Question on Every k-cell is compact (Rudin Thm 2.40) A similar question is Every k- cell is compact (Walter Rudin)
I intuitive understand the complete proof as a whole (I do not have a rigorous math background). I am stuck on this specific assumption of the proof from Rudin Thm 2.40
"At least one of these set $Q_i$, call it $I_1$, cannot be covered by any finite subcollection of $\{G_\alpha\}$ (otherwise $I$ could be so covered)."
My doubts/questions are 


*

*I cannot grasp the otherwise part of the statement. If there exists a finite cover for $I_1$, say $I_1\subset \cup_{i\in S} \{G_{\alpha_i}\}$ for some finite set $S$ of indices. Then how do we know that $I$ is also covered by a finite sub-collection: $I=(\cup_{i=2}^{2^k}I_i )\cup I_1\subset (\cup_{j\notin S} \{G_{\alpha_j}\})\cup (\cup_{i\in S}\{G_{\alpha_i}\})$. We still cannot show that $\cup_{j\notin S} \{G_{\alpha_j}\}$ is a finite cover.

*Is my justification correct: Every $2^k$ partition is also k-cell, by the contradiction used in the proof of the theorem, we starts with no k-cell is compact. Hence, $I_1$ being a k-cell cannot have a finite subcover. But I feel this justification is wrong, as every $2^k$ k-cell will not have a finite subcover by this reasoning. What is the justification that there exists at least one set $Q_i$ with no finite subcover.


I am pretty sure I am missing something (may be obvious). Kindly help my understanding.
 A: Indeed by definition each of the $Q_i$ is a $k$-cell. But you have misunderstood the negation of the statement being proved; you want to show that:

Every $k$-cell is compact.

This is shown by a proof by contradiction; assuming the statement is false, a contradiction is derived. The negation of the statement above is:

There exists a $k$-cell that is not compact.

This is not the same as 'no $k$-cell is compact' as you seem to claim.
Then from the existence a $k$-cell $I$ that is not compact, the author derives a contradiction by constructing a sequence $\{I_n\}_{n\in\Bbb{N}}$ of $k$-cells with some desired properties:
Because $I$ is not compact, there exists a cover $\{G_\alpha\}$ of $I$ that does not have a finite subcover. If all of the $Q_i$ have a finite subcover of $\{G_\alpha\}$, then these $2^k$ finite subcovers together form a finite subcover of $I$, a contradiction. Hence at least one of the $Q_i$ does not have a finite subcover; let $I_1$ be one of those $Q_i$. Then $I_1$ is another $k$-cell that is not compact, and we can repeat this construction.
