the model-independence of P
A definition is only as useful as it is stable. If 'solvable in polynomial time' meant one thing for Turing machines and a different thing for a real computer, P would be an academic curiosity rather than a meaningful claim about the world. The model-independence of P is the happy fact that this never happens: whether a problem is in P does not depend on which reasonable deterministic model of computation you use to define it. P is the same class no matter how you wire up the computer.
The reason is the polynomial-overhead simulation already met in the Cobham-Edmonds thesis. Switch from a single-tape Turing machine to a multitape one and you might save time, but going back costs at most a squaring. Move to a random-access machine, or to ordinary code, and the translations in both directions cost only a polynomial factor. Now stack these: a polynomial running time, run through a polynomial-overhead simulator, is still polynomial, because a polynomial of a polynomial is a polynomial. So a problem that has an n^3 algorithm on one model gets some n^c algorithm on any other, and membership in P is preserved. (Smaller distinctions are not model-independent, which is exactly why we union over all polynomials: the class TIME(n^2) can shift between models, but P does not.)
This robustness is what licenses the whole practice of complexity theory: we prove a problem is in P by exhibiting a polynomial-time algorithm in whatever clean model we like, confident the verdict transfers. It is also why P, not a narrower bound, is the standard meaning of tractable. The one celebrated possible exception lies outside classical determinism: quantum computation (BQP) is not known to be simulable by classical machines with only polynomial overhead, so the model-independence is a statement about reasonable classical deterministic models, and the quantum question is held open separately.
Sorting is O(n log n) on a random-access machine. Simulate that machine on a single-tape Turing machine and the running time rises to some polynomial, perhaps O(n^2) or O(n^3), but it stays polynomial. So 'sorting is in P' is true on every reasonable model, even though the precise exponent depends on the model.
The exact exponent shifts between models, but membership in P does not.
Model-independence holds for P (a union over all polynomials), NOT for a single bound like TIME(n^2), which can differ across models. The classical guarantee says nothing about quantum machines, which are not known to be polynomially simulable.