I have some confusion in RUDIN book Theorem $2.40$ :Every $k-$cell is compact

My confusion written in red colour given below

Here is an outline of Rudin's proof(Theorem $2.40$):

Every k-cell is compact.

Proof. Let $I$ be a k-cell consisting of all points $x = (x_1, \dots, x_k)$ such that $a_j \leq x_j \leq b_j$ for $1 \leq j \leq k$. Put $\delta = \{ \sum\limits_1^k(b_j - a_j)^2\}^{1/2}$ Then $|x-y| \leq \delta$ if $x, y \in I$.

Suppose there is an open cover $\{G_\alpha \}$ which contains no finite subcover. Put $c_j = \frac{a_j + b_j}{2}$. The intervals $[a_j, c_j]$ and $[c_j, b_j]$ determine $2^k$ k-cells whose union is $I$. At least one of these subsets of $I$, say $I_1$, cannot be covered by any finite subcollection of $\{ G_\alpha \}$. So we begin again with the k-cell $I_1$ and subdivide further to achieve a sequence of k-cells such that

$(a)$ $I \supset I_1 \supset I_2 \supset I_3 \supset \dots$

$(b)$ $\color{red}{ I_n \text{is not covered by any finite subcollection of}\ G_\alpha}$

$(c)$ If $x \in I_n$ and $y \in I_n$ then $|y-x| \leq 2^{-n}\delta$

by $( a)$ and theorem $2.39$ , there is a point $x^*$ which lie in every $I_n$. For some $\alpha , x^* \in G_\alpha.$ Since $G_{\alpha}$ is open , there exist $r >0$ such that $|y-x^*|<r $ implies that $y \in G_{\alpha}$.If $n$ is so large that $2^{-n} \delta < r$(there is such an $n$ , for otherwise $2^n \le \frac{\delta}{r}$ for all positive integr $n$ , which is absurd since $R$ is archimedean , then $(c)$ implies that $\color{red}{I_n \subset G_{\alpha}}$ , which $\color{red}{\text{contradict} (b)}$

This completes the proof

My doubt : in b) it already said that $I_n$ is not covered by any finite subcollection of $G_{\alpha}$

This implies that $I_n \subset \bigcup_{\alpha}G_{\alpha}$ doesn't have a finite subcover

Then why $I_n \subset G_\alpha$ contradicts $(b)?$


1 Answer 1


Take neighborhood $N_r(x^*) \subset G_\alpha$. If $n$ is large enough that $2^{-n}\delta < r$ then given $p \in I_n$ $|x^* - p | < 2^{-n}\delta < r \implies p \in N_r(x^*) \implies I_n \subset N_r(x^*)\subset G_\alpha$

Here $I_n \subset N_r(x^*)$ where $N_r(x^*)$ is bounded open ball this implies that $I_n $is covered by a finite subcover so it contradicts $(b)$

  • 1
    $\begingroup$ +1. If $n$ is large enough then $I_n$ is covered by a single member of $\{G_{\alpha}\}.$ $\endgroup$ Dec 1, 2020 at 20:18

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