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Jan 25, 2023 at 17:10 comment added FPV Dear @Satan's Minion, I had considered an argument like this. The only point I was missing is that if $V(f)$ is nowhere dense it cannot contain one of the Shilov boundary points. Thanks for your help
Jan 25, 2023 at 17:09 vote accept FPV
Jan 25, 2023 at 6:15 history answered Satan's Minion CC BY-SA 4.0