# Why is axiom of dependent choice necesary here? (noetherian space implies quasicompact)

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.

• Since the definition of Noetherian splits without dependent choice, you'll have to tell us what is your definition for Noetherian. – Asaf Karagila Jul 12 '16 at 14:43
• @Johnny T. I don't understand how you are using the axiom of choice – Armando j18eos Jul 12 '16 at 15:02
• I don't understand how the axiom of dependent choice is being used here either... – Johnny T. Jul 12 '16 at 15:06
• If you'll tell me how you define Noetherian, I might be able to tell you how you're using the axiom of choice. – Asaf Karagila Jul 12 '16 at 15:08
• I defined it. Thank you! – Johnny T. Jul 12 '16 at 15:09

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.

• Let's try it your way. How would you do that? – Asaf Karagila Jul 14 '16 at 16:09
• But is the choice of $U_2$ canonical? Or for that matter, is any choice that you made is canonical here? Can you describe the process without using the word "pick" (which is the same as the word "choose")? – Asaf Karagila Jul 15 '16 at 4:51
• Because induction only tells you "if you could have gone all the way up to $N$, then you can take one more step". There is nothing, anywhere, when you construct something by induction that tells you that you can take infinitely many steps. For example, you can prove that if $N$ is finite, then $N+1$ is also finite, and that's great; but that doesn't mean that you can prove that $\Bbb N$ itself is finite. The axiom of choice allows you to make your choices uniformly by saying "fix a choice function on such and such" and then when you say "proceed from $N$ to $N+1$" you are appealing to that ... – Asaf Karagila Jul 15 '16 at 19:33
• ... choice function for saying how you are proceeding. This reduces all your seemingly arbitrary choices into one arbitrary choice: the choice function with which you are making all of them. And this is what allows you to condense all your choices into a single sequence. And this is exactly what the principle of dependent choice is saying: if you could define something by induction for every $n$, then you can find an actual sequence which gives you a coherent choices for the inductive definition. – Asaf Karagila Jul 15 '16 at 19:35
• Yes, that is correct. If you want to really get confused, Dependent Choice is equivalent over ZF to the statement "If T is a tree where every node has finitely many predecessor, and at least one successor, then T has an infinite branch". So if Dependent Choice fails, there is a tree without maximal elements and without maximal chains. It can get worse, and you can get a full binary tree (every node has exactly two successors) without any branches---although this requires more than just the failure of DC. – Asaf Karagila Jul 15 '16 at 19:48