If $p(x)<1,$ then why does there exists $\alpha \in (0,1)$ such that $\alpha^{-1}x\in M$?

Question

Definition: Given that $E$ is a normed linear space. Let $M\subseteq E$ be an open, convex set with $0\in M.$ For all $x\in E$, define

\begin{align} p(x)=\inf\{\alpha>0:\,\alpha^{-1}x\in M \} \end{align} Then, $p$ is called a guage of $M$.

My question: I am going through a proof which claims that if $p(x)<1,$ then there exists $\alpha \in (0,1)$ such that $\alpha^{-1}x\in M.$

MY TRIAL: From my own understanding, for fixed $x\in E$ \begin{align} p(x)=\inf\{\alpha>0:\,\alpha^{-1}x\in M \} \end{align} if and only if $\forall\,\epsilon>0,\;\exists\,\alpha_{\epsilon}>0$ such that \begin{align} p(x)\leq \alpha_{\epsilon}<p(x)+\epsilon\,. \end{align} Please, can anyone explain why the claim if $p(x)<1,$ then there exists $\alpha \in (0,1)$ such that $\alpha^{-1}x\in M$ holds?

Answer 1

Since $p(x)<1$, take $p(x)<\alpha<1$. By definition of p: $\alpha ^{-1} x \in M$.

Note that an interesting question is proving that if $p(x)<1$ then x is and internal point of M.

Answer 2

I found the answer and would love to share:

If $p(x)<1$ for a fixed $x\in E$, then by Archimedean principle, there exists $\alpha\in (0,1)$ such that $p(x)<\alpha<1$ and $\alpha^{-1}x\in M.$