Proof that every normed vector space is a topological vector space 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.
 A: The first point is fine. For the second, fix $(v_0,\alpha_0)\in V\times K$ and $\varepsilon >0$. We have to find $\delta>0$ such that if $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$ then $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$.  We have 
\begin{align}
\lVert \alpha_0v_0-\alpha v\rVert&\leq \lVert \alpha_0v_0-\alpha v_0\rVert+
\lVert \alpha v_0-\alpha v\rVert\\
&=|\alpha_0-\alpha|\lVert v_0\rVert+|\alpha|\lVert v-v_0\rVert\\
&\leq |\alpha_0-\alpha|(\lVert v_0\rVert+\lVert v-v_0\rVert)+|\alpha_0|\lVert v-v_0\rVert.
\end{align}
We take $\delta$ such that $\delta^2+\delta(\lVert v_0\rVert+|\alpha_0|)\leq \varepsilon$ (which is possible).
In this case, $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$ when $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$.
A: Let fix $\alpha$ and $ x$  such that, 
$$\|x-x_0\|<1 $$ then, $$\|x\| \le \|x-x_0\|+\|x_0\|\le \|x_0\|+1$$
\begin{align}\|\alpha x -\alpha_0x_0\| &= \|\alpha x -\alpha_0 x+\alpha_0 x-\alpha_0x_0\| \\&\le \|x\||\alpha -\alpha_0| +|\alpha_0| \|x-x_0\|\\&< (\|x_0\|+1)|\alpha -\alpha_0| +(|\alpha_0|+1) \|x-x_0\|\\&\le  \max\left[(\|x_0\|+1),(|\alpha_0|+1)\right]\color{red}{\left[|\alpha -\alpha_0| + \|x-x_0\|\right]}\\&\le 2\max\left[(\|x_0\|+1),(|\alpha_0|+1)\right]\color{red}{\max\left[|\alpha -\alpha_0| , \|x-x_0\|\right]} \end{align}
for any $\varepsilon>0$ if you take 
$$\color{red}{\delta= \min\left(1, \frac{\varepsilon}{ 2\max(\|x_0\|+1),(|\alpha_0|+1)}\right)}$$
 Then for any $x$ and $\lambda$ such that, $$\color{red}{\max\left[|\alpha -\alpha_0| , \|x-x_0\|\right] <\delta}$$
you get, $$\color{blue}{ \|\alpha x -\alpha_0x_0\|<\varepsilon}$$
