Open, convex set of TVS I'm studying LCS using Conway's book. And I had a question about a proof of Proposition 3.2 in chapter 4.
The author said, the proof of this proposition is similar to that of proposition 1.14 (If V is a nonempy convex, balanced subset of a vector space X that is absorbing at each of its points, then there is the unique seminorm p on X such that V={x| p (x)<1})
I tried to prove this, but I failed the first step of the proof, that is, for any element x of X, the set {t>=0 | x is contained in tG} is nonempty.
(Here, X=TVS, G=open, convex subset)
Please give me any hint or answer about this.
Thank you in advance.
 A: Consider a convex set $A \subseteq X$ which is absorbing. The $\textit{Minkowski functional}$ $\mu_A$ of $A$ is defined by
\begin{align*}
\mu_A(x) &= \inf\{\alpha > 0 : \frac{x}{\alpha} \in A\} & & (x \in X).
\end{align*}
$\textbf{Lemma:} $Suppose $A$ is a convex absorbing set in a vector space $X$. Then
(a) $\mu_A(x + y) \leq \mu_A(x) + \mu_A(y)$.
(b) $\mu_A(\alpha x) = \alpha \mu_A(x)$, if $\alpha \geq 0$.
(c) $\mu_A$ is a seminorm if $A$ is balanced.
$\textit{Proof}:$ Associate with each $x \in X$ the set
\begin{equation}
H_A(x) = \{\alpha > 0 : \frac{x}{\alpha} \in A\}.
\end{equation}
Suppose $t \in H_A(x)$ and $s > t$. $0 \in A$ as $A$ is absorbing, and convexity of $A$ implies that 
\begin{equation}
\frac{t}{s}\Bigg(\frac{x}{t}\Bigg) + \Bigg(1 - \frac{t}{s}\Bigg)0 \in A.
\end{equation}
In turn $s \in H_A(x)$. Thus, each $H_A(x)$ is a half line whose left end point is $\mu_A$.
(a) Suppose $\mu_A(x) < s, \mu_A(y) < t, u = s + t$. Then $\frac{x}{s}, \frac{y}{t} \in A$. Since $A$ is convex,
\begin{equation}
\frac{x + y}{u} = \Bigg(\frac{s}{u}\Bigg)\Bigg(\frac{x}{s}\Bigg) + \Bigg(\frac{t}{u}\Bigg)\Bigg(\frac{y}{t}\Bigg)
\end{equation}
lies in $A$. Hence $\mu_A(x + y) \leq u = s + t = \mu_A(x) + \mu_A(y)$. This proves (a).
(b) This follows immediately from the definitions.
(c) Clearly, $\mu_A(x) \geq 0$ for all $x \in X$. So, the only thing we need to prove is that
\begin{align*}
& \mu_A(\alpha x) = |\alpha| \mu_A(x) & (\alpha \in \mathbb{K}).
\end{align*}
Notice that it is enough to show that $\mu_A(\frac{\alpha}{|\alpha|} x) = \mu_A(x)$ for all $\alpha \in \mathbb{K}$, because then it will follow that $\mu_A(\alpha x) = |\alpha|\mu_A(\frac{\alpha}{|\alpha|} x) = |\alpha|\mu_A(x)$. But this is immediate from the definition and the extra balanced condition on $A$ as
\begin{align*}
\mu_A(\frac{\alpha}{|\alpha|}x) &= \inf\{t > 0 : \frac{\alpha}{|\alpha|}\frac{x}{t} \in A\}\\
&= \inf\{t > 0 : \frac{x}{t} \in A\}. & \Bigg(\text{$\frac{\alpha}{|\alpha|}\frac{x}{t} \in A \Leftrightarrow \frac{x}{t} \in A$ as A is balanced}\Bigg)
\end{align*}
