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Nov 6, 2019 at 5:24 comment added abx Yes, but you don't even need to repeat: just take the intersection of $(X\cap\Lambda)^{[n]}\subset X^{[n]}$ with the previous open subset.
Nov 5, 2019 at 21:09 comment added modnar If we want $Z \in X^{[n]}$ to determine some fixed linear subspace $\Lambda$ of dimension $n - 1$, is it enough to repeat the steps above with $X \cap \Lambda$ replacing $X$ to find the subset of elements of $X^{[n]}$ determining $\Lambda$?
Nov 5, 2019 at 16:54 comment added abx Yes. More precisely perhaps, imposing any upper bound on that dimension gives an open condition.
Nov 5, 2019 at 15:21 comment added modnar Just to clarify, does this mean that specifying some arbitrary dimension other than $m - n + 1$ gives a locally closed condition?
Nov 5, 2019 at 8:30 history answered abx CC BY-SA 4.0