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Jan 30, 2013 at 12:07 comment added Martin It seems easier to argue that $X^\ast$ has uncountable dimension as a vector space. Every weak*-neighborhood contains a linear subspace of finite codimension, so the intersection of countably many $0$-neighborhoods contains a subspace of countable codimension, in particular it can't be reduced to $\{0\}$.
Jan 30, 2013 at 11:32 history answered Jochen Wengenroth CC BY-SA 3.0