[proofplan] We prove continuity at an arbitrary point for each operation using basic product neighborhoods. For addition, the triangle inequality shows that controlling the two input coordinates separately controls the norm of the sum difference. For scalar multiplication, we first restrict the vector variable to a unit ball around the base point so that its norm is locally bounded, and then use homogeneity and the triangle inequality to control the scalar and vector errors separately. [/proofplan]

[step:Control addition by controlling each coordinate separately] Fix $(x,y)\in V\times V$ and let $\varepsilon>0$. Define open norm balls \begin{align*} U:=\{u\in V:\|u-x\|<\varepsilon/2\} \end{align*} and \begin{align*} W:=\{v\in V:\|v-y\|<\varepsilon/2\}. \end{align*} Then $U\times W$ is an open neighbourhood of $(x,y)$ in the [product topology](/page/Product%20Topology) on $V\times V$. For any $(u,v)\in U\times W$, the triangle inequality gives \begin{align*} \|A(u,v)-A(x,y)\|=\|(u+v)-(x+y)\|\leq \|u-x\|+\|v-y\|. \end{align*} Since $\|u-x\|<\varepsilon/2$ and $\|v-y\|<\varepsilon/2$, it follows that \begin{align*} \|A(u,v)-A(x,y)\|<\varepsilon. \end{align*} Thus $A$ is continuous at $(x,y)$. Since $(x,y)$ was arbitrary, $A$ is continuous on $V\times V$. [/step]

[step:Bound scalar multiplication near an arbitrary point]Fix $(\lambda,x)\in \mathbb{F}\times V$ and let $\varepsilon>0$. Define \begin{align*} \alpha:=\frac{\varepsilon}{2(\|x\|+1)} \end{align*} and \begin{align*} \beta:=\min\left\{1,\frac{\varepsilon}{2(|\lambda|+1)}\right\}. \end{align*} Let \begin{align*} S:=\{\mu\in\mathbb{F}:|\mu-\lambda|<\alpha\} \end{align*} and \begin{align*} U:=\{u\in V:\|u-x\|<\beta\}. \end{align*} Then $S\times U$ is an open neighbourhood of $(\lambda,x)$ in the product topology on $\mathbb{F}\times V$. For any $(\mu,u)\in S\times U$, the choice $\beta\leq 1$ gives \begin{align*} \|u\|\leq \|u-x\|+\|x\|<\|x\|+1. \end{align*} Using linearity of scalar multiplication in $V$, the triangle inequality, and homogeneity of the norm, \begin{align*} \|M(\mu,u)-M(\lambda,x)\|=\|\mu u-\lambda x\|=\|(\mu-\lambda)u+\lambda(u-x)\|\leq |\mu-\lambda|\,\|u\|+|\lambda|\,\|u-x\|. \end{align*} By the definitions of $S$ and $U$, \begin{align*} |\mu-\lambda|\,\|u\|<\frac{\varepsilon}{2(\|x\|+1)}(\|x\|+1)=\frac{\varepsilon}{2}. \end{align*} Also, since $\beta\leq \varepsilon/(2(|\lambda|+1))$, \begin{align*} |\lambda|\,\|u-x\|<|\lambda|\frac{\varepsilon}{2(|\lambda|+1)}\leq \frac{\varepsilon}{2}. \end{align*} Therefore \begin{align*} \|M(\mu,u)-M(\lambda,x)\|<\varepsilon. \end{align*}[/step]

[guided]We prove continuity of scalar multiplication at the fixed point $(\lambda,x)\in\mathbb{F}\times V$. The main issue is that the expression $\mu u-\lambda x$ contains both the scalar variable $\mu$ and the vector variable $u$. We separate these two sources of error by adding and subtracting $\lambda u$ in the equivalent form \begin{align*} \mu u-\lambda x=(\mu-\lambda)u+\lambda(u-x). \end{align*} Let $\varepsilon>0$. We must find a product neighbourhood of $(\lambda,x)$ whose image under $M$ lies in the norm ball of radius $\varepsilon$ around $\lambda x$. Define \begin{align*} \alpha:=\frac{\varepsilon}{2(\|x\|+1)} \end{align*} and \begin{align*} \beta:=\min\left\{1,\frac{\varepsilon}{2(|\lambda|+1)}\right\}. \end{align*} Now define the scalar neighbourhood \begin{align*} S:=\{\mu\in\mathbb{F}:|\mu-\lambda|<\alpha\} \end{align*} and the vector neighbourhood \begin{align*} U:=\{u\in V:\|u-x\|<\beta\}. \end{align*} Because $\mathbb{F}$ has the absolute-value topology and $V$ has the norm topology, $S$ and $U$ are open neighbourhoods of $\lambda$ and $x$, respectively. Hence $S\times U$ is an open neighbourhood of $(\lambda,x)$ in the product topology. Take any $(\mu,u)\in S\times U$. The first term in the estimate will contain $\|u\|$, so we need a bound on $\|u\|$ that is valid throughout the chosen neighbourhood. This is why $\beta$ was chosen to be at most $1$. By the triangle inequality, \begin{align*} \|u\|\leq \|u-x\|+\|x\|<1+\|x\|. \end{align*} Using the decomposition of $\mu u-\lambda x$, the triangle inequality, and homogeneity of the norm, we obtain \begin{align*} \|M(\mu,u)-M(\lambda,x)\|=\|\mu u-\lambda x\|=\|(\mu-\lambda)u+\lambda(u-x)\|\leq |\mu-\lambda|\,\|u\|+|\lambda|\,\|u-x\|. \end{align*} The scalar error is controlled by the definition of $\alpha$: \begin{align*} |\mu-\lambda|\,\|u\|<\frac{\varepsilon}{2(\|x\|+1)}(\|x\|+1)=\frac{\varepsilon}{2}. \end{align*} The vector error is controlled by the definition of $\beta$: \begin{align*} |\lambda|\,\|u-x\|<|\lambda|\frac{\varepsilon}{2(|\lambda|+1)}\leq \frac{\varepsilon}{2}. \end{align*} Adding the two bounds gives \begin{align*} \|M(\mu,u)-M(\lambda,x)\|<\varepsilon. \end{align*} Thus every point sufficiently close to $(\lambda,x)$ in the product topology is sent by $M$ into the $\varepsilon$-ball around $M(\lambda,x)$.[/guided]

[step:Conclude continuity of scalar multiplication everywhere] The preceding step shows that for every $(\lambda,x)\in\mathbb{F}\times V$ and every $\varepsilon>0$, there is an open product neighbourhood $S\times U$ of $(\lambda,x)$ such that \begin{align*} M(S\times U)\subset \{z\in V:\|z-M(\lambda,x)\|<\varepsilon\}. \end{align*} This is precisely continuity of $M$ at $(\lambda,x)$ with respect to the product topology on $\mathbb{F}\times V$ and the norm topology on $V$. Since $(\lambda,x)$ was arbitrary, $M$ is continuous on $\mathbb{F}\times V$. Together with the first step, this proves that both [vector space](/page/Vector%20Space) operations are continuous. [/step]