Countable basis for function space Show that the space of functions $f:\Bbb{N}\to\Bbb{R}$ does not have a countable basis.
If the domain were a finite set instead of $\Bbb{N}$, then the set of functions that takes the value $1$ at a single point and vanishes elsewhere forms a basis. Now that the domain is countably infinite we get countable infinitely many functions $f_n, n\in\Bbb{N},$ defined by $f_n(m)=\delta_{mn}$. These are clearly linearly independent, but don't span the entire space. For example, the constant function $f(n)=1$ is not in their span. 
But, we can extend that set of functions to a bigger countable collection.
Can anyone give some hint about how to start the problem?
 A: $\def\R{\Bbb R}
\def\N{\Bbb N}
\def\F{\mathcal F}
\def\B{\mathcal B}$
Suppose $\B=\{b_n:n\in\N\}$ were a countable basis for the space $\mathcal F$ of all functions $f:\N\to\R$. That is, each $b_n\in\F$ and each $f\in\F$ is a linear combination with real coefficients of finitely many functions in $\B$. For convenience, assume $0\notin\N=\{1,2,3,...\}$. 
For each $n$ let $M_n=n\cdot\sum_{k=1}^n|b_k(n)|$. 
Define $g\in\F$ by $g(n)=M_n+1$. (This is what I refer to as "diagonalize", in my comment.) We will show that 
$g\notin span(\B)$. 
Suppose $g=a_1b_{n_1}+\cdots+a_jb_{n_j}$ for some $j\in\N$, some $a_1,...,a_j\in\R$, and some $n_1<\dots<n_j$. 
Let $A=\max\{|a_1|,\dots,|a_j|\}$. 
Take $n>\max\{n_j,A\}$, and so 
then $g=a_1b_{n_1}+\cdots+a_jb_{n_j}+0b_n$. 
But then 
$|g(n)|\le|a_1b_{n_1}(n)|+\cdots+|a_jb_{n_j}(n)|+|0b_n(n)|\le$
$A\cdot\bigl(|b_{n_1}(n)|+\cdots+|b_{n_j}(n)|+|b_n(n)|\bigr)\le$
$A\cdot\sum_{k=1}^{n}|b_k(n)|\le$
$n\cdot\sum_{k=1}^n|b_k(n)|=M_n<M_n+1=g(n)$, a contradiction. 
A: Vandermonde determinants to the rescue.
For each $x\in\Bbb{R}$ define the function $f_x:\Bbb{N}\to\Bbb{R}$ by the rule $f_x(n)=x^n$. 
Claim. The set $\{f_x\mid x\in\Bbb{R}\}$ is linearly independent.
Proof. Assuming contrariwise that some non-trivial linear combination of $f_{x_i}, i=1,2,\ldots,n$, $x_i\neq x_j$ whenever $i\neq j$, is the constant function zero, i.e. for all $k\in\Bbb{N}$
$$\sum_{i=1}^nc_if_{x_i}(k)=0.\qquad(*)$$ 
The identity $(*)$ must hold in particular, when $k=0,1,2,\ldots,n-1$. Then the resulting linear system of equations on the unknowns $c_i$ should also have a non-trivial solution. But, the matrix of the system is a Vandermonde matrix with distinct columns, so its determinant is non-zero. Implying that all zeros is the only solution. A contradiction.
Given that we displayed an explicit uncountable linearly independent collection, no countably infinite basis is possible.
