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I am reading through Vakil and got to exercise 11.1.B. This exercise states that a scheme $X$ has dimension $n$ iff every element of an open cover has dimension at most $n$ and equality is achieved for an element of the cover. This exercise is trivial in the case $n$ is finite. Furthermore, the forward implication is also trivial if you remove the requirement that equality be achieved on some element of the cover. Both this and this seem to assume $n$ is finite for the forward implication.

The case of $n=-\infty$ is of course trivial, but I am not sure whether this is true or false for $n=\infty$. The proof is again trivial if you can find one infinite descending chain of irreducible closed subsets whose limit is nonempty or if you can find an infinite increasing chain, but this is not true in general (take for example any infinite dimensional Noetherian scheme). It is also trivial for a quasicompact scheme. My question is, does there exist a scheme such that the dimension is infinite, but this is not achieved on any element of some open affine cover?

EDIT: Further thought reveals that instead of looking for any element of some open cover, I instead want any element of any open cover. This is the correct negation of the right hand side of the theorem.

Josiah
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    $\coprod_{n=1}^\infty \Bbb A^n$, with the obvious open cover. – KReiser Feb 17 '22 at 19:09
  • Like this example, are there any examples with an infinite chain of irreducible closed subsets rather than just arbitrarily long? – Josiah Feb 17 '22 at 19:13
  • Also, the proof for finite $n$ shows the result for all affine open covers, can we find a counterexample where this fails for all affine open covers? – Josiah Feb 17 '22 at 19:23

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