Prove that there is a unique seminorm $p$ on $V$ s.t $\{x \in V : p(x) < \alpha \} \subset U \subset \{x \in V : p(x) \leq \alpha \}$ Let $V$ be a vector space over $\mathbb{F}$ and let $U \subset V$ be a convex, balanced and absorbing set.
Prove that there is a unique seminorm $p$ on $V$ s.t $\{x \in V : p(x) < \alpha \} \subset U \subset \{x \in V : p(x) \leq  \alpha \}$.
I need a hint
EDIT
Using the Minkowski functional.
Define 
$$
p(x) = \inf \{ \alpha > 0 : x \in \alpha U \}
$$
Let $v \in \{x \in V : p(x) < \alpha\}$. Now we have that $\frac{p(x)}{\alpha}$ < 1. Thus using that $U$ is balenced then $\frac{p(x)}{\alpha} U \subset U$.
What can I do next?
 A: Hint on where to start: Minkowski Functional (and, of course, your inclusions are reversed). 
A: Recall that $$p(x)=\inf\{t:\ x\in tU\}.$$
Because $U $ is absorbing, $p $ is well-defined. Because it is convex and balanced, it is a seminorm. 
If $x\in U$, then $x\in 1\cdot U$, and so $p_U(x)\leq1.$ That is, $$U\subset\{x\in V:\ p(x)\leq1\}.$$
If $p(x)<1$, this means that there exists $t\in[0,1)$ such that $x\in tU$. But $U$ is balanced, so $tU\subset U$ and so $x\in U$. Then
$$
\{x\in V:\ p(x)<1\}\subset U.
$$
The uniqueness occurs for a fixed $\alpha$. Since everything is scalable, we may assume $\alpha=1$. So assume that $q$ is a seminorm and that $$\{x\in V:\ q(x)<1\}\subset U\subset\{x\in V:\ q(x)\leq1\}.$$
If $x\in U$, then $\beta=q(x)<1$. Then, for any $\varepsilon>0$, $\frac1{\beta+\varepsilon}\,x\in U$, since $q(\frac1{\beta+\varepsilon}\,x)<1$. But then $p(\frac1{\beta+\varepsilon}\,x)<1$, so $$p(x)<\beta+\varepsilon=q(x)+\varepsilon.$$
As we can do this for any $\varepsilon>0$, we conclude that $p(x)\leq q(x)$. As the roles of $p$ and $q$ are reversible, we obtain that $p(x)=q(x)$ for any $x\in U$. 
Finally, if $x\not\in U$, there exists $\gamma>0$ such that $\gamma x\in U$. Then
$$
\gamma p(x)=p(\gamma x)=q(\gamma x)=\gamma q(x),
$$
so $p(x)=q(x)$. 
