In the proof of noetherian space implies quasicompact I am reading, it goes as follows: Let $X$ be a noetherian space.
Let $U$ be the collection of open subsets of $X$ that can be expressed as a finite union of opens sets in the covering. If $U$ does not contain $X$, then there exists an infinite ascending chain of sets in $U$ (axiom of dependent choice), hence contradiction.
I know it is similar to Why is axiom of choice needed? (Equivalent conditions for Noetherian) , but I am just not seeing why the axiom of dependent choice is necessary here... I would greatly appreciate some explanation. Thank you!
Ps by Noethrian space, I mean Noetheiran topological space: Every descending chain of closed subsets of $X$ eventually becomes constant.
The issue is that without Dependent Choice stating that every chain is finite does not imply the existence of a maximal element.
It is a weird situation, because it means that every finite chain can be extended. However if you have an infinite sequence given, then you do know it is eventually stable.
The proof relies on being able to choose some arbitrary closed sets which is smaller, and that the induction must go through to define an infinite sequence, rather than just arbitrarily long finite sequences. And this is where the axiom of choice (or rather, dependent choice) is used.