Proving that scalar multiplication is continuous Let $\mathbb{K} \in \{ \mathbb{R} , \mathbb{C} \}$ and $s= \mathbb{K}^{\omega}$ be the usual sequence set with entries on $\mathbb{K}$. I proved that $\mathbb{K}$ induces a $\mathbb{K}$-vector space structure on $s$ and that the function $\rho : s \times s \to \mathbb{R}$ given by $$\displaystyle \rho(x,y) = \sum_{n=1}^{\infty} \frac{1}{2^n} \frac{|x_n - y_n|}{(1+|x_n - y_n|)} \ \ , $$ for all $x = (x_n) \in s$ and all $y = (y_n) \in s$, is a translation-invariant metric on $s$. Then I proved that addition $A: (s \times s , \tau_*) \to (s,\tau)$ is continuous, where $\tau$ is the topology on $s$ induced by $\rho$ and  $\tau_*$ is the product topology of $(s,\tau)$ with itself. For that purpose I used the metric $\sigma : (s \times s) \times (s \times s) \to \mathbb{R}$ such that $\sigma \big( (x,y) , (a,b) \big) = \rho(x,a) + \rho(y,b)$, $\forall x,y,a,b \in s$, which induces $\tau_*$ on $s \times s$.
Then I was trying to prove that the scalar multiplication $M: (\mathbb{K} \times s , \tilde{\tau}) \to (s, \tau)$ is continuous and I failed, where $\tilde{\tau}$ is the product topology on $\mathbb{K} \times s$ of $(\mathbb{K}, \eta)$ with $(s,\tau)$ and $\eta$ is the topology on $\mathbb{K}$ induced by the norm given by the absolute value function $| \cdot | : \mathbb{K} \to \mathbb{R}$.
I don't know which metric to consider and I don't know how to do it. I need help.
Any hint is appreciated. Thanks in advance.
 A: The following is a hint, and reflects my thought process when thinking about your question.
Of course continuity depends on the topologies in question. This is why for addition you consider the linear space $s\times s$ with the product topology, and so you use the metric $\sigma$ which induces this topology $\tau_*$, gives a nice measure of how far apart points in $s\times s$ are, and allows you to get your hands on computing the product metric based on the given metric on $s$.
If we parallel this idea to scalar multiplication, what should be the natural topology that makes sense given the space $\mathbb{K}\times s$? Specifically, it shouldn't make sense that $\mathbb{K}\times s$ should be endowed with $\tau_*$, since $\tau_*$ is the product topology on $s\times s$. The most natural topology on $\mathbb{K}\times s$ would be the product of $\delta\times\tau$ where $\delta$ is the standard topology generated by the standard metric $|\lambda-\mu|$ on $\mathbb{K}$. 
Thus $\tilde{\sigma}\colon (\mathbb{K}\times s,\delta\times\tau)^2\rightarrow\mathbb{R}^{\geq0}$ defined by $\tilde{\sigma}((\lambda,x),(\mu,y))=|\lambda-\mu|+\rho(x,y)$ should do the trick.
A: I think now I have an answer for my question.
Let $ \ \ \tilde{\sigma} : (\mathbb{K} \times s) \times (\mathbb{K} \times s) \to \mathbb{R} \ \ $ be the function given by $$\tilde{\sigma} \big( (\lambda,x) , (\alpha,a) \big) = | \lambda - \alpha | + \rho(x,a) \ , $$ for all $ \ \ (\lambda,x) , (\alpha,a) \in \mathbb{K} \times s$. It is already known that $ \ \tilde{\sigma} \ $ is a metric on $\ \mathbb{K} \times s \ $ and that this metric induces the product topology $\ \tilde{\tau} \ $ of $(\mathbb{K},\eta)$ with $(s,\tau)$, where $\eta$ is the topology on $\mathbb{K}$ induced by the norm given by the absolute value function $ \ |⋅|:\mathbb{K} \to \mathbb{R}$.
Let $ \ M : \mathbb{K} \times s \to s \ $ be the scalar multiplication function, that is $ \ M(\lambda,x) = \lambda \cdot x = \lambda x$, $\forall (\lambda,x) \in \mathbb{K} \times s$.
Let $ \ (\alpha , a) \in \mathbb{K} \times s \ $ and $ \ \varepsilon > 0 \ $ be fixed. Consider $ \ (\lambda , x) \in \mathbb{K} \times s$. We have $$\rho(\lambda x , \alpha a) = \sum_{n=1}^{\infty} \frac{1}{2^n} \frac{| \lambda x_n - \alpha a_n|}{(1+| \lambda x_n - \alpha a_n|)} = \lim_{j \to \infty} S_j \ ,$$ where $ \ (S_j)_{j =1}^{\infty} \in \mathbb{R}^{\omega} \ $ is the sequence of partial sums, given by $ \ \displaystyle S_j = \sum_{n=1}^{j} \frac{1}{2^n} \frac{| \lambda x_n - \alpha a_n|}{(1+| \lambda x_n - \alpha a_n|)} \, ,$ for all $ \ j \in \mathbb{N}^*$. There exists some $ \ k \in \mathbb{N}^*$ such that, $\forall n \in \mathbb{N}^*$, if $ \ n \geqslant k \ $, then $0 \leqslant \rho( \lambda x , \alpha a) - S_n = | \rho( \lambda x , \alpha a) - S_n | < \varepsilon/2$. In particular, we have $$ 0 \leqslant \rho( \lambda x , \alpha a) - S_k = | \rho( \lambda x , \alpha a) - S_k | < \frac{\varepsilon}{2} \ , $$ because $(S_n)$ is a monotonic nondecreasing real sequence.
