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Let $X$ be an irreducible finite dimensional Jacobson scheme (i.e. the closed points lie dense and the underlying topological space is sober). If one chooses for every closed point $x \in |X|$ an open neighbourhood $U_x$, is the union $\bigcup_{x\in|X|}U_x = X$ necessarily the whole space?

This is e.g. true if $\dim{X} = 1$ since in this case the only non-closed point is the generic point, which lies in every non-empty open subset.

I am already interested in the case $\dim{X} = 2$ if this simplifies things. For this, one only has to show that for $x$ of codimension $1$ (i.e. $\dim\overline{\{x\}} = 1$), there is a closed point $y$ such that $x \in U_y$. Isn't this automatic by the same argument as with the generic point and $\dim{X} = 1$?

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To translate this for "normal topologist": $X$ is just a space, $|X|$ is the set of $x$ such that $\{x\}$ is closed, and we just assume that $|X|$ is dense in $X$? No hidden assumptions in the words "scheme" etc.? –  Henno Brandsma May 3 at 11:49
    
"Scheme" implies that the topological space is sober, but I think this is the only assumption. –  user5262 May 3 at 14:20
    
what does sober translate to topologically? Just that irreducible closed sets are closures of singletons? –  Henno Brandsma May 3 at 14:27
    
Yes, see also en.wikipedia.org/wiki/Sober_space –  user5262 May 3 at 14:31

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