Timeline for Proving neighborhood of a compact product space contains a sub-neighborhood formed by taking product [closed]
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 4, 2022 at 3:59 | vote | accept | tgeng | ||
| Sep 26, 2020 at 11:53 | history | closed |
Piotr Hajlasz abx user44191 Andreas Blass Jochen Wengenroth |
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| Sep 26, 2020 at 6:08 | answer | added | Martin Väth | timeline score: 1 | |
| Sep 25, 2020 at 20:58 | comment | added | tgeng | Thanks! I see it now. | |
| Sep 25, 2020 at 18:56 | comment | added | Algernon | ... and this argument works in general. | |
| Sep 25, 2020 at 18:54 | comment | added | Algernon | Trying this in the Euclidean plane would help. You have a closed rectangle $R=[a,b]\times[c,d]$ which is covered by finitely many open rectangles $S_1, S_2, ..., S_n$. How would you find an open rectangle $T$ which includes $R$ and is included in $S:=\bigcup_{i=1}^n S_i$? For each $x\in[a,b]$, the set $V_x:=\{y: (x,y)\in S\}$ is open. Furthermore, varying $x$ over $[a,b]$, there are only finitely many distinct sets $V_x$. Hence, $V:=\bigcap_{x\in[a,b]} V_x$ is open. Similarly, you can construct $U$. | |
| Sep 25, 2020 at 18:32 | comment | added | tgeng | Thanks! I am actually stuck right here. Let $I$ be the finite set indexing the finite rectangles $U_i \times V_i$. But how does one get the cover $U \times V$ from knowing $A \times B \subseteq \bigcup_{i \in I} U_i \times V_i \subseteq W$? | |
| Sep 25, 2020 at 18:23 | comment | added | Algernon | Start by observing that $W$ is a union of (possibly infinitely many) open rectangles $U_i\times V_i$. Use compactness to reduce the problem to the simpler case in which $W$ is a union of finitely many open rectangles. | |
| Sep 25, 2020 at 17:35 | review | Close votes | |||
| Sep 26, 2020 at 11:55 | |||||
| Sep 25, 2020 at 17:04 | history | edited | tgeng | CC BY-SA 4.0 |
added 1 character in body
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| Sep 25, 2020 at 16:52 | history | asked | tgeng | CC BY-SA 4.0 |