Let $$c = \min \left\{ \frac{\varepsilon}{4(|a_1| +1)} , ... , \frac{\varepsilon}{4(|a_k| +1)} \right\}$$ and $$\delta = \min \left\{ c \, , \frac{\varepsilon}{2^k [4(| \alpha | +c) + \varepsilon]} \right\} \ .$$ Then $ \ c>0 \ $ and $ \ \delta >0$.
Supose that $ \ \tilde{\sigma} \big( (\lambda,x) , (\alpha,a) \big) < \delta$. Therefore $$| \lambda | - | \alpha | \leqslant \big| | \lambda | - | \alpha | \big| \leqslant | \lambda - \alpha | \leqslant |\lambda - \alpha | + \rho(x,a) = \tilde{\sigma} \big( (\lambda,x) , (\alpha,a) \big) < \delta \leqslant c \ \Rightarrow \ $$ $$ \ \Rightarrow \ 0 \leqslant | \lambda | \leqslant | \alpha | +c \ \Rightarrow \ 0 \leqslant \frac{| \lambda |}{| \alpha | +c} < 1$$ and, $\forall n \in \{ 1 , ... , k \}$, we have $ \ \displaystyle 0 \leqslant |a_n| < |a_n|+1 \ \Rightarrow \ 0 \leqslant \frac{|a_n|}{|a_n|+1} < 1 \ $, $$ |\lambda - \alpha | < c \leqslant  \frac{\varepsilon}{4(|a_n| +1)}$$ and $$\frac{1}{2^n} \frac{| x_n - a_n|}{(1+| x_n - a_n|)} \leqslant \sum_{n=1}^{\infty} \frac{1}{2^n} \frac{| x_n - a_n|}{(1+| x_n - a_n|)} = \rho(x,a) \leqslant | \lambda - \alpha | + \rho(x,a) =$$ $$= \tilde{\sigma} \big( (\lambda,x) , (\alpha,a) \big) < \delta \ \Rightarrow \ \frac{| x_n - a_n|}{1+| x_n - a_n|} < 2^n \delta \leqslant 2^k \delta \leqslant \frac{\varepsilon}{4(| \alpha | +c) + \varepsilon} \ \Rightarrow \ $$ $$ \ \Rightarrow \ | x_n - a_n| \cdot \left\{ 1 - \frac{\varepsilon}{[4(| \alpha | +c) + \varepsilon]} \right\} < \frac{\varepsilon}{4(| \alpha | +c) + \varepsilon} \ \Rightarrow \ $$ $$ \ \Rightarrow \ | x_n - a_n| \cdot \left[ \frac{4(| \alpha | +c)}{4(| \alpha | +c) + \varepsilon} \right] < \frac{\varepsilon}{4(| \alpha | +c) + \varepsilon} \ \Rightarrow \ | x_n - a_n| < \frac{\varepsilon}{4(|\alpha| + c)} \ .$$ That gives $$| \lambda x_n - \alpha a_n | = | \lambda x_n - \lambda a_n + \lambda a_n - \alpha a_n | \leqslant | \lambda x_n - \lambda a_n | + | \lambda a_n - \alpha a_n | = $$ $$ = | \lambda (x_n - a_n) | + | (\lambda - \alpha) a_n | = | \lambda| \cdot |x_n - a_n| + |\lambda - \alpha| \cdot |a_n| <$$ $$< \frac{| \lambda| \cdot \varepsilon}{4(|\alpha| + c)} + \frac{\varepsilon \cdot |a_n|}{4(|a_n| +1)} = \frac{\varepsilon}{4} \cdot \left( \frac{|\lambda|}{|\alpha|+c} + \frac{|a_n|}{|a_n|+1} \right) < \frac{\varepsilon}{4} \cdot (1+1) = \frac{\varepsilon}{2} \ ,$$ for all $ \ n \in \{ 1, ... , k \}$.
Then we have $$ S_k = \sum_{n=1}^{k} \frac{1}{2^n} \frac{| \lambda x_n - \alpha a_n |}{(1 + | \lambda x_n - \alpha a_n |)} < \sum_{n=1}^{k} \frac{1}{2^n} \frac{(\varepsilon/2)}{(1 + | \lambda x_n - \alpha a_n |)} =$$ $$= \frac{\varepsilon}{2} \cdot \sum_{n=1}^{k} \frac{1}{2^n} \frac{1}{(1 + | \lambda x_n - \alpha a_n |)} < \frac{\varepsilon}{2} \cdot \sum_{n=1}^{k} \frac{1}{2^n} \leqslant \frac{\varepsilon}{2} \cdot \sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{\varepsilon}{2} \cdot 1 = \frac{\varepsilon}{2} \ .$$ Finally $$\rho \big( M(\lambda,x) , M(\alpha,a) \big) = \rho(\lambda x , \alpha a) = S_k + [\rho(\lambda x , \alpha a) - S_k] =$$ $$= S_k + |\rho(\lambda x , \alpha a) - S_k| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \ .$$
Then $M$ is continuous at $ \ (\alpha , a) \in \mathbb{K} \times s$.
