Timeline for Proof that a compact subset of a CW complex is contained in a finite subcomplex
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 15, 2024 at 7:04 | comment | added | Haran | Oh, my bad! Yes, you need compactness to see that each closed cell is contained a union of finitely many (open) cells. Now the weak topology will do. | |
| Jan 15, 2024 at 6:51 | comment | added | nomadicmathematician | Ok I think I need to use (C) condition as well. So for each cell $e$, $A \cap e$ is at most singleton. Since $\bar{e}$ is contained in a union of finitely many cells, so is $\bar{e} - e$. Therefore, $A \cap \bar{e}= (A \cap e) \cup (A \cap \bar{e}-e)$ is at most a union of finite points, hence closed in $\bar{e}$. | |
| Jan 15, 2024 at 6:46 | comment | added | nomadicmathematician | I need $A \cap \bar{e}$ to be closed in $\bar{e}$. | |
| Jan 15, 2024 at 6:35 | comment | added | Haran | Aren't you done from showing that $A \cap e$ is a singleton? You only need to be closed in each cell. | |
| Jan 15, 2024 at 6:33 | comment | added | nomadicmathematician | Yes the definition I have is $A$ is closed iff $A\cap \bar{e}$ is closed in $\bar{e}$ for each (open) cell $e$. but the difficulty I have is well if $A \cap e$ is a singleton for each chosen cell $e$, how do I guarantee that $A \cap \bar{e}$ will still be singleton? | |
| Jan 15, 2024 at 6:15 | history | undeleted | Haran | ||
| Jan 15, 2024 at 6:15 | history | deleted | Haran | via Vote | |
| Jan 15, 2024 at 6:14 | history | answered | Haran | CC BY-SA 4.0 |