# Basis for vector space of real sequences

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?

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 $$0$$.