TITLE: Dimension of vector space, countable, uncountable?
QUESTION [6 upvotes]: In set theory, when we talk about the cardinality of a set we have notions of finite, countable and uncountably infinite sets.
Main Question
Let's talk about the dimension of a vector space. In linear algebra I have heard that vector spaces are either of finite dimension (for example $\mathbb{R}^n$) or infinite dimension (for example $C[0,1]$).
Why don't we have notions of countably infinite dimensional and uncountably infinite dimensional vector spaces?
Maybe, I am missing the bigger picture.
Extras
P.S. A long time ago, I attended a talk given on enumerative algebraic geometry and the professor said, "I always think of a positive natural number as the dimension of some vector space".
Can this idea then be extended to vector spaces of uncountably infinite dimension by considering transfinite numbers as denoting the dimension of some vector space?

REPLY [2 votes]: The following theorem is an example where we need to distinguish between countably/uncountably infinite dimensional vector spaces. Some important theorems, such as Hilbert Nullstallensatz, can be deduced from it.

Let $A$ be an associative, not necessarily commutative, $\mathbb{C}-$algebra with unit. For $a \in A$ define 
  $$\text{Spec } a = \{\lambda \in \mathbb{C} | a-\lambda \text{ is not invertible}\}$$
  Assume that $A$ has no more than countable dimension over $\mathbb{C}$. Then 
(a) If $A$ is a division algebra, then $A=\mathbb{C}$
(b) For all $a \in A$ we have $\text{Spec } A \neq \emptyset$; furthermore, $a \in A$ is nilpotent if and only if $\text{Spec } A = \{0\}$

(Adapted from Representation Theory and Complex Geometry by Chriss and Ginzburg, theorem 2.1.1.)
The proof uses the fact that for any $a \in A$,
$$\{(a - \lambda)^{-1} | \lambda \in \mathbb{C}\}$$
is an uncountable family of elements of $A$. But $A$ has only countable dimension over $\mathbb{C}$, so they are not linearly indenpendent over $\mathbb{C}$.
A weak version of Hilbert Nullstellensatz:

Let $A$ be a finitely generated commutative algebra over $\mathbb{C}$. Then any maximal ideal of $A$ is the kernel of an algebra homomorphism $A \to \mathbb{C}$

This follows directly from the above theorem: $A$ finitely generated $\Longrightarrow$
$A$ has countable dimension over $\mathbb{C}$ $\Longrightarrow$ $A/\mathfrak{m}$ has countable dimension over $\mathbb{C}$ $\Longrightarrow$ $A/\mathfrak{m}=\mathbb{C}$ 
(We can also deduce the strong version of Nullstellensatz from it but need more argument.)