Can we prove that a quasi-compact locally noetherian space is noetherian without Axiom of Choice?

Question

I will state some definitions to clarify my question.

Definition 1 Let $X$ be a topological space. If every open cover of $X$ has a finite subcover, $X$ is called quasi-compact.

Definition 2 Let $X$ be a topological space. If every non-empty set of open subsets of $X$ has a maximal element, $X$ is called noetherian.

Definition 3 Let $X$ be a topological space. If every point of $X$ has an open neighborhood which is noetherian, $X$ is called locally noetherian.

My question Can we prove that a quasi-compact locally noetherian space is noetherian without Axiom of Choice?

Remark I'm particularly interested in the following question. Let $k$ be a field. Let $X$ be a scheme of finite type over $k$. Can we prove that the underlying topological space of $X$ is noetherian without Axiom of Choice?

If the title questionj is affirmative, the question is also affirmative. See the David Speyer's answer to this question.

As for why I think this question is interesting, please see(particularly Pete Clark's answer): Why worry about the axiom of choice?

@TylerLawson You need the axiom of dependent choice to prove that the ascending chain condition implies the existence of a maximal element. — Makoto Kato, Mar 15 '14 at 02:33

score 1 · Answer 1 · answered Mar 15 '14 at 11:31

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Yes, we can.

Consider $\mathcal U$ as the open cover whose elements are all the open neighborhoods witnessing the local noetherian property. Since every $x\in X$ has such neighborhood, this is certainly an open cover. By quasi-compactness, we have some finite subcover, $V_1,\ldots,V_n$.

Now let $\mathcal V$ be any non-empty family of open sets. Write $\mathcal V_i=\{U\cap V_i\mid U\in\mathcal V\}$. Then there is at least one $i$ such that $\mathcal V_i$ is not empty, and by the noetherian property of $V_i$ there is a maximal open set in $\mathcal V_i$. Of course, there might be more than just $j$ such that $\mathcal V_j$ is non-empty.

For each $U\in\mathcal V$ write $m(U)=|\{i\mid U\cap V_i\text{ is maximal}\}|$, that is $m(U)$ is the number of $i$'s in which $U\cap V_i$ is maximal in $\mathcal V_i$. Since the values of $m(U)$ are between $0$ and $n$ there is some maximal value $m$, and it has to be at least $1$.

Finally, consider $\{U\in\mathcal V\mid m(U)=m\}$. Somewhere in this set we have a maximal open set in $\mathcal V$. Pick some $U_0$, if it is maximal then we are done. Otherwise, consider $\mathcal U_0=\{V\in\mathcal V\mid U_0\subseteq V\}$, and note that for each such $V$ we have $m(V)=m(U_0)=m$, and in particular where $U_0\cap V_i$ is maximal, $V\cap V_i=U_0\cap V_i$.

Let $i$ be the least index such that $U\cap V_i$ is not maximal, and there is some $V$ such that $U\cap V_i\subsetneq V\cap V_i$ (such index exists, otherwise $U$ is maximal), now consider $\{V\cap V_i\mid V\in\mathcal U_0\}$ and let $\mathcal U_1$ be the set of $V$ such that $V\cap V_i$ is maximal. Pick some $U_1\in\mathcal U_1$, if it is maximal, then we are done.

Otherwise we proceed to repeat the process on the least index $i$ such that for some $V\in\mathcal U_1$ we have $U_1\cap V_i\subsetneq V\cap V_i$. We observe that $U_0\subseteq U_1$, and the induction process guarantees that we have an increasing chain of open sets.

However the induction can only proceed $n-m$ steps, so it must halt and we must have a maximal element.

