Every uncountable set of vectors in a countable -dimensional space is linearly dependent I'm reading book 'Course in Algebra' and i'm stuck on this problem:
Prove that every uncountable set of vectors in a countable- dimensional space is linearly dependent (hence, every basis of such a space is countable).
Any hints?
 A: Fix a countable basis $B=\{b_i: i\in\mathbb{N}\}$ and suppose $U$ is an uncountable set of vectors.
Every element $u$ of $U$ can be written uniquely as a linar combination of vectors in $B$. Write "$length(u)$" for the largest $i$ appearing in this combination; e.g. $$length(b_1-2b_{17}+b_{18})=18.$$


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*Can you show that for some $k$, there are infinitely many $u\in U$ with $length(u)=k$?

*Now think about the subspace generated by $b_1,...,b_k$. This is $k$-dimensional, so can all the vectors in $U$ with $length=k$ possibly be linearly independent?

EDIT: Above, I've assumed that the definition of "countable dimension" you've been given is "has a countable basis." As Andres points out below, another reasonable definition you could be working with is "has a countable spanning set."
This changes the proof only slightly. We now define $length(u)$ as the least $k$ such that $u$ can be written as a linear combination of the first $k$ vectors in this countable spanning set (since we're not starting with a basis, there might be many such linear combinations). As before, we must have some $k$ which is the length of infinitely many vectors in $U$. The space spanned by these $k$ vectors has dimension at most $k$, so we still get linear dependence as desired.

If you're familiar with some basic set theory, it's a good exercise to generalize this argument to show that dimension is well-defined in general - if $V$ is a vector space with a basis of cardinality $\kappa$, then every basis of $V$ has cardinality $\kappa$, for any cardinal $\kappa$ (finite, countable, uncountable, whatever).
The key to this generalization of the argument above is that for any infinite set $A$, the set of finite subsets of $A$ has the same cardinality as $A$ itself.
