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.