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

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} 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?

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.
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 $$\alpha^{-1}x\in M.$$