TITLE: An uncountable linearly independent set
QUESTION [12 upvotes]: I've been taking a course in linear algebra and one of the first things we defined was linear independence. It made me wonder how big a linearly independent set can be, in particular whether we can find uncountable ones. 
After asking my professor he said I could consider the set $S$ of sequences whose entries lie in some arbitrary field $\mathbb{F}$. $S$ is not linearly independent so I think he is hinting that its basis is uncountable.
Since then I've managed to find an uncountable linearly independent set (and prove that it is so): the set of sequence of the form $x^n,\;0<x<1$. But the problem my professor gave me still escapes me. 
Any ideas how I could show that the set $S$ has uncountable dimension?
(thank you @egreg for your kind answer, it's made the concept clearer. It doesn't really answer my question sadly)

REPLY [4 votes]: Any set can be thought as the basis of a vector space over whatever field $F$ you like.
Consider a set $X$ and the set $F^{(X)}$ consisting of all maps
$$
f\colon X\to F
$$
such that $\operatorname{supp}(f)=\{x\in X:f(x)\ne0\}$ is finite.
Then $F^{(X)}$ is in a very natural way a vector space over $F$: if $f,g\in F^{(X)}$ and $\alpha \in F$, define
$$
f+g\colon x\mapsto f(x)+g(x),\qquad
\alpha f\colon x\mapsto \alpha f(x)
$$
The vector space axioms are easily verified. Define, for $y\in X$,
$$
e_y\colon x\mapsto
\begin{cases}
1 & \text{if $x=y$}\\
0 & \text{if $x\ne y$}
\end{cases}
$$
Then the set $\hat{X}=\{e_x:x\in X\}$ is linearly independent and there's an obvious bijection $X\to\hat{X}$. More precisely, $\hat{X}$ is a basis of $F^{(X)}$.
So you see that there's no limit in the cardinality of linearly independent sets.