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

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 $$|y-x^*| implies that $$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 $$(b)?$$

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