Let $F\hspace{.03 in}$ be a field, and let $E\hspace{.03 in}$ be an ordered subfield of $F$.
Let $\;\; |\hspace{-0.03 in}\cdot\hspace{-0.03 in}| \: : \: F \: \to \: E \;\;$ be such that for all elements $x$ of $E\hspace{.02 in}$, for all elements $y$ and $z$ of $\hspace{.02 in}F$,
$0\leq x \; \implies \; |\hspace{.025 in}x\hspace{.02 in}| = x$
and
$0\leq \left|\hspace{.03 in}y\hspace{.015 in}\right| \;\;\;\;\;$ and $\;\;\;\;\; \left|\hspace{.03 in}y\hspace{.015 in}\right| \cdot \left|\hspace{.02 in}z\hspace{.01 in}\right| \:\: = \:\: \left|\hspace{.03 in}y\hspace{-0.05 in}\cdot\hspace{-0.05 in}z\hspace{.02 in}\right| \;\;\;\;\;$ and $\;\;\;\;\; \left|\hspace{.03 in}y\hspace{-0.04 in}+\hspace{-0.04 in}z\hspace{.02 in}\right| \:\: \leq \:\: \left|\hspace{.03 in}y\hspace{.01 in}\right|+\left|\hspace{.02 in}z\hspace{.01 in}\right|$
.
Additionally, assume that $F$ is complete with respect to $\:|\hspace{-0.03 in}\cdot\hspace{-0.03 in}|\:$. $\;\;$ Does it follow that the standard results for Banach spaces over $\mathbb{R}$ and $\mathbb{C}$ (open mapping theorem, closed graph theorem, and the version of the uniform boundedness principle that just uses boundedness, rather than suprema) hold for "Banach spaces over $\hspace{.02 in}\langle F,\hspace{-0.02 in}|\hspace{-0.04 in}\cdot\hspace{-0.04 in}|\hspace{.02 in}\rangle$"$\hspace{.03 in}$?