every $k$ cell is compact

Question

I did not understand the proof from the word "and some sets in this infinite sequence are from $\{G\}$....." to the end. Thanks for helping me to understand.

$k$ cell in $\mathbb{R}^k$ is defined as $\{x=(x_1,\dots, x_k):a_i\le x_i\le b_i\}=[a_1,b_1]\times\dots[a_k,b_k]$

By its definition, the $k$-cell is a product of compact spaces, so ... — Hagen von Eitzen, Sep 08 '17 at 05:32
Indeed, the sentence you complain about does not make sense. All sets in the constructed infinite sequence are "rectangular" and possible none of the sets in ${G}$ is. — Hagen von Eitzen, Sep 08 '17 at 05:35
I am teaching undergraduate first 2nd yr students they doesnt know topology and hence tychonof theorem etc ..only metric space — Myshkin, Sep 08 '17 at 05:35
I'd still go for Tychonov, which is not hard for the finite case, just induction. The proof at hand seems to redo the proof that $[0,1]$ is compact instead — Hagen von Eitzen, Sep 08 '17 at 06:06

score 0 · Answer 1 · answered Sep 08 '17 at 10:37

It's not clear what the reasoning is. In my opinion the idea is that you have a sequence of subcells each covered by an infinite cover with no infinite subcover which is itself a subcover of $G$. This sequence converges to a point and this point is contained in an open set $U$ of $G$. But then $U$ must entirely contain a subcell and here it's the contradiction

score -1 · Answer 2 · edited Dec 22 '17 at 01:39

In that prove we have infinite subcell all of it contain only one point and each of it can be covered by one open set of cover but still $G$ covered by infinite open cover, I think we must say if we can't cover $G$ by finite sub cover we split $G$ into $2^k$ of $k$ cells each of them covered by finite sub cover (one) if one of it can't cover take the same manner and split it (split up finite turn).

every $k$ cell is compact

2 Answers2