Why position operator is non-degenerated? In quantum mechanics one can assume position operator $\hat{X}$ must have continuous spectrum, as experiments say it is possible to find a quantum particle at any point of the space. The question is why it is always assumed to be "non-degenerate" for a spinless particle? I mean one could have 
$$
\hat{X}|x,i\rangle=x|x,i\rangle
$$
for several $i's$. However this is always taken to be 
$$
\hat{X}|x\rangle=x|x\rangle
$$
which leads to the position representation $\langle x|\psi\rangle=\psi(x)$ for the wavefunction of a spinless particle. 
 A: It isn't assumed to be non degenerate:
For example, for the position operator in X-Direction, you have: 
$$
\hat{x} |x, y, z \rangle  = x |x, y, z \rangle
$$
But aswell:
$$
\hat{x} |x, \tilde{y}, \tilde{z} \rangle = x |x, \tilde{y}, \tilde{z} \rangle
$$
In this example, we see that degeneracy of the X-position operator means just that there are additional degrees of freedom (in that case, the Y- and the Z- position). Those degrees of freedom either do contribute to the dynamics of the system (which means that they show up in the hamiltonian operator), or they don't. If they don't, you can look at the system in a smaller Hilbert space, in which those degrees of freedom don't show up. 
Looking at my example, this smaller Hilbert space could be the space of square integrable functions defined over $\Re$, compared to the space of square integrable functions defined over $\Re^3$, that could be choosen to represent non-degenerated eigenstates of $\hat{x}$, $\hat{y}$, and $\hat{z}$. 
So in short: You make the position operator non-degenerated by choosing an appropriate Hilbert space, suitable for your problem. 
Edit: Eigenstates of the position operator are strictly no part of the hilbert space (since in the said representation, they wouldn't be square integrable functions, but instead Dirac $\delta$-distributions). I still think it is clear what I meant there. 
