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Oct 1, 2020 at 0:33 comment added Black Excellent! Thank you. I guess this also shows that there is no subspace $H$ whose intersection with any closed hyperplane is dense. If there was one, you could put $H$ inside some hyperplane, see it as the kernel of a functional $F$ and then use the same argument.
Oct 1, 2020 at 0:29 vote accept Black
Sep 30, 2020 at 22:54 history answered Bill Johnson CC BY-SA 4.0