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It is commonly stated that one cannot write down explicitly (i.e. constructively) a basis for the vector space of all real sequences, $\mathbb{R}^\mathbb{N}$. But what about the following: for $k \geq 1$, let $e^{(k)}$ be the sequence where all terms are $0$ except the one in position $k$, which is $1$. Then it seems obvious that the set $\{e^{(k)}: k \geq 1\}$ is linearly independent and spanning, so it should be a basis. What is wrong with this argument?

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The set is linearly independent, but it does not span $\mathbf{R}^{\mathbf{N}}$. Any finite linear combination of the vectors in the set has all but finitely many of its terms equal to $0$.

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The non-zero constant sequences are not in the span of your "basis".

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