[proofplan]
We first verify directly that the two displayed functions satisfy the norm axioms on the product [vector space](/page/Vector%20Space) $V\times W$. We then compare the two norms by the inequalities $\|\cdot\|_\infty\leq \|\cdot\|_1\leq 2\|\cdot\|_\infty$, which gives the same open sets for the two norm topologies. Finally, we identify the $\|\cdot\|_\infty$-balls as exact products of the component norm balls, so the $\|\cdot\|_\infty$-topology is precisely the [product topology](/page/Product%20Topology).
[/proofplan]
[step:Verify that the sum and maximum formulas define norms on $V\times W$]
Let $0_V\in V$ and $0_W\in W$ denote the zero vectors. Let $0_{V\times W}:=(0_V,0_W)$ denote the zero vector in $V\times W$.
For every $(v,w)\in V\times W$, both quantities $\|v\|_V+\|w\|_W$ and $\max\{\|v\|_V,\|w\|_W\}$ are non-negative because $\|\cdot\|_V$ and $\|\cdot\|_W$ are norms. Moreover,
\begin{align*}
\|(v,w)\|_1=0
\end{align*}
if and only if $\|v\|_V=0$ and $\|w\|_W=0$, which holds if and only if $v=0_V$ and $w=0_W$. Thus $\|(v,w)\|_1=0$ if and only if $(v,w)=0_{V\times W}$. Similarly,
\begin{align*}
\|(v,w)\|_\infty=0
\end{align*}
if and only if $\|v\|_V=0$ and $\|w\|_W=0$, so $\|(v,w)\|_\infty=0$ if and only if $(v,w)=0_{V\times W}$.
Let $\lambda\in\mathbb{F}$ and let $(v,w)\in V\times W$. Using homogeneity of the component norms,
\begin{align*}
\|\lambda(v,w)\|_1=\|(\lambda v,\lambda w)\|_1=|\lambda|\,\|v\|_V+|\lambda|\,\|w\|_W=|\lambda|\,\|(v,w)\|_1.
\end{align*}
Likewise,
\begin{align*}
\|\lambda(v,w)\|_\infty=\max\{|\lambda|\,\|v\|_V,|\lambda|\,\|w\|_W\}=|\lambda|\,\|(v,w)\|_\infty.
\end{align*}
Let $(v_1,w_1),(v_2,w_2)\in V\times W$. By the triangle inequalities in $V$ and $W$,
\begin{align*}
\|(v_1,w_1)+(v_2,w_2)\|_1=\|v_1+v_2\|_V+\|w_1+w_2\|_W\leq \|v_1\|_V+\|v_2\|_V+\|w_1\|_W+\|w_2\|_W.
\end{align*}
Rearranging the right-hand side gives
\begin{align*}
\|(v_1,w_1)+(v_2,w_2)\|_1\leq \|(v_1,w_1)\|_1+\|(v_2,w_2)\|_1.
\end{align*}
For the maximum norm, the component triangle inequalities give
\begin{align*}
\|v_1+v_2\|_V\leq \|v_1\|_V+\|v_2\|_V\leq \|(v_1,w_1)\|_\infty+\|(v_2,w_2)\|_\infty
\end{align*}
and
\begin{align*}
\|w_1+w_2\|_W\leq \|w_1\|_W+\|w_2\|_W\leq \|(v_1,w_1)\|_\infty+\|(v_2,w_2)\|_\infty.
\end{align*}
Taking the maximum of the two component bounds,
\begin{align*}
\|(v_1,w_1)+(v_2,w_2)\|_\infty\leq \|(v_1,w_1)\|_\infty+\|(v_2,w_2)\|_\infty.
\end{align*}
Hence $\|\cdot\|_1$ and $\|\cdot\|_\infty$ are norms on $V\times W$.
[/step]
[step:Compare the two product norms by uniform inequalities]
For every $(v,w)\in V\times W$, since both $\|v\|_V$ and $\|w\|_W$ are non-negative,
\begin{align*}
\max\{\|v\|_V,\|w\|_W\}\leq \|v\|_V+\|w\|_W.
\end{align*}
Also,
\begin{align*}
\|v\|_V+\|w\|_W\leq 2\max\{\|v\|_V,\|w\|_W\}.
\end{align*}
Thus, for every $(v,w)\in V\times W$,
\begin{align*}
\|(v,w)\|_\infty\leq \|(v,w)\|_1\leq 2\|(v,w)\|_\infty.
\end{align*}
Let $z\in V\times W$ and let $r>0$. Define the open balls
\begin{align*}
B_1(z,r):=\{u\in V\times W:\|u-z\|_1<r\}
\end{align*}
and
\begin{align*}
B_\infty(z,r):=\{u\in V\times W:\|u-z\|_\infty<r\}.
\end{align*}
The inequalities above imply the ball inclusions
\begin{align*}
B_\infty(z,r/2)\subset B_1(z,r)\subset B_\infty(z,r).
\end{align*}
Indeed, if $u\in B_\infty(z,r/2)$, then $\|u-z\|_1\leq 2\|u-z\|_\infty<r$, so $u\in B_1(z,r)$. If $u\in B_1(z,r)$, then $\|u-z\|_\infty\leq \|u-z\|_1<r$, so $u\in B_\infty(z,r)$.
It follows that every $\|\cdot\|_1$-[open set](/page/Open%20Set) is $\|\cdot\|_\infty$-open and every $\|\cdot\|_\infty$-open set is $\|\cdot\|_1$-open. Therefore the two norms induce the same topology on $V\times W$.
