# A closed point in the closure of any point in the closure of any point of an irreducible scheme

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.

• 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? – Wojowu Apr 17 at 18:41
• @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? – user137767 Apr 17 at 18:45
• 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. – Wojowu Apr 17 at 19:29
• @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! – user137767 Apr 17 at 19:30

I'll recap the example anyways: take an infinite number of affine lines $$\Bbb A^1_k$$ and glue them along the open subschemes $$\Bbb G_m$$, to get a "line with infinite origins". It has dimension 1 because it is covered by copies of $$\Bbb A^1_k$$, it is irreducible because it has a cover by irreducible affine opens which have pairwise non-empty intersection, and it is not quasicompact because any finite refinement of the open cover by copies of $$\Bbb A^1_k$$ will be missing infinitely many of the "origins".