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Aug 5, 2020 at 20:42 vote accept CommunityBot
Aug 3, 2020 at 13:52 vote accept CommunityBot
Aug 5, 2020 at 20:42
Jul 5, 2020 at 21:12 vote accept CommunityBot
Aug 3, 2020 at 13:51
Jul 5, 2020 at 17:00 history edited Chris CC BY-SA 4.0
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Jul 5, 2020 at 16:37 comment added Chris Oops, you're right, I missed that $Y$ is allowed to be locally closed. I will think about it.
Jul 5, 2020 at 16:04 comment added user158636 actually why can't there be a Zariski open $Y\subset X$ such that $I=Y\cap S$? I suppose you could run the exact same argument for discontinuous functions but I feel like this should be explicitly mentioned.
Jul 5, 2020 at 16:02 vote accept CommunityBot
Jul 5, 2020 at 16:02
Jul 5, 2020 at 11:04 vote accept CommunityBot
Jul 5, 2020 at 16:01
Jul 5, 2020 at 10:54 history answered Chris CC BY-SA 4.0