Discussion
Let \(V_{k} = \{(a_{n}) \in l^{2} : a_{n} = 0\) for all \(n \gt k\}\). Then :
Answer: D
\(V_{k}\) is finite dimensional for all \(k\) and \(l^{2}\) is not a sum of \(\{V_{k}\}\).
\(V_{k}\) consists of sequences that vanish after the \(k\)-th term, so it is isomorphic to \(\mathbf{R}^{k}\) (or \(\mathbf{C}^{k}\)) and has dimension \(k\). The algebraic sum of the \(V_{k}\) is their union, the sequences with only finitely many non-zero terms.
That is a proper subspace of \(l^{2}\): \((1, \frac{1}{2}, \frac{1}{3}, \ldots)\) is square summable but has infinitely many non-zero terms. So \(l^{2}\) is not a sum (or direct sum) of the \(V_{k}\), only the closure of their union. A and B fail for that reason, and C because each \(V_{k}\) is finite dimensional.
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