Eigenstates of Spin Why are the eigenstates of spin vectors and not functions? Is this because the spin, $s$, and magnetic quantum number, $m$, take discrete values? My textbook in an earlier section used $Y_\ell ^m$ as the eigenfunction of the $L^2$ and $L_z$ operators. In a footnotes,he says that $Y_\ell ^m$ was used instead of $\left|\ell \; m \right>$ because in the context of that section (angular momentum), a function seemed more natural. 
Maybe I don't quite understand the difference between a function and a vector. A function is like an infinite dimensional vector. So since $\left|s \; m_s \right>$ only has certain limited values of $s$ and $m_s$, its a vector because the limited values implies it lives in a finite dimensional space? As a side note, can a function be called a function if it's domain is only a single point? 
 A: What you guessed is correct, the discrete values lead to a discrete number of eigenfunctions, but let me say the things a bit more precisely. 
The function $Y^m_ℓ$ for a GIVEN $ℓ$ and a GIVEN $m$ is indeed a FUNCTION, of the angles $θ$ and $φ$. However, for each $ℓ$ and for each $m$, we have ANOTHER function.
Now, let's take an example and see how we come to vectors. Let's take the hydrogen atom, the level $n = 2$. Let me disregard the spin, for simplicity. 
We have two possibilities for $ℓ$, i.e. $ℓ= 1$, and $ℓ= 0$.
Now, for $ℓ= 0$ there is only one possible value of $m$, i.e., $m = 0$, while for $ℓ= 1$ we have 3 possibilities $m = -1$, $m = 0$, and $m = +1$. 
In all, we have 4 FUNCTIONS of $θ$ and $φ$, i.e. $Y^0$, $Y^1_{-1}$, $Y^1_0$, and $Y^1_1$ (for details see Laplace's spherical harmonics).
Sometimes in our calculi, it is convenient to represent them in a simplified way, in vector form, without specifying their dependence on the angles:
$$Y^0        = (1, 0, 0, 0),$$
$$Y^1_{-1} = (0, 1, 0, 0), \qquad
Y^1_0    = (0, 0, 1, 0),  \qquad
Y^1_1    = (0, 0, 0, 1),$$
For instance, any wave-function suitable for the hydrogen level $n=2$ can be represented as a superposition of the four above vectors. I.e. in the space of wave-functions of the level $n = 2$, the four above vectors form a BASE.
To the difference of this example, the linear momentum admits CONTINUOUS, not discrete eigenvalues. Its eigenfunctions are of the form  $e^{ipx/ħ}$, where the eigenvalue $p$ has continuous values. It is not possible to arrange the eigenfunctons as discrete vectors as we did in the previous case (for details, see Fourier transform).
A: As you might have already notice that spin $\vec s$ commutes with many variables, especially coordinates $\vec x$. So the Hilbert space of wave functions can be expressed as the tensor product of two Hilbert spaces $\mathscr S$ and $\mathscr X$ spanned by eignvectors of $\vec s$ and $\vec x$. 
You should be familiar with $\mathscr X$ which is parametrized by $\vec x$. Similarly, $\mathscr S$ is parametrized by eigenvalues of $\vec s$, that is, $|s,m_s\rangle$. This is why $|s,m_s\rangle$ does not depend on $\vec x$. 
