha }_1,\dots ,{\alpha }_n$ and $U_i=X_i$ otherwise. Let $x\in U$ be given, and choose $B_{{\overline{d}}_i}(x_i,{𝜖}_i)\subseteq X_i$ for $i={\alpha }_1,\dots ,{\alpha }_n$, where each ${𝜖}_i\le 1$. Then, defining $𝜖=\min <!--\; FUNCTION\; APPLICATION\; -->\{{𝜖}_i∕i\mid i={\alpha }_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\alpha }_n\}$, we claim that $x\in B_D(x,𝜖)\subseteq U$. Let $y$ be a point of $B_D(x,𝜖)$. Then, for all $i$,

Now if $i={\alpha }_1,\dots ,{\alpha }_n$, then $𝜖\le {𝜖}_i∕i$, so that ${\overline{d}}_i(x_i,y_i)<{𝜖}_i\le 1$; it follows that $|x_i-y_i|<{𝜖}_i$, and so $y\in \prod <!--\; FUNCTION\; APPLICATION\; -->U_i$ as desired. We thus have that the metric topology is finer than the product topology; combined with the above this implies the topologies are equal. □

Proof. The proofs of both parts are quite simple. Regarding part (1) we have that $x_{\beta }\le y_{\beta }\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{y_{\alpha }\right\}$ for any $\beta \in J$ so that clearly $\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{y_{\alpha }\right\}$ is an upper bound of the set $\left\{x_{\alpha }\right\}$. It then follows by the definition of the supremum as the least upper bound that $\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{x_{\alpha }\right\}\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{y_{\alpha }\right\}$ as desired.

For part (2) consider any $\beta \in J$. Clearly $x_{\beta }\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{x_{\alpha }\right\}$ and $y_{\beta }\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{y_{\alpha }\right\}$ so that

Since $\beta$ was arbitrary, this shows that $\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{x_{\alpha }\right\}+\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{y_{\alpha }\right\}$ is an upper bound for the set $\left\{x_{\alpha }+y_{\alpha }\right\}$. Therefore $\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{x_{\alpha }+y_{\alpha }\right\}\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{x_{\alpha }\right\}+\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{y_{\alpha }\right\}$ as desired by the definition of the supremum as the least upper bound. □

Proof. First we must show that $\rho$ is a metric at all. In what follows suppose that $x,y,z\in X_1\times \cdots \times X_n$ and that $k$ and $m$ are elements of $\left\{1,\dots ,n\right\}$ such that

and

First, we have $\rho (x,y)=d_k(x_k,y_k)\ge 0$ since $d_k$ is a metric. If $x=y$ then of course $x_k=y_k$ so that $\rho (x,y)=d_k(x_k,y_k)=0$. Now suppose that $\rho (x,y)=0$ and consider any $l\in \left\{1,\dots ,n\right\}$. Then we have that $d_l(x_l,y_l)\ge 0$ and that $d_l(x_l,y_l)\le \rho (x,y)=0$, and hence it must be that $d_l(x_l,y_l)=0$ so that $x_l=y_l$ since $d_l$ is a metric. Since $l$ was arbitrary, this shows that $x=y$, which shows that $\rho$ satisfies part (1) of the definition of a metric.

As usual, part (2) of the definition is the easiest to show since

as we have that each $d_l(x_l,y_l)=d_l(y_l,x_l)$ since $d_l$ is a metric. Lastly, for part (3) we have

since of course $d_m$ is a metric. This completes the proof that $\rho$ is a proper metric.

Now we show that both topologies are the same using Lemma 13.3. So suppose that $x\in X_1\times \cdots \times X_n$ and $B_{\rho }(x,𝜖)$ is any basis element of the metric topology and let $B={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{B}_{d_i}(x_i,𝜖)$, which is clearly a basis element of the product topology that contains $x$ since each $B_{d_i}(x_i,𝜖)$ is open in the metric space $X_i$. Now suppose that $y\in B$ so that each $y_i\in B_{d_i}(x_i,𝜖)$. So, for every $i\in \left\{1,\dots ,n\right\}$, we have $d_i(y_i,x_i)<𝜖$ so that clearly

which shows that $y\in B_{\rho }(x,𝜖)$. This shows that $x\in B\subset B_{\rho }(x,𝜖)$ so that $B_{\rho }(x,𝜖)$ the product topology is finer than the metric topology by Lemma 13.3.

