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Feb 12, 2021 at 15:30 comment added D_S Oh yeah I guess so. So every subset of $V(k)$ is also discrete and closed in $V(\mathbb A)$
Feb 12, 2021 at 8:48 comment added Laurent Moret-Bailly Isn't it just that this is a subset of $V(k)$, and $V(k)$ is discrete and closed in $V(\mathbb{A})$?
Feb 12, 2021 at 4:44 history edited YCor
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Feb 12, 2021 at 2:43 history edited D_S CC BY-SA 4.0
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Feb 12, 2021 at 2:38 history asked D_S CC BY-SA 4.0