## A vector space of uncountable dimension

Let $V=\{ f: {\bf N}\rightarrow R\}$ be the real vector space of all functions from the natural numbers to the reals. Then $V$ has uncountable dimension. To see this, for each $a>0$, let $f_a$ be the function such that $f_a(n)=a^n$.

Claim: The $f_a$ are linearly independent.

Proof: It suffices to show that any finite collection $\{f_{a_i}\}_{i=1}^n$ with a strictly increasing sequence of $a_i$ are linearly independent.

We prove this by induction on $n$, the fact being clear for $n=1$. Suppose $\sum_{i=1}^nc_if_{a_i}=0$ with $n>1$. The equation means

$\sum_{i=1}^nc_ia_i^m=0$

for all $m$. Thus, $\sum_{i=1}^{n-1}c_i(a_i/a_n)^m+c_n=0$ for all $m$. Now let $m\rightarrow \infty$. This shows that $c_n=0$. Thus, by induction, all $c_i=0$.

We have displayed an uncountable linearly independent collection of functions in $V$. Now, let $\{b_i\}_{i\in I}$ be a basis for $V$. For each $f_a$ there is then a unique finite set $I(a)\subset I$ of indices such that $f_a$ can be written  as a linear combination with non-zero coefficients of $\{b_i\}_{i\in I(a)}$. The linear span of any given finite set of the $b_i$ is finite-dimensional. Hence, for any finite subset $S\subset I$, there are at most finitely many $a$ such that $I(a)=S$. That is, the map $a\mapsto I(a)$ is a finite-to-one map from the positive reals $R_{>0}$ to the finite subsets of $I$. Hence, the set of finite subsets of $I$ must be uncountable. But then $I$ 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 $f_a$‘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 $V$ is not isomorphic to $R[x]$. 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.