Which subsets of a given real linear space $V$ are open unit balls with respect to some norm on the space? Which subsets of a given real linear space $V$ are open unit balls with respect to some norm on the space?
My textbook (Metric Spaces - Michael Searcoid) gives the following hints to answer the above questions : 
Let $V$ be a given real linear space and let $U$ be an open ball in $V$. Then : 


*

*$U$ must be convex. This is true since any ball in a normed linear space must be convex.

*It must be balanced in the sense that, $~\forall~a \in U,$ we have $-a \in U$ because we want $-a$ to have the same norm as $a$.


However, I don't fully follow the next criterion given below :





*

*For $~\forall~x \in V \backslash \{0\},$ the set $\{t \in \mathbb R^+ ~|~tx \in U\}$ must be non empty to ensure that $x$ has some real norm, it's supremum $s$ must be real to ensure that $x$ does not have zero norm, and , to ensure that $U$ has a chance of being open when $V$ is endowed with an appropriate norm, we have $sx \not \in U$


The above criterion argues that $~\forall x \in V \backslash \{0\} : tx \in U$ for some $t >0$. I am not sure why this must be necessarily true.
Moreover, the supremum $s$ can be $\infty$ and still $x$ can have non zero norm.
I don't fully comprehend the last given condition that $s \not \in U$ either.
Could somebody please help me understand the need for these conditions to be true in order for $U$ to be open in the normed real linear space $V$.
Thank you for your help.
( Attaching a screen shot of the page for your reference as well )
 A: Suppose that $V$ is a vector space over $\mathbb R$, $\|\cdot\|$ is a norm on it, and $$U\equiv\{v\in V\,|\,\|v\|<1\}$$ is the open unit ball with respect to the norm. Fix $x\in V$ and $x\neq 0$.
Claim 1: There exists some $t>0$ such that $tx\in U$.
Proof: Since $x\neq 0$, one has $\|x\|>0$. Put $t\equiv 1/(2\|x\|)>0$. Then, $$\|tx\|=\frac{1}{2\|x\|}\times\|x\|=\frac{1}{2}<1,$$ so $tx\in U$. $\blacksquare$
Claim 2: Let $s\equiv\sup\{t>0\,|\,tx\in U\}.$ Then, $s<\infty$.
Proof: The previous claim already implies that the set whose supremum is taken is not empty, so the supremum is well-defined, though possibly infinite. To see that it is actually finite, note that if $t>0$ and $tx\in U$, then $t\|x\|=\|tx\|<1$ by the definition of $U$, so one must have $$t<\frac{1}{\|x\|}.$$ Hence, the set whose supremum is taken has a finite upper bound, implying that $$s\leq\frac{1}{\|x\|}<\infty,$$
completing the proof. $\blacksquare$
Claim 3: If $s$ is as before, then $sx\notin U$.
Proof: For the sake of contradiction, suppose that $sx\in U$. This means that $\|sx\|<1$, or $s<1/\|x\|$. Pick a number $t>0$ such that $s<t<1/\|x\|$. Then, one has that $\|tx\|=t\|x\|<1$, so that $tx\in U$. By the definition of $s$, one must then have $t\leq s$. But this contradicts $t>s$. $\blacksquare$
