Open mapping theorem for normed abelian groups A norm on an abelian group is a function valued in $\mathbb{R}_{\geq 0}$ which satisfies $|x|=0 \Leftrightarrow x=0$, $|{-}x|=|x|$, and $|x+y| \leq |x|+|y|$, not necessarily $|z x| = |z| |x|$ for integers $z$ (we only have $|zx | \leq |z| |x|$). A normed abelian group is called complete if the metric space with $d(x,y)=|x-y|$ is complete.
Let $A,B$ be two complete normed abelian groups, and let $f : A \to B$ be a bounded homomorphism of abelian groups (i.e. there is some constant $K$ with $|f(x)| \leq K |x|$ for all $x \in A$), which is surjective. Does it follow that $f$ is open?
If not, what additional assumptions do we need exactly so that this becomes true? For example, does it suffice that $A,B$ are $\mathbb{Z}[1/p]$-modules for some integer $p \geq 2$? Perhaps then might imitate the proof for normed vector spaces.
 A: It seems the following. 
If the group $A$ is separable then the group $B$ is separable too. So in this case the map $f$ is open as a surjective continuous homomorphism of Polish groups (see, for instance, [HM, Th. 1.5]).   
If the group $A$ is not separable then there is a counterexample (see, for instance, [HM, Ex. 0.1]). Let $A=B$ be the additive group $\Bbb R$ of real numbers and $f:A\to B$ be the identity map. Let $|\cdot|$ be the usual norm on the group $\Bbb R=B$. The norm $|\cdot|$ induces the natural topology on the group $\Bbb R=B$. Define a norm $\|\cdot\|$ on the group $A$ by putting $\|x\|=0$ if $x=0$ and $\|x\|=1+|x|$ otherwise. Then the norm $\|\cdot\|$ induces the discrete topology on the group $\Bbb R=A$. Clearly, that $|f(x)|\le\|x\|$ for each element $x\in A$, but the map $f$ is not open. 
It seems that the map $f$ is open provided there is some positive constant $k$ with $|f(x)| \geq k |x|$ for all $x \in A$.
References
[HM] Karl H. Hofmann and Sidney A. Morris. Open Mapping Theorem for Topological Groups.
