If x_k→x in ℓ¹ and Tx_k→y in ℓ¹, coordinate convergence gives y(n)=nx(n). Summability of y implies x∈X and Tx=y, proving a closed graph. Yet ‖e_n‖₁=1 and ‖Te_n‖₁=n, so T is unbounded. Although T is bijective, its inverse cannot be open, since that would make T continuous. **A**.
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