Prob. 6, Sec. 20 in Munkres' TOPOLOGY, 2nd ed: How is this set not open? 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) 
Then $U(x,\epsilon)$ is not equal to the open ball $B_{\tilde{\rho}}(x, \epsilon)$ of radius $\epsilon$ centered at $x$; in fact the former properly contains the latter: the point $x^\prime \colon= (x_n + \epsilon- \frac{\epsilon}{n})_{n \in \mathbb{N}}$ is in the former but not in the latter. 
Am I right? 
(b)
How to show that $U(x, \epsilon)$ is not even open in the uniform topology? 
For this purpose, we need to explicitly exhibit a point $y$ in $U(x, \epsilon)$ such that no ball centered at $y$ is contained in $U(x,\epsilon)$. Which point will do, I wonder? I would appreciate if the process could be shown clearly and in detail. 
(c) And, I have managed to show that 
$$B_{\tilde{\rho}}(x, \epsilon) = \bigcup_{0< \delta < \epsilon} U(x, \delta).$$
What is the situation in each of  (a), (b), and (c) above (i) if  $\epsilon= 1$? and  (ii) if $\epsilon > 1$? 
 A: Your answer to part (a) is correct. As Rolf Hoyer noted in the comments, you can use the point $x'$ from part (a) to answer part (b): you’ll find that no matter how small a $\delta>0$ you choose, $B_{\tilde\rho}(x',\delta)\setminus U(x,\epsilon)\ne\varnothing$, because, for instance, $x_k'+\frac{\delta}2>x_k+\epsilon$ for all sufficiently large $k$.
If $\epsilon>1$, $B_{\tilde\rho}(x,\epsilon)=\Bbb R^\omega$, since $\tilde\rho(x,y)\le 1$ for all $x,y\in\Bbb R^\omega$. $U(x,\epsilon)$, on the other hand, is definitely not all of $\Bbb R^\omega$, so in this case we have $U(x,\epsilon)\subsetneqq B_{\tilde\rho}(x,\epsilon)$. However, $U(x,\epsilon)$ is still not open in the uniform topology: you can use the same basic idea to prove it as in the case $0<\epsilon<1$, taking a point $x'\in U(x,\epsilon)$ such that $x_k'-x_k$ approaches $\epsilon$ as $k$ increases. As far as (c) goes, it’s clear that
$$B_{\tilde\rho}(x,\epsilon)=\Bbb R^\omega\supsetneqq\bigcup_{0<\delta<\epsilon}U(x,\delta)$$
when $\epsilon>1$. In fact, by considering a point $x'$ such that $x_k'=x_k+k$ for each $k$, we can see that
$$B_{\tilde\rho}(x,\epsilon)=\Bbb R^\omega\supsetneqq\bigcup_{\delta>0}U(x,\delta)\;.$$
Finally, the answers when $\epsilon=1$ are the same as when $0<\epsilon<1$, and essentially the same arguments work.
