An uncountable linearly independent set 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)
 A: One way of doing this is to use the identities $0$ & $1$ along with the following, rather neat fact:

Claim$\;[\dagger]$ There exists an uncountable collection $X$ of infinite subsets of $\mathbb{N}$ such that the intersection of any two is finite

Such a set provides a natural foundation for an uncountable, linearly independent subset of $\mathbb{F}^{\mathbb{N}}:$
For any $A\in X$ define $\phi_n(A)=1$ if $n\in A$ and $0$ otherwise. Let $\Phi=\big\{\big(\phi_n(A)\big)_n\;\big|\;A\in X\big\}\subseteq \mathbb{F}^{\mathbb{N}}$. To see that this is $\text{l.i.}$ consider some arbitrary finite collection $\{A_1\,\cdots \,A_k\}\subseteq X$. For each $1\leqslant j\leqslant k$ there exists $N_j$ $\text{s.t.}$ $\phi_{N_j}(A_k)=\delta_{jk}$, and hence if $\big(\lambda_1 \phi_n(A_1)+\cdots+\lambda_k \phi_n(A_k)\big)_n=(0)_n$  necessarily $\lambda_{j}=0$; so $\mathbb{F}^{\mathbb{N}}$ cannot have countable dimension.


 $[\dagger]\;\text{Pf.}:$ For any $s\in (0,1)$ define a reading sequence $\big(\alpha_n(s)\big)_n\in \mathbb{N}^{\mathbb{N}}$ where the digits of $\alpha_k(s)$ from left $\to$ right are $s_1\dots s_k$ where $s_i$ is the $i^{\text{th}}$ digit of $s$ after the decimal point. Collect these into $\mathscr{A}(s)=\{ \alpha_n(s)\;|\;n\in\mathbb{N}\}$, then $X=\{\mathscr{A}(s)\;|\; s\in (0,1)\}\subseteq \wp (\mathbb{N})$ and $\text{card}\,X=\text{card}\,\wp (\mathbb{N})$ but for any distinct $x,y\in (0,1)$ we must have $\mathscr{A}(x)\cap \mathscr{A}(y)$ finite.

A: This is not quite elegant, but it works: Split into cases according to the cardinality of $\mathbb F$.
First case: $\mathbb F$ is finite or countably infinite: In that case $S$ has cardinality $|\mathbb F|^{\aleph_0}=2^{\aleph_0}$. However the span of any countable set of vectors from $S$ would be itself countable, and therefore cannot be all of $S$.
We know (assuming the Axiom of Choice) that $S$ has a basis; since we have just argues that a countable set cannot span all of $S$, the basis must be an uncountable linearly independent set, as required.
Second case: $\mathbb F$ is uncountable. For every $\alpha\in \mathbb F$, let $\tilde\alpha$ mean the vector $(1,\alpha,\alpha^2,\alpha^3,\ldots)\in S$. Then the set
$$ \{ \tilde\alpha \mid \alpha \in \mathbb F \} $$
is uncountable (obviously) and linearly independent. To see that it is linearly independent, suppose we have some linear relation
$$ c_1\tilde\alpha_1 + c_2\tilde\alpha_2 + \cdots + c_n\tilde\alpha_n = 0 $$
for some $c_i$ and $\alpha_i$ from $\mathbb F$. This means that for each $k\ge 0$ we have
$$ c_1\tilde\alpha_1^k + c_2\tilde\alpha_2^k + \cdots + c_n\tilde\alpha_n^k = 0 $$
and in particular, looking only at the first $n$ $k$s,
$$ \begin{pmatrix}1 & 1 & \cdots & 1 \\
\alpha_1 & \alpha_2 & \cdots & \alpha_n \\
\alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2 \\
\vdots & \vdots & & \vdots \\
\alpha_1^{n-1} & \alpha_2^{n-1} & \cdots & \alpha_n^{n-1} \end{pmatrix}
\begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} = 0 $$
So if the $c_i$ are not all zero, then the $n\times n$ matrix is singular.
But then there is a nontrivial linear relation between its rows too, and such a relation would be a polynomial of degree at most $n-1$ with all of the $n$ $\alpha_i$s as roots. And that is not possible in a field, so the linear relation we started with must be trivial.
A: 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.
A: I don't know whether that's going to help you, but here's another example of a space with an uncountable linearly independent set. Consider $\mathbb{R}$ as a vector space over the field $\mathbb{Q}$. Choose its Hamel basis (you can choose one by transfinite induction). It should be uncountable, otherwise $\mathbb{R}$ would be countable (since any element of $\mathbb{R}$ corresponds uniquely to a finite set from the basis together with the finite set of the respective rational coordinates). (basically, it seems, what your professor was talking about bears some relation to this...)
