How exactly does a basic open set in uniform topology look like? How to prove that the uniform topology is different from both the product and the box topology?
Here in the link I don't see how V is an uniform-1/2 ball centred at 0.
 A: The original (and, unfortunately, accepted) answer was incorrect. This is a corrected version.
The set in question is
$$V=\left\{x\in\Bbb R^{\Bbb N}:d(x,\mathbf{0})<\frac12\right\}\,,$$
where
$$d:\Bbb R^{\Bbb N}\times\Bbb R^{\Bbb N}\to\Bbb R:\langle x,y\rangle\mapsto\sup_{n\in\Bbb N}\min\{1,|x_n-y_n|\}\,.$$
Since $d$ is the uniform metric on $\Bbb R^{\Bbb N}$, $V$ is by definition the open ball of radius $\frac12$ centred at $\mathbf{0}$ with respect to the uniform metric on $\Bbb R^{\Bbb N}$, just as
$$\left\{x\in\Bbb R:|x|<\frac12\right\}$$
is the open ball of radius $\frac12$ centred at $0$ with respect to the usual metric on $\Bbb R$.
If one unpacks the definition, one can see that
$$\begin{align*}
V&=\left\{x\in\Bbb R^{\Bbb N}:\sup_{n\in\Bbb N}|x_n|<\frac12\right\}\\
&=\left\{x\in\Bbb R^{\Bbb N}:\exists r\in\left(0,\frac12\right)(|x_n|<r\text{ for all }n\in\Bbb N)\right\}\,.
\end{align*}$$
Note that this is not the same as
$$W=\left\{x\in\Bbb R^{\Bbb N}:|x_n|<\frac12\text{ for all }n\in\Bbb N)\right\}\,:$$
the point $x=\left\langle\frac12-\frac1{2^{n+1}}:n\in\Bbb N\right\rangle$ is in $W$ but not in $V$, since
$$\sup_{n\in\Bbb N}\left(\frac12-\frac1{2^{n+1}}\right)=\frac12\,.$$
