1).why $U(\mathbf{x}, \epsilon)$ is open in box product, ? 2) why $U(\mathbf{x}, \epsilon)$ not open in the uniform topology? 
Given $\mathbf{x} = (x_1, x_2, \ldots) \in \mathbb{R}^{\omega}$ and given $0 < \epsilon < 1$, let $$U(\mathbf{x,\epsilon}) = (x_1 - \epsilon, x_1 + \epsilon) \times \cdots \times (x_n - \epsilon, x_n + \epsilon) \times \cdots.$$

My  Question  is  that
1).why $U(\mathbf{x}, \epsilon)$ is open in box product, ?
2) why $U(\mathbf{x}, \epsilon)$ is  not  open in the uniform topology ?
I see the munkre topology book page no. 124 section 20  ,,,as Munkre  have not  exaplain this in detail, as  im very confused now ..
so I need  detail  exaplanation  ....with simple language...
Thanks  u
 A: $U(\mathbf{x},\varepsilon)$ is open in the box topology, because each set $(x_i - \varepsilon, x_i + \varepsilon)$ is open in $\mathbb{R}$ and so $U(\mathbf{x},\varepsilon)$ is a product of infinitely many open sets and so a basic open set in the definition of the box topology on $\mathbb{R}^\omega$. So in particular it's open in the box topology.
Now define $p_i = x_i + (1-\frac{1}{2^i})\varepsilon$ for $i=1,2,\ldots$. Note that $p_i \in (x_i - \varepsilon, x_i + \varepsilon)$, for all $i$, and thus $\mathbf{p} \in U(\mathbf{x},\varepsilon)$, as $p_i < x_i + \varepsilon$, but also note that $p_i$ gets closer and closer to the boundary of the set, as $i$ increases. 
The claim is that $\mathbf{p} = (p_1, p_2, \ldots)$ is not an interior point of $U(\mathbf{x},\varepsilon)$ in the uniform metric topology. Suppose it were:
then there would be an $r>0$ (and we can also assume $r<1$ WLOG) such that $$B(\mathbf{p},r) \subseteq U(\mathbf{x},\varepsilon)$$
where the ball is taken in the uniform metric.
Now, define $p'_i = p_i + \frac{r}{2}$. Then $|p'_i - p_i| = \frac{r}{2}< 1$, so that $d= \sup_i \min(|p'_i - p_i|, 1) = \frac{r}{2} < r$, and thus $\mathbf{p'} = (p'_1, p'_2, \ldots,) \in B(\mathbf{p}, r)$.
But if we pick $i$ so large that $\frac{1}{2^i}\varepsilon < \frac{r}{2}$, then $$p'_i = p_i + \frac{r}{2} = x_i + \varepsilon(1-\frac{1}{2^i}) + \frac{r}{2} = x_i + \varepsilon + (\frac{r}{2} - \frac{1}{2^i}\varepsilon) > x_i + \varepsilon$$ and so $p'_i \notin (x_i - \varepsilon, x_i + \varepsilon)$ and we see that $\mathbf{p'} \notin U(\mathbf{x},\varepsilon)$, contradicting the above inclusion we were supposed to have. So indeed $\mathbf{p}$ is not an interior point of $U(\mathbf{x},\varepsilon)$.