Now consider again any $x\in X_1\times \cdots \times X_n$ and any basis element $B={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{U}_i$ of the product topology. Then of course each $U_i$ is open in $X_i$ and $x_i\in U_i$ so that there is a ball $B_{d_i}(x_i,{𝜖}_i)$ such that $B_{d_i}(x_i,{𝜖}_i)\subset U_i$ by Lemma 20.4.1. Let $𝜖=\min <!--\; FUNCTION\; APPLICATION\; -->\left\{{𝜖}_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{𝜖}_n\right\}$ and consider the basis element $B_{\rho }(x,𝜖)$ in the metric space induced by $\rho$. Clearly $x\in B_{\rho }(x,𝜖)$ so consider any $y\in B_{\rho }(x,𝜖)$ so that

Then, for any $i\in \left\{1,\dots ,n\right\}$, we have

so that $y_i\in B_{d_i}(x_i,{𝜖}_i)\subset U_i$. Since this is true of any $i\in \left\{1,\dots ,n\right\}$, it follows that $y\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{U}_i=B$. We can then conclude that $x\in B_{\rho }(x,𝜖)\subset B$ since $y$ was arbitrary. This shows that the $\rho$-metric topology is finer than the product topology, again by Lemma 13.3. Hence the two topologies are the same as desired. □

Proof. First we show that the metric $D$ is well-defined and is in fact a metric. In what follows suppose that $x,y,z\in \prod <!--\; FUNCTION\; APPLICATION\; -->X_i$. For any $j\in {\mathbb{Z}}_{\text{+}}$ we have that $0<1\le j$ so that $1∕j\le 1$. We also have that ${\overline{d}}_j(x_j,y_j)=\min <!--\; FUNCTION\; APPLICATION\; -->\left\{d_j(x_j,y_j),1\right\}\le 1$ so that ${\overline{d}}_j(x_j,y_j)∕j\le 1∕j\le 1$. Hence $1$ is an upper bound for the set $\left\{{\overline{d}}_i(x_i,y_i)∕i\right\}$ since $j$ was arbitrary, so that

exists and hence $D$ is well defined.

To show part (1) of the definition of a metric, pick any $j\in {\mathbb{Z}}_{\text{+}}$ so that clearly ${\overline{d}}_j(x_j,y_j)\ge 0$ and hence ${\overline{d}}_j(x_j,y_j)∕j\ge 0$ also since $j\ge 1>0$. Thus we have

If $x=y$ then we have that $x_j=y_j$ for any $j\in {\mathbb{Z}}_{\text{+}}$. Thus ${\overline{d}}_j(x_j,y_j)=0$ since ${\overline{d}}_j$ is a standard bounded metric, and therefore

Since $j$ was arbitrary, this shows that

On the other hand, if $D(x,y)=0$ then, for any $j\in {\mathbb{Z}}_{\text{+}}$, we have

This shows that ${\overline{d}}_j(x_j,y_j)∕j=0$ so that clearly ${\overline{d}}_j(x_j,y_j)=0$ as well, from which it follows that $x_j=y_j$ since ${\overline{d}}_j$ is a standard bounded metric. Since $j$ was arbitrary, this shows that $x=y$, which completes the proof of part (1).

Showing part (2) is quite easy:

since ${\overline{d}}_i$ is a standard bounded metric for every $i\in {\mathbb{Z}}_{\text{+}}$.

For part (3) consider any $j\in {\mathbb{Z}}_{\text{+}}$. Then

since ${\overline{d}}_j$ is a standard bounded metric. Then of course

since $j\ge 1>0$. Then, since $j$ was arbitrary this shows that

by both parts of Lemma 1 . This of course completes the proof that $D$ is a well defined metric.

