Write a Compact Set as a Union of Open Cells in $\mathbf{R}^n$
Solution 1:
For each $x \in F_k$, define $I_x ^k = \prod_{i = 1} ^n (x_i - \frac{1}{4nk^2}, x_i + \frac{1}{4nk^2})$. Then $\operatorname{diam}(I_x ^k) = \sqrt{\sum_{i = 1} ^n \frac{1}{2nk^2}} = \sqrt{\frac{1}{2k^2}} = \frac{1}{\sqrt{2}k} < \frac{1}{k}$ for all $k \geq 1$. Now we have $F_k \subseteq \bigcup_{x \in F_k} I_x ^k$. Note moreover, by the definition of $F_k$, we have $I_x ^k \subseteq E$ automatically.