For $n\ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_{\mathrm{\infty }}$ on ${\mathbb{R}}^n$, and hence also of $d_1$ and $d_2$, so ${\mathbb{R}}^n$ as a product and ${\mathbb{R}}^n$ as a metric space are one space

For $y\in {\mathbb{R}}^n$: $d_{\mathrm{\infty }}(x,y)<r$ if and only if $\mathrm{\mid }{x}_k-y_k\mathrm{\mid }<r$ for every $k<n$, since by [L3] the maximum is one of the values $\mathrm{\mid }{x}_k-y_k\mathrm{\mid }$ and is an upper bound for all of them.

For $t\in \mathbb{R}$ and $r>0$: $\mathrm{\mid }t-y_k\mathrm{\mid }<r$ if and only if $y_k\in (t-r,t+r)$, by [L1].

Write $t_k:=\mathrm{\mid }{x}_k-y_k\mathrm{\mid }$ and $M:=d_{\mathrm{\infty }}(x,y)=\max \{t_k:k<n\}$. Then $t_j\le M$ for every $j<n$ and $M=t_{j_0}$ for some $j_0<n$, by [L3].

$d_2(x,y{)}^2={\sum }_{k<n}(x_k-y_k{)}^2$ by [L5], and $(x_k-y_k{)}^2=t_k^2\ge 0$ by [L5].

$n\le n^2$ as reals: for $n\ge 1$ the canonical natural satisfies $\iota (1)=1\le \iota (n)$ by [L7], so either $\iota (n)=1$, in which case $\iota (n)=\iota (n{)}^2=1$, or $1<\iota (n)$, in which case multiplying that strict inequality by $\iota (n)>0$ gives $\iota (n)<\iota (n{)}^2$ by [L7].

Conversely let $B={\prod }_{k<n}{U}_k$ be a box with every $U_k$ open in $\mathbb{R}$ and let $x\in B$. For each $k<n$ the set $\{\rho \in \mathbb{R}:\rho >0,(x_k-\rho ,x_k+\rho )\subseteq U_k\}$ is nonempty by [L1], so [L6] supplies ${\rho }_k$ in it for every $k<n$; put $r:=\min \{{\rho }_k:k<n\}$, which exists and is positive by [L3].

$d_1(x,y)={\sum }_{k<n}{t}_k\le {\sum }_{k<n}M=nM$, using $t_k\le M$ from step 1.3 and [L4].

$M=t_{j_0}\le {\sum }_{k<n}{t}_k=d_1(x,y)$, since every $t_k\ge 0$ by [L5] and a single nonnegative term is at most the sum, by [L4].

$B_{d_{\mathrm{\infty }}}(x,r)={\prod }_{k<n}(x_k-r,x_k+r)$: by step 1.1 a point $y$ lies in the ball exactly when $\mathrm{\mid }{x}_k-y_k\mathrm{\mid }<r$ for every $k<n$, and by step 1.2 that says exactly $y_k\in (x_k-r,x_k+r)$ for every $k<n$.

$M^2=t_{j_0}^2\le {\sum }_{k<n}{t}_k^2=d_2(x,y{)}^2$ by steps 1.3 and 1.4 with [L4], and both $M$ and $d_2(x,y)$ are nonnegative by [L2] and [L5], so $M\le d_2(x,y)$ by [L5].

$d_2(x,y{)}^2={\sum }_{k<n}{t}_k^2\le {\sum }_{k<n}{M}^2=n{M}^2\le n^2{M}^2=(nM{)}^2$, using $t_k\le M$ with [L5] and [L4], then step 1.5 with $M^2\ge 0$; since $d_2(x,y)\ge 0$ and $nM\ge 0$, [L5] gives $d_2(x,y)\le nM$.

Every $d_{\mathrm{\infty }}$-ball is a box with open factors, by step 2.3 and [L1], hence a basic open set of the product topology by [A2]; so every $d_{\mathrm{\infty }}$-open set is product-open, by [L2] and [A2].

With $r$ as in step 1.6: $B_{d_{\mathrm{\infty }}}(x,r)={\prod }_{k<n}(x_k-r,x_k+r)\subseteq {\prod }_{k<n}(x_k-{\rho }_k,x_k+{\rho }_k)\subseteq B$, since $r\le {\rho }_k$ for every $k$ by [L3].

Steps 2.1, 2.2, 2.4 and 2.5 give $d_{\mathrm{\infty }}\le d_1\le n{d}_{\mathrm{\infty }}$ and $d_{\mathrm{\infty }}\le d_2\le n{d}_{\mathrm{\infty }}$ at every pair of points, which is claim 2, the constants $1$ and $n$ being positive by [L7].

By steps 1.6 and 3.2 every basic open set of the product topology is $d_{\mathrm{\infty }}$-open by [L2], hence every product-open set is $d_{\mathrm{\infty }}$-open; with step 3.1 this gives claim 1.

By step 3.3 and [L8] the metrics $d_1$, $d_2$ and $d_{\mathrm{\infty }}$ have the same metric topology, which by step 4.1 is the product topology; so all three induce it and ${\mathbb{R}}^n$ with the product topology is metrizable. This is claim 3, and with steps 4.1 and 3.3 all three claims are proved.