Let be the real vector space of all functions from the natural numbers to the reals. Then has uncountable dimension. To see this, for each , let be the function such that .

Claim: The are linearly independent.

Proof: It suffices to show that any finite collection with a strictly increasing sequence of are linearly independent.

We prove this by induction on , the fact being clear for . Suppose with . The equation means

for all . Thus, for all . Now let . This shows that . Thus, by induction, all .

We have displayed an uncountable linearly independent collection of functions in . Now, let be a basis for . Each can be written as a linear combination of for some finite set of indices, where we choose the set to be minimal with this property. Since the linear span of any finite set of the is finite-dimensional, for any finite subset , there are at most finitely many such that . That is, the map is a finite-to-one map from the positive reals to the finite subsets of . Hence, the set of finite subsets of must be uncountable. But then itself must be uncountable. (I leave it as an exercise to show that the set of finite subsets of a countable set is itself countable. You should really write out the proof if you’ve never done it before.)

I might point out that before the tutorials, I was a bit confused myself. That is, the first bit about the ‘s being an uncountable linearly independent set is rather easy. However, I started hesitating: Still, why can’t there be a countable set of elements in terms of which we can express all of them? After all, the set of coefficients we can use for the expressions is uncountable… So think through again clearly: how is this resolved above?

As a final remark, note that this proves that is not isomorphic to . This is perhaps the first example you’ve seen where you can prove that two vector spaces of *infinite* dimensions are not isomorphic by simply counting the dimensions and comparing them.