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The discussion in Convergence in topologies, especially the comments of GEdgar, led me to another (converse) question concerning convergence. In the paper

G. A. Edgar, A long James space, in: Measure Theory, Oberwolfach 1979, Lectures Notes in Math. 794, Springer-Verlag (1980) p. 31-37

he points out that the weak and weak* topologies in $J(\omega_1)^*$ have different Borel $\sigma$-fields. Do they have the same convergent sequences? Heuristically, they should as $J(\omega_1)$ just "differs" with a reflexive space "by one dimension".

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2 Answers 2

Impossible, since this space contains a copy of James space which is separable but not reflexive (every separable Grothendieck space, that is, space with this property must be reflexive).

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Tomek's answer enlarged: Even in $J = J(\omega)$ there is a sequence that converges weak* but not weakly. Explictly: in the notation of my paper cited, sequence $e_n$ converges weak* to $e_\omega$, but not weakly. So if we alternate $e_1, e_\omega, e_2, e_\omega, e_3, e_\omega, \dots$ we get a sequence that converges weak* but not weakly.

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