Timeline for A closed point in the closure of any point in the closure of any point of an irreducible scheme

Current License: CC BY-SA 4.0

9 events

when toggle format	what		by	license	comment
S Apr 21, 2019 at 15:06	history	suggested	user74900	CC BY-SA 4.0	clarified where to search for counter-examples
Apr 21, 2019 at 14:38	review	Suggested edits
S Apr 21, 2019 at 15:06
Apr 18, 2019 at 9:33	history	edited	user137767	CC BY-SA 4.0	added 199 characters in body
Apr 17, 2019 at 19:37	history	edited	user137767	CC BY-SA 4.0	deleted 276 characters in body
Apr 17, 2019 at 19:30	comment	added	user137767		@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!
Apr 17, 2019 at 19:29	comment	added	Wojowu		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.
Apr 17, 2019 at 18:45	comment	added	user137767		@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?
Apr 17, 2019 at 18:41	comment	added	Wojowu		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?
Apr 17, 2019 at 17:56	history	asked	user137767	CC BY-SA 4.0