A question about the theorem:
Suppose $p$ is a seminorm on a vector space $X$. Then
(a) $p(0) = 0$
(b) $\left|p(x) - p(y) \right| \leq p(x - y)$
(c) $p(x) \geq 0$
(d) $\left\{x : p(x) = 0 \right\}$ is a subspace of $X$
(e) The set $B = \left\{x : p(x) < 1 \right\}$ is convex, balanced, absorbing, and $p = \mu_B$
The point $(e)$ is what my question is about, so I'll just write down the author's proof and then I'll ask the question
If $x \in X$ and $s > p(x)$ then $p(s^{-1}x) = s^{-1}p(x) < 1$. This shows that $B$ is absorbing and also that $\mu_B(x) \leq s$. Hence $\mu_B \leq p$. But if $0 < t < p(x)$ then $p(t^{-1}x) \geq 1$ and so $t^{-1}x$ is not in $B$. This implies $p(x) \leq \mu_B(x)$ and completes the proof.
I was going through this proof last night and I'll give now my interpretation.
Suppose $x \in X$ and pick any arbitrary $s > 0$ such that $p(x) < s$, since $s$ is arbitrary but greater than $p(x)$ we can write $s = p(x) + \epsilon$, where $\epsilon > 0$. The author also states that $\mu_B(x) \leq s$, but this specifically follows from the definition of $\mu_B(x)$ to be precise. Hence we have the inequality $$ \mu_B(x) < p(x) + \epsilon $$ From this inequality it follows $\mu_B(x) \leq p(x)$, I've seen the proof many times on this website.
For the other part of the inequality I'm not sure what the author does, because to me
$$ \mu_B(x) = \inf \left\{ t > 0 : t^{-1}x \in B \right\} = \inf \left\{ t > 0 : p(t^{-1}x) < 1 \right\} = \inf \left\{ t > 0 : p(x) < t \right\} $$
And from this definition it should follow that for any $\epsilon > 0$ we have
$$ p(x) < \mu_B(x) + \epsilon $$
And as the former case it follows $p(x) \leq \mu_B(x)$.
Is my proof correct?