$\Bbb{R^\omega}=\{(x_n)_{n\in \mathbb{N}}: x_n \in \Bbb{R}\}$
$d(x, y) =\sum_{j\in\mathbb{N}}{(a_j)} \frac{|x_j -y_j|}{1+|x_j -y_j|}$
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?
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If no, then is there any linear metric space in which every open ball is not convex?
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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.
Best Answer
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.