Now we show that the metric topology induced by $D$ is the same as the product topology on $\prod <!--\; FUNCTION\; APPLICATION\; -->X_i$, which we do using Lemma 13.3. So consider any $x\in \prod <!--\; FUNCTION\; APPLICATION\; -->X_i$ and any basis element of the metric topology $B_D(x,𝜖)$ centered at $x$. Clearly there is a positive integer $N$ large enough such that $N>2∕𝜖$, and so $1∕N<𝜖∕2$. Now define the sets

for $i\in {\mathbb{Z}}_{\text{+}}$, and the set $B=\prod <!--\; FUNCTION\; APPLICATION\; -->U_i$. Clearly $B$ is a basis element of the product topology since each $U_i$ is open in $X_i$ and $U_i\ne ℝ$ for only finitely many $i$, namely when $i<N$. Clearly also $x\in B$ since each $x_i\in U_i$. Now suppose that $y\in B$ and consider any $j\in {\mathbb{Z}}_{\text{+}}$. If $j<N$ then we of course have that $y_j\in U_j=B_{d_j}(x_j,𝜖∕2)$ so that

since $j\ge 1$. If $j\ge N$ then we have $1∕j\le 1∕N$ so that

since ${\overline{d}}_j(y_j,x_j)\le 1$. Since $j$ was arbitrary, this shows that

so that $y\in B_D(x,𝜖)$. Therefore $x\in B\subset B_D(x,𝜖)$ since $y$ was arbitrary. Hence the product topology is finer than the metric topology by Lemma 13.3.

Now again consider any $x\in \prod <!--\; FUNCTION\; APPLICATION\; -->X_i$ and any basis element $B=\prod <!--\; FUNCTION\; APPLICATION\; -->U_i$ of the product topology where $x\in B$. Then of course each $U_i$ is open in $X_i$ and there is a finite subset $J\subset {\mathbb{Z}}_{\text{+}}$ where $U_i=ℝ$ for every $i\notin J$. Of course also $x\in U_i$ for each $i\in {\mathbb{Z}}_{\text{+}}$. For any $j\in J$ we have that $x\in U_j$ and $U_j$ is open in $X_i$ so that there is a basis element $B_{d_j}(x_j,{𝜖}_j)$ such that $B_{d_j}(x_j,{𝜖}_j)\subset U_j$. So let $𝜖=\min <!--\; FUNCTION\; APPLICATION\; -->(\left\{{𝜖}_j\mid j\in J\right\}\cup \left\{1\right\})$ and $k=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{j\mid j\in J\right\}$, which both exists since $J$ is finite, and also clearly $𝜖>0$.

Now consider the set $B_D(x,𝜖∕k)$, which is a basis element of the metric topology that clearly contains $x$. Suppose that $y\in B_D(x,𝜖∕k)$ so that $D(y,x)<𝜖∕k$. Then, for any $j\in {\mathbb{Z}}_{\text{+}}$, clearly $y_i\in ℝ=U_i$ if $j\ne J$. On the other hand, if $j\in J$ then we have that $k\ge j$ so that $1∕k\le 1∕j$. We also have

since $1∕k\le 1∕j$ and $𝜖\le 1$. Therefore ${\overline{d}}_j(y_j,x_j)<1$ so that it must be that ${\overline{d}}_j(y_j,x_j)=d_j(y_j,x_j)$. Hence we have $d_j(y_j,x_j)={\overline{d}}_j(y_j,x_j)<𝜖\le {𝜖}_j$ so that $y_j\in B_{d_j}(x_j,{𝜖}_j)\subset U_j$. Therefore $y_i\in U_i$ for all $i\in {\mathbb{Z}}_{\text{+}}$ so that $y\in \prod <!--\; FUNCTION\; APPLICATION\; -->U_i=B$. Since $y$ was arbitrary, this shows that $x\in B_D(x,𝜖)\subset B$, which in turn proves that the metric topology is also finer than the product topology by Lemma 13.3. Thus the two topologies must be the same. □