# Prob. 3 (a), Sec. 21, in Munkres' TOPOLOGY, 2nd ed: A Finite Cartesian Product Of Metrizable Spaces Is Metrizable

Here is Prob. 3, Sec. 21, in the book Topology by James R. Munkres, 2nd edition:

Let $X_n$ be a metric space with metric $d_n$, for $n \in \mathbb{Z}_+$.

(a) Show that $$\rho ( \mathbf{x}, \mathbf{y} ) = \max \left\{ \ d_1 \left( x_1, y_1 \right), \ldots, d_n \left( x_n, y_n \right) \ \right\}$$ is a metric for the product space $X_1 \times \cdots \times X_n$.

(b) Let $\bar{d}_i = \min \left\{ \ d_i, 1 \ \right\}$. Show that $$D ( \mathbf{x}, \mathbf{y} ) = \sup \left\{ \ \bar{d}_i \left( x_i, y_i \right) / i \ \right\}$$ is a metric for the product space $\prod X_i$.

My Attempt:

Part (a)

Let $B \colon= \prod_{i=1}^n U_i$ be an arbitrary basis element for the product topology on $\prod_{i=1}^n X_i$, where, for each $i \in \{ \ 1, \ldots, n \ \}$, the set $U_i$ is a subset of $X_i$ such that $U_i$ is open in $\left( X_i, d_i \right)$. Let $\mathbf{p} \colon= \left( p_1, \ldots, p_n \right)$ be an arbitrary element of $B$.

Then, for each $i \in \{ \ 1, \ldots, n \ \}$, $p_i \in U_i$ and $U_i$ is open in the metric space $\left( X_i, d_i \right)$, which implies that there is a real number $\delta_i > 0$ such that the set $$B_{d_i} \left( p_i, \delta_i \right) \colon= \left\{ \ x \in X_i \colon \ d_i \left( x, p_i \right) < \delta_i \ \right\}$$ is contained in $U_i$.

Let $\delta \colon= \min \left\{ \ \delta_1, \ldots, \delta_n \ \right\}$. Then $\delta > 0$ of course. Moreover, if $\mathbf{x} \colon= \left( x_1, \ldots, x_n \right)$ is any point in $\prod_{i=1}^n X_i$ such that $$\mathbf{x} \in B_{\rho} ( \mathbf{p}, \delta ) \colon= \left\{ \ \mathbf{y} \in \prod_{i=1}^n X_i \ \colon \ \rho ( \mathbf{y}, \mathbf{p} ) < \delta \ \right\},$$ then for each $i \in \{ \ 1, \ldots, n \ \}$, we have $$d_i \left( x_i, p_i \right) \leq \rho ( \mathbf{x}, \mathbf{p} ) < \delta \leq \delta_i,$$ which implies that $x_i \in B_{d_i} \left( p_i, \delta_i \right)$ and hence $x_i \in U_i$. So $\mathbf{x} \in B$. Moreover, $\mathbf{p} \in B_{\rho} ( \mathbf{p}, \delta)$.

Thus, we have shown that, for any basis set $B$ for the product topology on $\prod_{i=1}^n X_i$ and for any element $\mathbf{p} \in B$, there is a basis element $B^\prime \colon= B_{\rho} ( \mathbf{p}, \delta)$ for the $\rho$-metric topology on $\prod_{i=1}^n X_i$ such that $$\mathbf{p} \in B^\prime \subset B.$$ Therefore the $\rho$-metric topology is finer than the product topology on $\prod_{i=1}^n X_i$.

Now let $B_{\rho}(\mathbf{p}, \epsilon )$ be an arbitrary basis element for the $\rho$-metric topology on $\prod_{i=1}^n X_i$, where $\mathbf{p} \colon= \left( p_1, \ldots, p_n \right)$ is an arbitrary point of $\prod_{i=1}^n X_i$ and $\epsilon$ is a positive real number. Now for each point $\mathbf{x} \in B_{\rho} ( \mathbf{p}, \epsilon)$, we need to find a basis element $B_{\mathbf{x}}$ for the product topology on $\prod_{i=1}^n X_i$ such that $$\mathbb{x} \in B_{\mathbf{x}} \subset B_{\rho}( \mathbf{p}, \epsilon). \tag{1}$$

But $$B_{\rho} (\mathbf{p}, \epsilon) = \prod_{i=1}^n B_{d_i} \left( p_i, \epsilon \right). \tag{2}$$ This we show as follows. If $\mathbf{x} \colon= \left( x_1, \ldots, x_n \right) \in B_{\rho} ( \mathbf{p}, \epsilon)$, then $\mathbf{x} \in \prod_{i=1}^n X_i$ and $\rho ( \mathbf{x}, \mathbf{p} ) < \epsilon$. So for each $i = \{ \ 1, \ldots, n \ \}$, we have $$d_i \left( x_i, p_i \right) \leq \rho ( \mathbf{x}, \mathbf{p} ) < \epsilon,$$ and hence $x_i \in B_{d_i} \left( p_i, \epsilon \right)$. Therefore $\mathbf{x} \in \prod_{i=1}^n B_{d_i} \left( p_i, \epsilon \right)$. Thus $B_{\rho} ( \mathbf{p}, \epsilon) \subset \prod_{i=1}^n B_{d_i} \left( p_i, \epsilon \right)$. On the other hand, if $\mathbf{x} \in \prod_{i=1}^n B_{d_i} \left( p_i, \epsilon \right)$, then for each $i \in \{ \ 1, \ldots, n \ \}$ we can conclude that $x_i \in B_{d_i} \left( p_i, \epsilon \right)$, which implies that $x_i \in X_i$ and $d_i \left( x_i, p_i \right) < \epsilon$; so $$\rho ( \mathbf{x}, \mathbf{p} ) = \max \left\{ \ d_1 \left( x_1, p_1 \right), \ldots, d_n \left( x_n, p_n \right) \right\} < \epsilon$$ also, which implies that $\mathbf{x} \in B_{\rho} ( \mathbf{p}, \epsilon)$ and hence $\prod_{i=1}^n B_{d_i} \left( p_i, \epsilon \right) \subset B_{\rho} ( \mathbf{p}, \epsilon)$. Therefore (2) holds.

But the set on the right-hand-side of (2) is a basis element for the product topology on $\prod_{i=1}^n X_i$, because for each $i \in \{ \ 1, \ldots, n \ \}$ the set $B_{d_i} \left( p_i, \epsilon \right)$ is open in $\left( X_i, d_i \right)$. Thus the requirements of (1) are satisfied by the set $B_{\rho} (\mathbf{p}, \epsilon)$ itself. So the product topology on $\prod_{i=1}^n X_i$ is finer than the $\rho$-metric topology.

Hence these two topologies on $\prod_{i=1}^n X_i$ are the same.

Is this proof correct? Is every step of this proof correct and clear enough for someone who is merely an elementary learner?