Let the $E \subseteq \mathbf{R}^n$ be an open set. Define the compact set $$ F_k := \{ x \in E: \operatorname{dist}(x, E^c) \geq \frac{1}{k}, \| x \| = d(0, x) \leq k \}. $$ How do we show the compact set $F_k$ can be covered by union of open cells $I \subseteq E$ (product of real open intervals) with diameter less than $1/k$?

I suppose the place where I got stuck is understanding how the diameter of an open cell is evaluated. We should be able to construct the open cells once knowing how to ensure the diameter of it to be less than $1/k$ and then just union over all elements of $F_k$, and we will be done (I think?). The diameter of a set $A \subseteq \mathbf{R}^n$ is defined as $$ \operatorname{diam}(A) := \sup_{x, y \in A} d(x, y) $$ with $d$ the usual Euclidean distance.

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    $\begingroup$ You just want any cover of $F_k$? What about just taking a sufficiently small cells centred at each point in $F_k$? Any set, compact or not, can be covered in this way by cells of arbitrarily small diameter. $\endgroup$ Jan 5 at 1:14
  • $\begingroup$ @TheoBendit Yes, any cover. I would like to see what the covering open cells should look like exactly. $\endgroup$ Jan 5 at 1:22
  • $\begingroup$ @TheoBendit Following your comment, I just wrote an answer to my question below. Could you check if it is correct? $\endgroup$ Jan 5 at 2:30

1 Answer 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.

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    $\begingroup$ +1 This is good. I just fixed a minor typo. $\endgroup$ Jan 5 at 3:32

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