TITLE: Is there any example of a sequentially-closed convex cone which is not closed?
QUESTION [5 upvotes]: I am interested in showing that a sequentially-closed convex cone is closed in order to prove a representation theorem for a pre-ordered preference relation. Thank you in advance!

REPLY [3 votes]: Consider the space $(\ell_\infty)^*$ endowed with the weak*-topology. The canonical image of $\ell_1$ in that space is sequentially closed by the Schur property of $\ell_1$, however it is also dense by Goldstine's theorem.