[guided]
The goal of this step is to show that the two norms have exactly the same small-neighborhood structure. We do this by comparing their balls.
For any $(v,w)\in V\times W$, the two component norms are non-negative [real numbers](/page/Real%20Numbers). For non-negative real numbers $a$ and $b$, one has
\begin{align*}
\max\{a,b\}\leq a+b\leq 2\max\{a,b\}.
\end{align*}
Applying this with $a=\|v\|_V$ and $b=\|w\|_W$ gives
\begin{align*}
\|(v,w)\|_\infty\leq \|(v,w)\|_1\leq 2\|(v,w)\|_\infty.
\end{align*}
Now fix a point $z\in V\times W$ and a radius $r>0$. Define
\begin{align*}
B_1(z,r):=\{u\in V\times W:\|u-z\|_1<r\}
\end{align*}
and
\begin{align*}
B_\infty(z,r):=\{u\in V\times W:\|u-z\|_\infty<r\}.
\end{align*}
If $u\in B_\infty(z,r/2)$, then $\|u-z\|_\infty<r/2$. The upper comparison inequality gives
\begin{align*}
\|u-z\|_1\leq 2\|u-z\|_\infty<r.
\end{align*}
Therefore $u\in B_1(z,r)$, and hence
\begin{align*}
B_\infty(z,r/2)\subset B_1(z,r).
\end{align*}
Conversely, if $u\in B_1(z,r)$, then $\|u-z\|_1<r$. The lower comparison inequality gives
\begin{align*}
\|u-z\|_\infty\leq \|u-z\|_1<r.
\end{align*}
Therefore $u\in B_\infty(z,r)$, and hence
\begin{align*}
B_1(z,r)\subset B_\infty(z,r).
\end{align*}
These inclusions prove equality of the two norm topologies. If a set is open for $\|\cdot\|_1$ and contains a point $z$, then it contains some $B_1(z,r)$; the inclusion $B_\infty(z,r/2)\subset B_1(z,r)$ gives a $\|\cdot\|_\infty$-ball around $z$ inside the same set. If a set is open for $\|\cdot\|_\infty$ and contains $z$, then it contains some $B_\infty(z,r)$; the inclusion $B_1(z,r)\subset B_\infty(z,r)$ gives a $\|\cdot\|_1$-ball around $z$ inside the same set. Hence the open sets are identical.
[/guided]
[/step]
[step:Identify maximum norm balls with products of component norm balls]
Let $\tau_V$ denote the norm topology on $V$, let $\tau_W$ denote the norm topology on $W$, and let $\tau_\times$ denote the product topology on $V\times W$ induced by $\tau_V$ and $\tau_W$. For $v\in V$ and $r>0$, define
\begin{align*}
B_V(v,r):=\{x\in V:\|x-v\|_V<r\}.
\end{align*}
For $w\in W$ and $r>0$, define
\begin{align*}
B_W(w,r):=\{y\in W:\|y-w\|_W<r\}.
\end{align*}
For every $(v,w)\in V\times W$ and every $r>0$,
\begin{align*}
B_\infty((v,w),r)=B_V(v,r)\times B_W(w,r).
\end{align*}
Indeed, for $(x,y)\in V\times W$,
\begin{align*}
(x,y)\in B_\infty((v,w),r)
\end{align*}
if and only if
\begin{align*}
\max\{\|x-v\|_V,\|y-w\|_W\}<r.
\end{align*}
This is equivalent to the two inequalities $\|x-v\|_V<r$ and $\|y-w\|_W<r$, which is equivalent to $(x,y)\in B_V(v,r)\times B_W(w,r)$.
Therefore every $\|\cdot\|_\infty$-ball is a product of open sets in $V$ and $W$, so every $\|\cdot\|_\infty$-open set belongs to $\tau_\times$.
[/step]
[step:Show that every product-neighborhood contains a maximum norm ball]
Let $O\in\tau_\times$, and let $(v,w)\in O$. By the definition of the product topology, there exist sets $U\in\tau_V$ and $S\in\tau_W$ such that
\begin{align*}
(v,w)\in U\times S\subset O.
\end{align*}
Since $U$ is open in the norm topology on $V$ and $v\in U$, there exists $r_V>0$ such that
\begin{align*}
B_V(v,r_V)\subset U.
\end{align*}
Since $S$ is open in the norm topology on $W$ and $w\in S$, there exists $r_W>0$ such that
\begin{align*}
B_W(w,r_W)\subset S.
\end{align*}
Define
\begin{align*}
r:=\min\{r_V,r_W\}.
\end{align*}
Then $r>0$, and the ball identity from the previous step gives
\begin{align*}
B_\infty((v,w),r)=B_V(v,r)\times B_W(w,r)\subset B_V(v,r_V)\times B_W(w,r_W)\subset U\times S\subset O.
\end{align*}
Thus every point of every product-open set has a $\|\cdot\|_\infty$-ball contained in that set, so every product-open set is $\|\cdot\|_\infty$-open.
[/step]
[step:Conclude that both standard product norms induce the product topology]
The previous two steps show that the topology induced by $\|\cdot\|_\infty$ is equal to $\tau_\times$, the product topology on $V\times W$. The norm comparison step shows that $\|\cdot\|_1$ and $\|\cdot\|_\infty$ induce the same topology. Therefore the topology induced by $\|\cdot\|_1$ is also equal to $\tau_\times$. Hence both $\|\cdot\|_1$ and $\|\cdot\|_\infty$ induce the product topology on $V\times W$.
[/step]
