show that $U(x,\epsilon)$ is not even open in the uniform topology. Given  $x = (x_1,x_2,x_3,\ldots)$ in $\mathbb{R}^\omega$. Let $\mathbb{R}^\omega$ denote the set of all sequences of real numbers, let $\tilde{\rho}$ denote the uniform metric on $\mathbb{R}^\omega$ defined as 
$$ 
\begin{align*}
\tilde{\rho}(x,y) \colon= \sup \left\{ \ \min \{ \ \vert x_n - y_n \vert, 1 \} \ \colon \ n = 1, 2, 3, \ldots \ \right\} &\\
&\mbox{ for all } \ x \colon= (x_n)_{n \in \mathbb{N}}, \ y \colon= (y_n)_{n \in \mathbb{N}} \in \mathbb{R}^\omega. 
\end{align*}
$$
Let $0 < \epsilon < 1$, and let 
$$ 
U(x,\epsilon) \ \colon= \ (x_1 - \epsilon, x_1 + \epsilon) \times (x_2 - \epsilon, x_2 + \epsilon) \times (x_3 - \epsilon, x_3 + \epsilon) \times \ldots
$$
$a)$  show that $ U(x,\epsilon)$ is not equal to the  $\epsilon - $ball $B_{\tilde{\rho}}(x, \epsilon)$
$b)$ show that  $U(x,\epsilon)$ is not even open in the uniform topology.
$c) $ show that $B_{\tilde{\rho}}(x, \epsilon) = \bigcup_{0< \delta < \epsilon} U(x, \delta)$
Here  i got the  answer:
  Prob. 6, Sec. 20 in Munkres' TOPOLOGY, 2nd ed: How is this set not open?
But I didn't understand the answer:
as I didn't understand the  the  red circle,  why  it has been taken.
 A: We have
$$\tilde{\rho}(x,x') = \sup_{n\in\mathbb{N}}\min\left\{\left|x_n -x_n - \varepsilon\left(1-\frac{1}{n}\right)\right|, 1\right\} = \sup_{n\in\mathbb{N}} \varepsilon\left(1-\frac{1}{n}\right) = \varepsilon$$
so $x' \notin B_{\tilde\rho}(x,\varepsilon)$.
However $$\left|x_n -x_n - \varepsilon\left(1-\frac{1}{n}\right)\right| = \varepsilon\left(1-\frac{1}{n}\right) < \varepsilon \implies x_n + \varepsilon\left(1-\frac{1}{n}\right) \in \langle x_n - \varepsilon, x_n + \varepsilon\rangle$$
so we conclude $x' \in U(x, \varepsilon)$.
The point is, there exist sequences $(y_n)_n$ such that $|x_n - y_n| < \varepsilon, \forall n \in \mathbb{N}$ and yet $\sup_{n\in\mathbb{N}} |x_n - y_n| = \varepsilon$.

Edit:
For (c), you wish to prove
$$B_{\tilde{\rho}}(x, \varepsilon) = \bigcup_{0< \delta < \varepsilon} U(x, \delta)$$
Take $y \in B_{\tilde{\rho}}(x, \varepsilon)$. Then $\tilde{\rho}(x,x') = \delta < \varepsilon$ so $|x_n - y_n| \le \delta < \frac{\delta + \varepsilon}2 < \varepsilon, \forall n \in \mathbb{N}$ so $y \in U\left(x,\frac{\delta + \varepsilon}2\right)$.
Conversely, if $y \in \bigcup_{0< \delta < \epsilon} U(x, \delta)$ then $y \in U(x,\delta)$ for some $\delta \in \langle 0, \varepsilon\rangle$. Thus, $|x_n - y_n| < \delta, \forall n \in \mathbb{N}$ so $\tilde\rho(x,y) = \sup_{n\in\mathbb{N}} |x_n - y_n| \le \delta < \varepsilon$. Therefore, $y \in B_{\tilde{\rho}}(x, \varepsilon)$.
