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!
Is there any example of a sequentially-closed convex cone which is not closed?
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functional-analysis
weak-convergence
topological-vector-spaces
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0Are you okay with my answer? – 2017-02-12
1 Answers
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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.