The topology induced by the norm of a normed vector space is such that the space is a topological vector space.
Can you tell me if my proof is correct? Of course we have to show that addition and scalar multiplication are continuous with respect to the product topology (induced by the norm).
(1) To show that $(x,y) \mapsto x + y$ is continuous let $\varepsilon > 0$. I can show that the norm $\|\cdot\|_{V \times V}: V \times V \to \mathbb R$ defined as $\|(x,y) - (x_0, y_0) \| = \|x-x_0\| + \|y_0 - y\|$ induces the same topology as the product topology on $V \times V$. Hence we can choose $\delta = \varepsilon$ to get $ \| (x+y) - (x_0+y_0)\| \leq \|x-x_0\| + \|y-y_0\| < \delta = \varepsilon$
(2) To show that $V \times K \to V$, $(v, \alpha) \mapsto \alpha v$ is continuous at $(v,\alpha)$, observe that $\| \alpha v - \beta w\| = \| \alpha v - \beta w + \alpha w - \alpha w\| = \|\alpha(v-w) + (\alpha - \beta) w\| \leq |\alpha| \|v-w\| + |\alpha - \beta| \|w\|$
Hence $\| \alpha v - \beta w\| < \varepsilon$ if $\|v-w\| < \frac{\varepsilon}{2 |\alpha|}$ and $|\alpha - \beta| < \frac{\varepsilon}{2 \|w\|}$. Unfortunately, the second inequality depends on $w$. How do I make it independent of $w$? Thanks.