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Let $X$ be the underlying space of an irreducible scheme. In particular, $X$ is non-empty.

Note that the closure of any point is a closed irreducible subset. We say that a point has codimension $n$ if its closure has finite codimension equal to $n$. So, the unique point of codimension 0 is the generic point.

Given a non-negative integer $n$, is it possible that there is a closed point in the closure of any point of codimension$\leq n$ but there is a point of codimension $n+1$ that has no closed points in its closure?

Note that if there is such an $X$, then $X$ is neither quasi-compact nor of finite Krull dimension (because then there is a closed point in the closure of any point).

P.S. a sub-question asked whether the fact that every point has a closed point in its closure implies the quasi-compactness of $X$. The answer is "no", as shown in the answer by A. Mathers.

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  • $\begingroup$ For $n>0$ I think the question is somewhat trivial - a closure of a point $x$ is also a closure of a point $x$ in the closure of the point $x$, and closure of a point $y$ in a closure of a point $x$ is the closure of a point $x$, which would make $P(n)$ equivalent to any other $P(m)$ for $n,m\geq 1$. Are you sure you have stated the condition correctly? $\endgroup$
    – Wojowu
    Commented Apr 17, 2019 at 18:41
  • $\begingroup$ @Wojowu could you clarify "and the closure of a point $y$..."? I take $x$ to be the generic point. The closure is the whole scheme. Then you say that the closure of any point in an irreducible scheme is the whole scheme? $\endgroup$
    – user137767
    Commented Apr 17, 2019 at 18:45
  • $\begingroup$ I mixed up $x,y$ in there, so let me clarify. Correct me if I am misunderstanding the question, but this is how I read it. $P(2)$ says that for any point $x\in X$ and for any $y\in\overline{\{x\}}$, there is a closed point in $\overline{\{y\}}$. But if we assume $P(1)$, then there is a closed point in $\overline{\{y\}}$. So $P(1)$ should imply $P(2)$ this way. $\endgroup$
    – Wojowu
    Commented Apr 17, 2019 at 19:29
  • $\begingroup$ @Wojowu yeah, you are totally right. I probably want to say something different but actually expressing it rigorously is not so easy. I will edit it out, for the time being. Thank you! $\endgroup$
    – user137767
    Commented Apr 17, 2019 at 19:30

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