connectedness of uniform topology This question can also be found in this link, but because of dull brain I could not understood that One line proof.

Let $\mathbb{R}^{\omega}$ be the set of all (infinite) sequences of real numbers. Determine whether $\mathbb{R}^{\omega}$ is connected or not in the uniform topology.
Basis elements of uniform topology are $B_{\rho}(x,\epsilon)=\{ y\mid \rho(x,y)&lt;\epsilon \}$, where $\rho(x,y)=\sup\{ \min(d(x_\alpha,y_\alpha),1)\mid \alpha \in \mathbb{Z} \}$

From the answer in the link, I understood that it is not connected.
What I found:
Note that $B_{\rho}(x,1)=\overline{B_{\rho}(x,1)}$. So I need to find $x_1,x_2\in \mathbb{R}^{\omega}$ such that $B_{\rho}(x_1,1)\cap B_{\rho}(x_2,1)=\phi$
How to find such $x_1$ and $x_2$?
So we consider on $\mathbb{R}^\omega$ what Munkres calls the uniform topology, i.e. the topology induced by the metric 
 $$d(x,y) = d((x_n)_n,(y_n)_n) = \sup_n \{\min(|x_n - y_n|,1): n \in \mathbb{N}\}$$

Define the following relation $\sim$ on $X$: $x \sim y$ iff the sequence $(x_n - y_n)_n$ is bounded in $\mathbb{R}$.

This is an equivalence relation on $X$: 


$x \sim x$ because the $0$-sequence is bounded.
$x \sim y$ implies $y \sim x$ because $(x_n -y_n)_n$ is bounded iff 
its negative $(y_n - x_n)_n$ is.
$x \sim y$ and $y \sim z$ imply $x \sim z$ because if the sequences 
$(x_n - y_n)_n$ and $(y_n - z_n)_n$ are bounded then their sum 
$(x_n -z_n)_n$ is bounded too.


So, we have a partition into equivalence classes. 
Denote the equivalence class of $x=(x_n)\in \mathbb{R}^\omega$ by $[x]$.
Also note that all classes are mutually homeomorphic:

If we have a fixed $x \in \mathbb{R}^\omega$, then the map $f_x$ that sends $(y_n) \in [x]$ to
$(y_n - x_n)$ in $[0]$ (the class of the $0$-sequence, i.e, all 
bounded sequences in $X$) is a homeomorphism.
($X$ is a topological group under $+$, or check continuity 
with the definition; it's even an isometry.)

Also, every set $[x]$ is open in $X$. For, let $y \in [x]$.
Then the open ball $B_d(y, {1 \over 2}) \subset [x]$: if $z  \in B_d(y,{1 \over 2})$ this implies that for
all $n$,  $|y_n - z_n| &lt; {1 \over 2}$, so $y_n - z_n$ is bounded, so $z$ is in $[y]=[x]$.

So we have a partition of open sets, so they are all closed as well, as the complement of a class is
the union of the other classes, hence also open.

Note also that $[0]$ is a topological vector space, it is homeomorphic to
$\ell_{\infty}$; or see in another way that $[0]$ (and hence all classes) are path-connected:

Every $x$ in $[0]$ is connected by a path $F_x:[0,1] \to [0]$ from $0$ to $x$:
$F(t) = ((t*x_n)$, for $t \in [0,1]$. This is continuous on $[0]$
because there $d(F_x(t_1),F_x(t_2)) = \sup_n \{\min(1, |t_1-t_2|*|x_n|)\}$;
now let $M$ be a bound on all the $|x_n|$ (by boundedness of $x$), and let $|t_1-t_2| &lt; {1 \over M}$. Then: 
$d(F_x(t_1),F_x(t_2)) &lt; M\cdot |t_1-t_2|$, which shows directly that $F_x$ is continuous.

So every point in $[0]$ is connected by a path to $0$, hence $[0]$ is path-connected, 
and as said, so are all other classes as well. Because the classes are clopen, there can be
no larger path-connected or connected set that has points in more than one class.
So sets of the form $[x]$ are maximal (path-)connected subsets.
So the $[x]$ ,$x \in X$ form the (path-)components of $X$.

As to your last part: take one sequence $x_n=0$, and the other $x_n =n$. Their balls of radius $\frac{1}{2}$ or $1$ do not intersect. These sequences clearly lie in different (and hence disjoint clopen) classes.
Not connected,  because there is set of bounded sequences that is both open and closed, in the uniform topology. 
If $A$ is this set of bounded sequences and $X - A$ be the set of unbounded sequence, then both seem to be open in uniform topology. 
So, that makes a separation on  $\mathbb{R}^\omega$ in the uniform topology.