Every k-cell is compact / alternative proof

Question

I have become really interested in trying to prove directly things that are more easily proved by contradiction. The below seems to be a good example of this. Rudin's proof makes perfect sense to me. In fact I think it is a very elegant proof. But nonetheless I am very much interested in a constructive proof.

Essentially Rudin proves that every k-cell is compact by assuming the opposite: that there is an open cover of a k-cell $I$ which does not contain a finite subcover. I would like to assume $\{G_\alpha \}$ is an arbitrary open cover of $I$ and show directly that $G_\alpha$ must have a finite subcover.

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) $I_n$ is not covered by any finite subcollection of $G_\alpha$

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

Hence there is a point $x^* \in \cap I_n$ and for some $\alpha$ $x^* \in G_\alpha$. Since $G$ is open there is a neighborhood $N_r(x^*) \subset G_\alpha$. If $n$ is large enough that $2^{-n}\delta < r$ then given $p \in I_n$ $|x^* - p | < 2^{-n}\delta < r \implies p \in N_r(x^*) \implies I_n \subset G_\alpha$ contradicting the fact that $I_n$ cannot be covered by a finite subcollection of $\{ G_\alpha\}$

Answer 1

https://personal.math.ubc.ca/~malabika/teaching/ubc/fall18/math320/HW7-revised.pdf

This assignment provides an alternative way to prove it. Is this what you want?

Answer 2

I might be wrong here, but I think there is an issue with the proof above.

Now Rudin proves that there is no sequence $I_1 \supset I_2 \ldots$ of $k$-cells with properties b and c.

Unfortunately this doesn't prove that the original $k$-cell can be covered with a finite subset of $\{G_{\alpha}\}$. Here is why. Suppose you do the subdivision of the original $k$-cell into $2^n$ smaller cells and you do the same to these cells, etc. . This forms a tree where $k$-cells are connected with their direct subdivision elements. Each path from the root of the tree(original $k$-cell) defines an $I_1 \supset I_2 \supset \ldots$ sequence. What is proven by Rudin is that for each of these sequences there is an $l \in \mathbb{N}$ such that $I_l \subseteq G_{\alpha}$ for some $\alpha$. Let's not subdivide further the cells with such property. These are the leafs. Now what's proven is that each leaf has finite depth, but the depths are not bounded and it's easy to construct a tree where each leaf has finite depth, but the tree has $\mathbb{N}$ leafs.