answered Mar 15 '14 at 11:31

Asaf Karagila

405,794

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Dear Asaf Karagila, [Then there is at least one $i$ such that $\mathcal V_i$ is not empty,] $\mathcal V_i$ is not empty for every $i$. Regards, – Makoto Kato Mar 15 '14 at 22:01
Makoto, it might be empty for some $i$. It depends on $\mathcal V$. – Asaf Karagila Mar 15 '14 at 22:02
No. It never be empty. At least it might contain $\phi$. Regards, – Makoto Kato Mar 15 '14 at 22:21
Ah, yes. That is correct. It doesn't matter for the proof, anyway. – Asaf Karagila Mar 15 '14 at 22:28
I would appreciate if you would kindly edit the above statement I quoted. Regards, – Makoto Kato Mar 15 '14 at 23:18
Why? Is my statement wrong? Does it make the proof wrong? No, and no to both questions. – Asaf Karagila Mar 15 '14 at 23:20
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It's misleading. You also wrote [Of course, there might be more than just $j$ such that $\mathcal{V}_j$ is non-empty.] – Makoto Kato Mar 15 '14 at 23:27
You seem to have cleared that up in the comments above, so it's not misleading anymore. Problem solved. – Asaf Karagila Mar 15 '14 at 23:29
This is a public forum. – Makoto Kato Mar 15 '14 at 23:32
Yes, and the comments are visible to everyone. – Asaf Karagila Mar 15 '14 at 23:37
Dear Asaf Karagila, not everybody reads all the comments. – Makoto Kato Mar 16 '14 at 00:27
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True. I don't expect people to read all these nonsensical comments, but the first three (especially when you point out the issue in the first comment) are probably going to be read. I don't understand why you keep on posting comments. I find your insistence strange, borderline rude and quite frankly, annoying. – Asaf Karagila Mar 16 '14 at 00:38
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Dear Asaf Karagila, I'm just politely asking you to edit the misleading statement. Not everybody reads the comments at all. Regards, – Makoto Kato Mar 16 '14 at 02:22
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Well, your politeness is not coming across as politeness. I don't see the need to edit my answer and that is not going to change. – Asaf Karagila Mar 16 '14 at 07:29
Dear Asaf Karagila, [Let $i$ be the least index such that $U\cap V_i$ is not maximal,] Could you explain what you mean by $U$? Regards, – Makoto Kato Mar 16 '14 at 11:20

score 0 · Answer 2 · answered Mar 19 '14 at 08:44

Since $X$ is quasi-compact and locally noetherian, there exists a finite open cover $V_1,\ldots,V_n$ each of which is noetherian. Let $\mathcal V$ be any non-empy family of open sets. Write $\mathcal V_i=\{U\cap V_i\mid U\in\mathcal V\}$.

We define inductively an element $U_i \in \mathcal V_i$ and a non-empty subset $\mathcal U_i$ of $\mathcal V$ for each $i$. Since $V_1$ is noetherian, there exists a maximal element $U_1$ of $\mathcal V_1$. Let $\mathcal U_1 = \{U \in \mathcal V\mid U\cap V_1 = U_1\}$. Let $i$ be an integer such that $1 \lt i \le n$. Suppose $U_j$ and $\mathcal U_j$ are defined for all $j$ such that $1 \le j \lt i$ Let $U_i$ be a maximal element of $\{U\cap V_i\mid U\in\mathcal U_{i-1}\}$. Let $\mathcal U_i = \{U \in \mathcal U_{i-1}\mid U\cap V_i = U_i\}$. Notice that $\mathcal U_i$ is not empty since $V_i$ is noetherian.

Now that $U_i$ and $\mathcal U_i$ are defined for all $i(1 \le i \le n)$, it suffices to prove that any element $W$ of $\mathcal U_n$ is a maximal element of $\mathcal V$. Let $V$ be an element of $\mathcal V$ such that $W \subset V$. It suffices to prove that $W = V$. We first prove by induction that $V \in \mathcal U_i$ for all $i = 1, \cdots n$. Since $\mathcal U_n \subset \mathcal U_{n-1} \subset \cdots \subset \mathcal U_1 \subset \mathcal V$, $W \in \mathcal U_i$ for all $i = 1, \cdots, n$. Since $W\in \mathcal U_1$, $W\cap V_1 = U_1$. Since $U_1$ is a maximal element of $\mathcal V_1$ and $V\in \mathcal V$, $W\cap V_1 = V\cap V_1 = U_1$. Hence $V\in \mathcal U_1$. Let $i$ be an integer such that $1 \lt i \le n$. Suppose $V \in \mathcal U_j$ for all $j$ such that $1 \le j \lt i$. Since $W \in \mathcal U_i, W\cap V_i = U_i$. Since $U_i$ is a maximal element of $\{U\cap V_i\mid U\in\mathcal U_{i-1}\}$ and $V \in \mathcal U_{i-1}$, $V\cap V_i = U_i$. Hence $V \in \mathcal U_i$

Therefore $W\cap V_i = V \cap V_i = U_i$ for each $i$. Hence $W = V$ as desired. QED

Can we prove that a quasi-compact locally noetherian space is noetherian without Axiom of Choice?

2 Answers2