Are subspaces always enumerable with an order? I'm not quite sure how to ask this question so bear with me.

I know that $\mathbb{R}^0$ is a subspace of $\mathbb{R}^1$ which is a subspace of $\mathbb{R}^2$ ... etc.

And similarly, there is an isomorphism from $\mathbb{R}^n$ to polynomials of (n-1)th degree.

But now I've just read that the set of all functions from $\mathbb{R}$ to $\mathbb{R}$ forms a vector space, and moreover the subset of those functions that are continuous is a subspace of those functions, and the subset of differentiable functions lies within the subset of continuous ones as another subspace.

Since these function spaces are from $\mathbb{R}$ to $\mathbb{R}$, should I imagine the subspaces somehow lying between $\mathbb{R}^0$ and $\mathbb{R}^1$? Or are these function spaces isomorphic to $\mathbb{R}$ of infinite dimensions? I'm guessing the latter since polynomials can go up to any degree, and polynomials are a subset of differentiable functions.

If I'm on the right track with all of this so far, my final question I'm building up to is, can these vector spaces be enumerated?
These function spaces are absolutely gigantic. For instance, the space of polynomial functions - which is relatively tiny part of the space of (say) all differentiable functions - is already infinite-dimensional, since the following set of polynomials is linearly independent: $$1, x,x^2, x^3,x^4, ...$$ Now, as to how we can enumerate vector spaces, remember the following fact:


  Two vector spaces $V_1, V_2$ over a field $k$ (say, $k=\mathbb{R}$) are isomorphic iff they have the same dimension, that is, if there are bases $B_1,B_2$ of $V_1, V_2$ respectively with $\vert B_1\vert=\vert B_2\vert$.


In particular, this holds for infinite-dimensional vector spaces as well as finite-dimensional ones, by exactly the same proof (any bijection between bases induces an isomorphism in the usual way). So up to isomorphism, vector spaces are enumerated by their dimension. 

Calculating the dimension of an infinite-dimensional vector space requires knowing a bit of set theory, but assuming that background knowledge here are some specific numbers:


The set of polynomials above (= monic monomials) is a basis for the space of all polynomial functions, and is countably infinite. So the dimension of the space of polynomial functions is $\aleph_0$.
By contrast, the dimension of the space of all differentiable functions is $2^{\aleph_0}$. To see that this is a lower bound, note that for each positive real number $r$ the function $(x^2)^r$ (note that this is different in general from $x^{2r}$ - think about $x=-1$ and $r={1\over 4}$...) is differentiable, and the set of such functions is linearly independent. To see that it's an upper bound, it's a good exercise to show that there are only $2^{\aleph_0}$-many continuous functions, hence only $2^{\aleph_0}$-many differentiable functions, and the dimension of the space clearly can't be bigger than the number of points.
Perhaps surprisingly, the space of all continuous functions has the same dimension, $2^{\aleph_0}$! This follows since it contains the space of differentiable functions, hence has dimension at least $2^{\aleph_0}$ since that's the dimension of the space of differentiable functions, but (as said above) there are only $2^{\aleph_0}$-many continuous functions at all. This is true even though the space of differentiable functions is of course a proper *(indeed, very proper - and see below for what I mean by this)* subspace of the space of all continuous functions. This sort of thing happens whenever we think about infinite sets, the ur-example being Hilbert's hotel.
Finally, the space of all functions has dimension $2^{2^{\aleph_0}}$. Roughly speaking, this is because the number of functions is $2^{\aleph_0}$ times the dimension of the space of functions, and the number of functions is $2^{2^{\aleph_0}}$. (Making this precise takes a bit of thought.)




There is, however, still a problem with enumerating subspaces. In a vector space $V$, if $W_1,W_2$ are subspaces of $V$ with $dim(W_1)=dim(W_2)$ finite, then there is an automorphism of $V$ swapping $W_1$ and $W_2$; that is, any two subspaces with the same finite dimension don't just look the same "on their own," they "sit inside $V$" in the same way. 

What do I mean intuitively by "sit inside $V$ in the same way"? Well, take $V=\mathbb{R}^2$, $W_1=$ the $x$-axis, and $W_2=$ the $y$-axis. Obviously $W_1$ and $W_2$ are different subspaces, but just as obviously they "look the same" - I can distort the plane via a bijective linear transformation (= automorphism) and swap the $x$-axis and the $y$-axis. This is what I'm getting at, above.

This fails for infinite-dimensional vector spaces: just like the set of even numbers has the same cardinality as the set of natural numbers, but is a proper subset of the naturals, if $V$ is an infinite-dimensional vector space we can find a proper subspace $W\subsetneq V$ of the same dimension; taking $W_1=W, W_2=V$ yields a counterexample to the situation above, in the infinite-dimensional case (since there can be no automorphism of $V$ which takes all of $V$ to a proper subset of $V$). For a concrete example of this, think about the continuous vs. differentiable case above: the space of continuous functions, and the subspace of differentiable functions, have the same dimension, but clearly the way the space of all continuous functions sits inside the space of all continuous functions is different from the way the space of all differentiable functions sits inside the space of all continuous functions (the former "covers everything," the latter doesn't). So subspaces of a given vector space, in general, can't be classified by their dimension alone if we care about how they "sit inside" the big space.

This is resolved by the concept of codimension - basically, this measures how many dimensions a subspace "misses." Specifically, the codimension of a subspace $W$ of $V$ is the unique dimension of a subspace $W'$ such that $W\cap W'=\{0\}$ (they don't overlap nontrivially) but $W\oplus W'=V$ (together they get you all of $V$ - so, $W'$ is in some sense exactly what $W$ misses). The fact that codimension is well-defined is nontrivial, and a good exercise. (The definition I've given isn't the same as the definition in the wiki page, which for a lot of reasons turns out to be better - but the one I've given is in my opinion more concrete, and they turn out to be equivalent.)

Codimension is really a different thing from dimension: we saw above that we can have two subspaces with the same dimension but different codimensions, and similarly it's not hard to find examples with different dimensions but the same codimensions. It turns out, though, that codimension is the only new idea we need:


  Suppose $W_1, W_2$ are subspaces of $V$ with the same dimension and the same codimension. Then there is an automorphism of $V$ swapping $W_1$ and $W_2$. (The converse also holds.)


So to classify subspaces up to "sit inside the big space in the same way," I need to look at two different numbers - dimension and codimension - but these are the only two numbers I need to look at.
Hint:

$\mathbb{R}$ is a vector space over $\mathbb{Q}$. Think about the Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$. You will get some idea.