In the proof $I$ is a $k$-cell whose coordinates are bounded by $a_{j}\le x_{j}\le b_{j}$ where $1\le j\le k$.
From the proof: Suppose, to get a contradiction, that there exists an open cover $\{G_{\alpha}\}$ which has no finite subcover of $I$.
Put $c_{j}=(a_{j}+b_{j})/2$. The intervals $[a_{j},c_{j}]$ and $[c_{j},b_{j}]$ then determine $2^{k}$ $k$-cells $Q_{i}$ whose union is $I$.
One of these $Q_i$, call it $I_1$, will also have no finite subcover. Continuing this process we obtain a sequence $I_n$ with the following properties:
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$I_n \subset I_{n-1}$ for all $n$.
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$I_n$ is not covered by any finite subcollection of $\{ G_{\alpha}\}$.
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If $x,y \in I_n$, the $|x-y| \leq 2^{-n}\delta$
By the definition of sequence $\{I_n\}$, (1) is satisfied
My question is: How to prove that $\{I_n\}$ satisfies the (2) and (3) properties?
Please help me to clear my doubts.
Best Answer
For (2), each set $I_n$ is chosen specifically so that it has no finite subcover. Indeed, by our assumption that $\{G_\alpha\}$ has no finite subcover, upon dividing $I$ into $2^k$ cells $Q_i$, it must be the case that at least one of these $Q_i$, call it $I_1$, has no finite subcover.
Now suppose we divide $I_1$ into $2^k$ cells $R_j$. It must also be the case that at least one of these $R_j$, call it $I_2$, has no finite subcover, otherwise each $R_j$ is covered by finitely many $G_\alpha$, whereby $I_1$ is covered by finitely many $G_\alpha$, contradicting what we have deduced that $I_1$ has no finite subcover.
Proceeding in this manner, we deduce that each $I_n$ has no finite subcover, which gives us (2).
To prove (3), recall the definition of $\delta$ as the diameter of the set $I$: $$ \delta = \sup_{x,y\in I}|x-y| = \bigg(\sum_{j=1}^k(b_j-a_j)^2\bigg)^{1/2}. $$ Now, $I_1$ has half the diameter of $I$ since $I_1$ is one of the cells obtained from $I$ by dividing each edge of $I$ in half. Thus $\operatorname{diameter}(I_1) = 2^{-1}\delta$. Likewise, $I_2$ has half the diameter of $I_1$, so that $\operatorname{diameter}(I_2) = 2^{-1}\operatorname{diameter}(I_1) = 2^{-2}\delta$. Proceeding in this manner, if $n\ge 1$, we deduce that $\operatorname{diameter}(I_n) = 2^{-n}\delta$.
By definition of the supremum, if $x,y\in I_n$, then $$ |x-y|\le\sup_{x,y\in I_n}|x-y| = \operatorname{diameter}(I_n) = 2^{-n}\delta, $$ which finally gives us (3).