TITLE: Uncountable basis of vector space of infinite sequences in R
QUESTION [5 upvotes]: How does one go about showing that the dimension of a vector space of infinite sequences is uncountable? My methods was to try and show the existence of an uncountable, linearly independent of sequences which implies that the basis must be uncountable. I have tried the set of following sequences i.e. 1) An = n^r where r is any real number 2) set of convergent sequences. However, I cant seem to show that they are uncountably infinite dimensional. Any tips? 
P.S Other solutions are welcomed but i have read other similar posts and do not understand the solutions

REPLY [6 votes]: The set of sequences $\{(1,t,t^2,t^3,\ldots):t\in\mathbb R\}$ is a linearly independent set with the cardinality of $\mathbb R.$  
One way to show it is linearly independent is to show that each element is an eigenvector for the "backward shift," each with distinct eigenvalues, and eigenvectors for distinct eigenvalues of a linear transformation are always linearly independent.
Because the cardinality of real sequences is the same as that of $\mathbb R$, this lower bound is the exact dimension.

REPLY [4 votes]: If I were you, I'd consider the characteristic functions of subsets of $\mathbb{N}$ - that is, sequences of zeroes and ones. Because the basis is only permitted to "cover" other elements using finite addition, you can't reach (for example) a sequence with infinitely many $1$'s using only sequences with finitely many $1$'s.
I think you can probably use a variant of Cantor's diagonalization argument to argue that no countable collection of these characteristic functions spans the rest.