TITLE: Does there exists any non trivial linear metric space in which every open ball is not convex?
QUESTION [7 upvotes]: $\Bbb{R^\omega}=\{(x_n)_{n\in \mathbb{N}}: x_n \in \Bbb{R}\}$
$d(x, y) =\sum_{j\in\mathbb{N}}{(a_j)} \frac{|x_j -y_j|}{1+|x_j -y_j|}$
Then $(\Bbb{R^\omega}, d) $ is a metric space.
I know in a normed space any ball is convex. And it is easy to prove.
The space $(\Bbb{R^\omega}, d) $  is not a normed space, I mean no norm on $\Bbb{R^\omega} $ can induce the metric $d$.
So, I guess in that space, It may be possible to find an open ball which is not convex.
My question :1) Can I pick any open ball to test whether it is convex or not?

If no, then is there any linear metric space in which every open ball is not convex?

Can we get an example of a linear metric space (not a normed space) in which every open ball is convex?


For the last question can I take $(X, \|•\|)$ be any  normed space and then define a metric $d(x, y) =\sqrt\|x-y\|$. I think it works. Isn't it?
Here $d$ is not scaling equivalent, hence not induced by any norm.
$B_{d}(x_0, r) =\{x\in X : \|x-x_0\|<r^2 \}=B_{\|•\|} (x_0, r^2) $
Hence, every open ball is convex.

REPLY [4 votes]: A simpler example of an “$L_p$” space: $\mathbb R^2$ with the $l_p$ quasi-norm, where $p=\tfrac12$. The quasi-norm is $\|(x,y)\|_p = \sqrt{|x|}+ \sqrt{|y|}$.
I like this example because you can easily draw pictures of the open balls —- they all look like diamonds with concave sides. See here.