When first asked if $V= \mathbb{R}^{\mathbb{N}}$ has a countable basis as a vector space over $\mathbb{R}$ almost everyone will think for a second: “of course! (1,0,0,…),(0,1,0,…),… will do it”. Then most people realise this is total rubbish because (1,1,….) isn’t a finite linear combination of these. This then usually raises the suspicion that any basis must be uncountable.

A quick way to check this is just to consider the set of sequences of form $a_n = e^{\alpha \log n}$ where $\alpha$ is any positive real. This is clearly an uncountable linearly independent set.

However, actually constructing a basis seems to be impossible (though the Zorn argument I posted on this blog about 6 months ago shows it has to exist). If anyone knows how to do it, I’d love to hear it.

EDIT: It has been pointed out to me that it’s not entirely obvious that finding an uncountable linearly independent set constitutes a proof that there is no countable basis. I have assumed the dimension theorem, that with every vector space $V$ there is associated a fixed cardinal $\text{dim}V$ which gives the cardinality of any basis. Equivalently, any two bases for $V$ have the same cardinality, and any set of greater cardinality therefore has linear dependencies (since it cannot be extended to a basis). To prove the dimension theorem, we take two bases of different sizes, write out each element of the smaller one in terms of the larger one, and because we only use a finite subset of the elements from the large basis to write out each element of the smaller one, the union of the elements used in this process can’t possibly be equal to the entire large basis, giving the necessary contradiction.