Talk:Noetherian Topological Space is Compact/Proof 2

What is confusing?
''There should be a link somewhere to the proof that $\bigcup A$, is indeed maximal, I think it's in some of the basic set theory I was hacking through earlier this year ''

Why considering $\bigcup A$? I was only saying that $A$ has a maximal element, since $A$ is a non-empty set of open sets. This is exactly .--Usagiop (talk) 16:45, 29 August 2022 (UTC)


 * All it says is that a set of open sets has a maximal element, it doesn't say that $\set {U_1, \ldots, U_n}$ *is* that maximal element.


 * You misunderstood some definitions. Please point out where the confusion starts:
 * $\paren 1$: $A$ consists of all finite unions of elements from $\CC$
 * $\paren 2$: Since such unions are open, $A$ is indeed a set of open sets.
 * $\paren 3$: As $\CC$ is a cover of non-empty $X$, there is at least a $W \in \CC$.
 * $\paren 4$: As $W \in A$, $A \ne \O$.
 * $\paren 5$: says $A$ has a maximal element.
 * $\paren 6$: Let $\alpha$ denote this maximal element.
 * $\paren 7$: As $\alpha \in A$, $\alpha$ can be written as $\alpha = U_1 \cup \cdots \cup U_n$. ($U_1,\ldots,U_n \in \CC$ are introduced here)
 * $\paren 8$: In fact, we can prove $\alpha = X$.
 * $\paren 9$: That means, $\set {U_1,\ldots,U_n}$ is a finite over of $X$.

--Usagiop (talk) 19:14, 29 August 2022 (UTC)


 * Can this be explained in the body of the proof? --prime mover (talk) 19:23, 29 August 2022 (UTC)


 * I have now a better idea. I improve the proof again. Please wait a bit. --Usagiop (talk) 19:30, 29 August 2022 (UTC)


 * Is this now less confusing?--Usagiop (talk) 20:12, 29 August 2022 (UTC)

''I don't understand why $\set {U_1, \ldots, U_n}$ should be a cover for $\CC$. How do we know it will be? All we know is that $\bigcup U_i$ is maximal in $A$. Why does an arbitrary subset of a topology necessarily going to be a cover? All we know is that $\set {U_1, \ldots, U_n}$ is a cover for $A$. What have I missed? ''

$\set {U_1, \ldots, U_n}$ is a cover of $X$. It means simply that:
 * $X = U_1 \cup \cdots \cup U_n$

There is nothing deep.--Usagiop (talk) 16:45, 29 August 2022 (UTC)


 * Too deep for me. Where does it say that $\set {U_1, \ldots, U_n}$ is a cover of $X$? --prime mover (talk) 17:59, 29 August 2022 (UTC)


 * I don't understand why $A$ has to cover $X$ in the first place. --prime mover (talk) 19:23, 29 August 2022 (UTC)


 * Why do you think that $A$ has to cover $X$?--Usagiop (talk) 19:30, 29 August 2022 (UTC)


 * Because if $A$ does not cover $X$ then why should $U_1 \cup \cdots \cup U_n \cup V \in A$? All we know is that $\set {U_1, \ldots, U_n}$ covers $A$. And $\set {U_1, \ldots, U_n, V}$ covers more than $A$ but does not necessarily cover $\CC$. --prime mover (talk) 19:46, 29 August 2022 (UTC)


 * All covers considered in this proof are covers of $X$. I have never been thinking about covers of $\CC$ nor $A$. $\CC$ is an arbitrary cover of $X$, just the same $\CC$ used in Definition:Open Cover. On the other hand, $A$ is a set introduced by me only for a technical reason, i.e. to apply the Noetherian property to this set.--Usagiop (talk) 19:59, 29 August 2022 (UTC)