[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 $V\times W$.[/step]

[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: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]