Prove scalar product in a normed vector space is an open mapping. I have the feeling this is a really obvious question, but I'm having trouble with it, here it goes:
Let $(X, ||\,||)$ be a normed vector space over $K$, prove that $\odot:K\setminus\{0\}\times X \mapsto X$, defined by 
$$\odot(\alpha,x)=\alpha x,$$
is an open mapping. 
Since it was not stated in the problem, I'm assuming 
$$||(\alpha,x)||=|\alpha|+||x||.$$
Thanks in advance.
 A: If the space $X$ is banach it is an easy consequences of the open mappig theorem. 
Anyway with the norm induced topology over $ X $ you in fact are resizing a ball so it is a ball again and it is open by definition of the topology. So the map sends open ball in open ball therefore it is open. This reasoning heavily rely on the "absolute omogeneity" of the norm we are working with
Recall the property of the norm $\| kx\| =|k|\|x\|$ and that the open set of the norm induced topology are unions of open balls $B_x(r) :=\{ y \in X | \| x-y\| <r \}$ so in fact if we prove that the map $\odot$ sends open balls in open balls we conclude.
The eps-delta argument is a bit "heavy" here. It is only a matter of the property of the norm. What is $\odot(k,B_x(r)) $? By using the definition of multiplication It is obvious that $\odot(k,B_x(r))= |k|B_x(r)=B_x(|k|r)$ (use the double inclusion and the invertibility of $k$)
Another analogue approach is to prove that $\odot$ is bijective (if $k\neq 0$) and for every $k$ is continuous (you can argue by definition). Think why it is sufficient this reasoning
Addendum in the language of the Topological Vector Spaces the previous reasoning is obvious. In fact another approach is to verify that norm induced topology is consistent with the axioms of a topological vector space
