The uniform metric on $\mathbb{R}^\omega$ I'm working on a question from Topology by Munkres on p. 127 (Exercise #6). Here it is: 
Let $\bar{\rho}$ be the uniform metric on $\mathbb{R}^\omega$. Given $\mathbb{x}= (x_1, x_2, ...) \in \mathbb{R}^\omega$ and given $0 < \epsilon < 1$, let $U(\mathbb{x}, \epsilon):= (x_1 - \epsilon, x_1 + \epsilon) \times \cdots \times (x_n - \epsilon, x_n + \epsilon) \times \cdots$. 
The question is divided into three parts, of which I have been able to answer the first and last parts of the problem. For the second part, i.e. part (b), I think I'm on the right track but I'm not too sure about that. Part (b) asks to show that $U(\mathbb{x}, \epsilon)$ is not even open in the uniform topology. 
In my attempt to do this part, I let $\mathbb{x}= \mathbb{0}= (0, 0, 0, 0, ...)$ and $\mathbb{y}= (0, 1/2, 2/3, 3/4, ...)$. Then I computed the distance
$\displaystyle\bar{\rho}(\mathbb{x}, \mathbb{y})= \sup_{i \geq 0}(\min\{|x_i - y_i|, 1\})= 1$. So that means the point $\mathbb{y}$ would not be in the open ball of radius $1$, i.e. $B_\bar{\rho}(\mathbb{x}, 1)$. So wouldn't this mean that we can't fit an $\epsilon$-ball around some point $\mathbb{y}$ that stays in $U(\mathbb{x}, \epsilon)$, and hence, $U(\mathbb{x}, \epsilon)$ is not open in the uniform topology induced by $\bar{\rho}$? Any guidance or advice would be greatly appreciated. I'm just starting to learn topology, and still have a lot to learn! Interesting stuff, though.
 A: You have the right idea: it’s true that no open ball centred at your point $y$ is contained in $U(0,1)$. To finish the argument, you have to show this, and then you have to generalize the argument to all sets of the form $U(x,r)$.
Consider the ball $B_{\bar\rho}(y,r)$, where $r>0$; we want to show that it’s not a subset of $U(0,1)$ no matter how small $r$ is. Clearly $y$ is nudging the ‘top edge’ of $U(0,1)$, so we should try to pick something ‘bigger than’ $y$, but still ‘small’ enough to fit in $B_{\bar\rho}(y,r)$; since $$y=\left(\frac{n-1}{n}\right)_{n\in\mathbb{Z}^+}=\left(1-\frac1n\right)_{n\in\mathbb{Z}^+},$$ a natural idea is $$z=\left(\frac{n-1}{n}+\frac{r}2\right)_{n\in\mathbb{Z}^+} = \left(1-\frac1n+\frac{r}2\right)_{n\in\mathbb{Z}^+}.$$ Then $\bar\rho(y,z) = \sup\limits_{k\in\mathbb{Z}^+}\min\{|y_k-z_k|,1\}=\min\{r/2,1\}\le r/2$, so $z\in B_{\bar\rho}(y,r)$. But if $n\ge \dfrac2r$, then $d\frac1n \le \dfrac{r}2$, $z_n = 1 - \dfrac1n + \dfrac{r}2 \ge 1$, and hence $z\notin U(0,1)$.
This shows that $U(0,1)$ is not open, and the same idea generalizes very easily. To show that an arbitrary $U(x,\epsilon)$ is not open, just scale this example by a factor of $\epsilon$ and translate it from the origin to $x$. (This is what Davide was suggesting in his comment.) That is, instead of taking $$y=\left(\frac{n-1}{n}\right)_{n\in\mathbb{Z}^+},$$ let $$y=x+\left(\frac{n-1}{n}\epsilon\right)_{n\in\mathbb{Z}^+}=\left(x_n+\frac{n-1}{n}\epsilon\right)_{n\in\mathbb{Z}^+}.$$ Then $y\in U(x,\epsilon)$, but no ball centred at $y$ is a subset of $U(x,\epsilon)$: the argument used above works with only very minor modifications, which I’ll leave to you.
