In Dirac's The Principles of Quantum Mechanics he often uses the notion of "a complete set of commuting observables". This means a set of observables $\{\xi_1, \xi_2, \ldots, \xi_v, \xi_{v+1}, \ldots, \xi_u\}$ where $\{\xi_1, \xi_2, \ldots, \xi_v\}$ have discerete eigenvalues, and $\{\xi_{v+1}, \ldots, \xi_u\}$ have continous eigenvalues and each pair of observables commute, and to each set of eigenvalues only one simultanious eigenvector belongs.
The definiton of observables by Dirac is that each $\xi_i$ is a real linear operator, and the set of eigenvectors of the operator is a complete set. Now my question is: how is it even possible that obsevables with discrete and continous igenvalues are commuting? I mean if an opeator with discrete eigenvalues is an observable, that means that the set of basis vectors for the vector space is a countably infinite set. However if an operator with continous eigenvalues is also an observable (acting on the same vector space), that means that the set of basis vectors for the same vector space is an uncountably infinite set. For me, these two properties contradict with eachother.
Not to mention that Dirac also gives a proof of two commuting observables having a complete set of simultanious eigenvalues, the "side result" of his proof being that if two observables commute, their sets of eigenvectors mutually contain eachother, which means that the two sets are equal. To me this is also a contradiction.
Also can even two observables with discerete and continous eigenvlaues act on the same space?
Can somebody give me an explanation?