Does there exists any non trivial linear metric space in which every open ball is not convex?

$\Bbb{R^\omega}=\{(x_n)_{n\in \mathbb{N}}: x_n \in \Bbb{R}\}$

$d(x, y) =\sum_{j\in\mathbb{N}}{(a_j)} \frac{|x_j -y_j|}{1+|x_j -y_j|}$

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?

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

3. 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.

Yes, a standard example is $L^p([0,1])$ for $0 < p < 1$. It is a metric space for $$d(f,g):= \int_0^1|f(x)-g(x)|^pdx.$$

One can show that if $C$ is a convex open set in this space, then $C= L^p([0,1])$. In particular, not a single open ball is convex.

The example of quantumspace is of course the classical/elegant one. Let me give another one (just for fun). We can pick $\mathbb{R}^2$ and put the "Paris"-metric on it (that is how our topology prof called it 10 years ago). Everytime you want to go anywhere, you have to go to "Paris" first and then you can continue your travel. For us Paris will be the origin. Of course it will depend from where you go to "Paris", meaning on some half-ray we put a different metric.

Let $R=\{ (x,0) \in \mathbb{R}^2 \ : \ x\geq 0 \}$, then we define the following metric $$ d(x,y) := \begin{cases} \vert x \vert + \vert y \vert,& x,y\in \mathbb{R}^2 \setminus R, \\ 2\vert x \vert + 2 \vert y \vert.& x,y \in R, \\ 2\vert x \vert + \vert y \vert,& x\in R, y \in \mathbb{R}^2\setminus R, \\ \vert x \vert + 2 \vert y\vert,& x\in \mathbb{R}^2 \setminus R, y\in R. \end{cases} $$ I guess that all balls in this metric are non-convex, if not, just add one more half-ray with a different metric.