Let $V$ be a vector space with a metric $d$. Suppose that this metric satisfies:
(1) $d(v+z,w+z)=d(v,w)$ $\forall v,w,z\in V$
(2) $d(\lambda v,\lambda w)=|\lambda|d(v,w)$ $\forall v\in V, \lambda \in \mathbb R$
Prove that $f:\mathbb R \times V\to V$ $f(\lambda,v)=\lambda v$ is a continuous function:
What I have done so far:
Let $(\lambda_0,v_0)\in \mathbb R\times V$ and $\epsilon>0$ I want to find $\delta>0$ such that if $$d_{\Bbb R\times V}((\lambda,v),(\lambda,v_0))=d_V(v,v_0)+d_{\mathbb R}(\lambda,\lambda_0)<\delta$$ then $d_v(\lambda v,\lambda_0 v_0)<\epsilon.$ (I´m working with the "taxicab" metric in $\Bbb R\times V$.)
$$\begin{align} d_V(\lambda v,\lambda v_0) &\le d_V(\lambda v,\lambda v_0)+d_V(\lambda v_0,\lambda_0 v_0)\quad \text{[by triangle inequality]}\\ &\overset{(2)}{=} |\lambda|d_V(v,v_0)+ d_V(\lambda v_0,\lambda_0 v_0)\\ &\overset{(1)}{=} |\lambda|d_V(v,v_0)+d_V(\lambda v_0-\lambda v_0,\lambda_0 v_0-\lambda v_0)\\ &= |\lambda|d_V(v,v_0)+d_V(0,(\lambda_0-\lambda)v_0)\\ &\overset{(2)}{=} |\lambda|d_V(v,v_0)+|\lambda-\lambda_0|d_V(0,v_0)\\ &= |\lambda|d_V(v,v_0)+d_{\mathbb R}(\lambda, \lambda_0)d_V(0,v_0)\\ \end{align}$$
I think I'm almost there but the problem is the factors $|\lambda|$ and $d_V(0,v_0)$ but I don´t know what to do from this point.
I would really appreciate if you can help me with this problem
From where you left: If $|\lambda - \lambda_{0}| < 1$, then $\lambda_{0} - 1 < \lambda < |\lambda_{0}| + 1$, so $|\lambda| < \max \{ |\lambda_{0}-1|, |\lambda_{0}| + 1 \} =: \varepsilon'$, and hence $$ d_{V}(v,v_{0})|\lambda| + |\lambda - \lambda_{0}|d_{V}(0,v_{o}) < \varepsilon' d_{V}(v,v_{0}) + |\lambda - \lambda_{0}|[ d_{V}(0,v_{o}) + 1]; $$ taking any $\varepsilon > 0$, we have $$ \varepsilon'd_{V}(v,v_{0}) + |\lambda - \lambda_{0}|[ d_{V}(0,v_{o}) + 1] < \varepsilon $$ if $d_{V}(v,v_{0}) < \varepsilon/2\varepsilon'$ and if $|\lambda - \lambda_{0}| < \varepsilon/2[d_{V}(0,v_{0}) + 1]$; therefore, we have proved that for every $\varepsilon > 0$, if $d_{V}(v,v_{0}) < \varepsilon/2\varepsilon'$ and if $|\lambda - \lambda_{0}| < \min \{ 1, \varepsilon/2[ d_{V}(0,v_{0}) + 1]\},$ then $|f(\lambda,v) - f(\lambda_{0},v_{0})| < \varepsilon$.