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 Definition 4 of Noetherian Topological Space.--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 $W \in A$, $A \ne \O$.

$\paren 5$: Definition 4 of Noetherian Topological Space 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)

Maybe my mistake was that I should have written that $\set {U_1, \ldots, U_n}$ is a finite subcover of $\CC$ for $X$. I improved this point.--Usagiop (talk) 20:19, 29 August 2022 (UTC)

Nearly there.

Just need to explain one of your "trivial" steps.

Incidentally I point you towards Help:Editing/House Style/Linguistic Style#Empty Statements and Waffle where it is pointed out that we prefer not to add things like "This is trivial".

If and when you embark on teaching, you will find there are many cases where things you consider "trivial" are not actually trivial as such, just "too difficult to explain". But you're getting there. --prime mover (talk) 05:20, 30 August 2022 (UTC)