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It's the same question in Show that in a quasi-compact scheme every point has a closed point in its closure.

Show that in a quasi-compact scheme $X$, every point $p$ has a closed point in its closure

So here I want to ask for a solution verification. At first I did it wrong, without using the quasicompact properties. And after hours of trying I got a proof via Zorn's lemma, as following.

Proof: Suppose $p \in X$, define $\Sigma = \{Z: \text{nonempty, closed, and } \subseteq \overline{\{q\}} \}$. Then $\Sigma$ is nonempty for $\overline{\{q\}} \in \Sigma$. Define a partial order on $\Sigma$ by the subset containing relationship, i.e., $a \leq b$ if and only if $a \subseteq b$.

Then every descending chain chain has a lower bound: Suppose $\cdots \leq Z_n \leq \cdots \leq Z_1$, i.e, $\cdots \subseteq Z_n \subseteq \cdots \subseteq Z_1$. Then $\cap Z_n$ cannot be empty. Otherwise we have $X=X-\cap Z_n = \cup (X-Z_n)$. Then by $X$ being quasicompact, some subcover of $\{Z_n\}$ covers $X$. WLOG suppose it's $\{X-Z_1, \cdots, X-Z_n\}$. Then $X = \cup_{i=1}^{n} (X-Z_i) = X - Z_n$. Hence $Z_n$ is empty. It contradicts with the assumption that $Z_n$ is not empty. Hence $\cap Z_n$ is not empty and hence a lower bound for the descending chain.

Now every descending chain has a lower bound. By Zorn's lemma, $\Sigma$ has some minimal element $m$. We prove that $m$ is a single point set $\{q\}$ and hence $q$ is a closed point contained in $\overline{\{p\}}$:

Since $\overline{\{q\}} \subseteq m$. By $m$ being minimal, we have $\overline{\{q\}} = m$. If there is some other point $q' \in m$, for the same argument we have $\overline{\{q'\}} = m$. Then by Vakil's FOAG, 5.1.B Exercise, there is a bijection between irreducible closed subsets and points for scheme. We have $q = q'$. $\square$

Is this proof correct? In general I got the idea from the proof for affine scheme. It differs with the linked answer and uses the result of another Exercise(cf Exercise 5.1.B in Vakil's FOAG). I am afraid if I still make something wrong. more related context can be found in Vakil's FOAG August 29, 2022 Edition, Page 155. Thank you very much.

onRiv
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    Yeah, it works! (Modulo some confusion concerning the names of the points $p$ and $q$.) Alternatively, you can use the points in the closure of the point ordered by the specialization order. Notice that a point is closed iff it is minimal for the specialization order, so Zorn's lemma gives you the conclusion directly if you can show the hypothesis is satisfied. – Dabouliplop Nov 14 '22 at 20:03
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    (Just something that I'm not sure is clear from the way you wrote the proof: the chain has transfinite length in general, not countable.) – Dabouliplop Nov 14 '22 at 20:07
  • @Dabouliplop Ah here I did mean a countable chain. I have never noticed that the chain in the hypothesis of Zorn’s lemma can be transfinite. Luckily the proof seems requiring no change in the general case. The intersection still works. – onRiv Nov 15 '22 at 01:03
  • @Dabouliplop Thank you for pointing out via the specialization order. Taking any point from the intersection of the closures gives a lower bound. It’s much direct. And it seems even not requiring the space is a scheme, only requiring the quasicompact property? That’s saying, all points in a quasicompact (compact in other context) topological space has a closed point in its closure? – onRiv Nov 15 '22 at 01:12
  • I got an reference from https://stacks.math.columbia.edu/tag/005E for this. It stills requiring $X$ is $T_0$/Kolmogorov, which is holded by schemes. My last comment mistook the specialization order as a partial order, which in fact is just a preorder in general. – onRiv Nov 15 '22 at 04:03
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    Yeah, it's important that the length can be uncountable. Otherwise the first uncountable ordinal would be a counter-example to Zorn's lemma. You're welcome! – Dabouliplop Nov 15 '22 at 07:22

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