That's precisely what an instability is about... The Fermi surface of free particles is usually stable. But if you have attractive interaction between the free electrons it becomes instable. Cooper showed that the Fermi surface is unstable when some attractive electron-phonon interaction are present. So in fact the Fermi surface doesn't exist in superconductors.

The strategy is the following: you take a Hamiltonian whose ground state is known [the free electron gas for the Cooper problem, with known Fermi surface]. Then you switch on some interaction [the attractive electron-phonon interaction] and you ask whether the ground state of the two systems -- with or without the interaction -- is the same. In the case of the Cooper problem the two systems have not the same ground state.

It is nevertheless difficult to know how to write the ground state of the interacting system, so you're happy already knowing that the ground state of the interacting system has a lesser energy than the non-interacting one. It simply means that, whenever the attractive interaction is present, the system will favour a ground state which is not a Fermi surface [or a free electron gas in the Cooper problem].

In the Cooper problem, this new ground state is the Cooper fluid, or BCS ground state. Its approximate (= mean-field) form has been postulated by Bardeen, Cooper and Schrieffer a few months after Cooper understood that the Fermi surface is not the ground state of a superconductor.

Historically, the Cooper problem is important, since it clearly demonstrated that the Fermi surface is not always the ground state of interacting fermions. Before Cooper, it was commonly believed that the Fermi surface is a stable object, following some arguments by Luttinger, who showed that the Fermi surface can still be defined even with interaction. Of course, what Luttinger forget was the possibility for phase transition.

First of all, at zero temperature not all of the electrons in the BCS ground sate (GS) form Cooper pairs. One way to think about this is that deep inside the Fermi surface, there is no available states for scattering events, including the phonon-mediated ones, and thus we wouldn't have the effective attraction to form the Cooper pairs.

This can also be seen from the coefficients $u_k$ and $v_k$ in the postulated BCS GS given in the original post. When $k$ is way below the Fermi surface, then $|\xi_k|>>\Delta$, with $\xi_k<0$. In this case, $|u_k|\to 0$ and $v_k\to 1$, i.e. the states far below the Fermi surface are almost fully occupied by electrons created by $c^{\dagger}_{k\uparrow}c^{\dagger}_{-k\downarrow}$. These electrons are created in pairs, but they are NOT Cooper pairs!!! The pair creation operator does not tell us anything regarding whether the pair is a Cooper or just a normal pair. Only when it is close to the Fermi surface, and thus possible to have phonon-mediated scattering, is the Cooper pair creation possible.

In summary, the BCS GS looks like the following: deep down inside the Fermi surface, it's just normal electrons. Close to the Fermi surface, it's mostly a superposition $\left(u_k+v_kc^{\dagger}_{k\uparrow}c^{\dagger}_{-k\downarrow}\right)|0\rangle$ of empty state, with amplitude $u_k$, and occupied Cooper pairs, with amplitude $v_k$.

EDIT:

After I submitted the above answer, I realized that I missed out the other part of your question, which is about the excitations of the BCS GS and here it goes:

You mentioned that "the excitations of the BCS state must be created or destroyed in pairs". This is not correct. The excitations of the BCS GS are fundamentally different from Cooper pairs or the normal electrons. Instead, it is the quasi-particle created by
$$\gamma^{\dagger}_{p\uparrow}=u^*_pc^{\dagger}_{p\uparrow}-v^*_pc_{-p\downarrow}$$
Not surprisingly, this is just the Bogolyubov transformation used to diagonalize the BCS mean field hamiltonian and this why it is the legitimate excitations of the GS. The corresponding eigenvalue is exactly
$$E_k=\sqrt{\xi_k^2+\Delta^2}$$
This is where the gap $\Delta$ to the BCS excitation comes from. This kind of excitation is slightly less intuitive compared to objects like single electrons or Cooper pairs. It has spin-1/2, but it doesn't carry well-defined charge, whereas a Cooper pair has spin 0 and charge $2e$.

Also, You can easily check that
$$\gamma_{p\uparrow}|BCS\rangle=\gamma_{-p\downarrow}|BCS\rangle=0$$
which makes perfect sense since the annihilation operator for the excitation annihilates the GS.

## Best Answer

The original Cooper instability as treated by Cooper in his seminal paper is indeed an approximation for only two electrons on top of the Fermi sea. Nevertheless, you can make the notion precise for the whole gas using mean-field approximation and renormalisation procedure.

The Cooper pairs form only in the vicinity of the Fermi level. Then this region is depopulated and the BCS condensate is gapped (picturesquely said). In the BCS mechanism, when the attractive interaction is driven by the phonon field, it seems consistent to use the Debye energy $\hbar \omega_{D}$ as a cut-off for your theory, since this is the characteristic energy of the phonons.

The remaining of the Fermi sea, as for the transport in normal metals, do not participate in the superconducting mechanism, except they define the Fermi energy of course. This role is important, the electrons deep inside the Fermi sea are not properly speaking superconducting.

The BCS ground state is variational exact for a point-contact interaction between the electrons in the range of energy given by (twice) the Debye energy, as it is discussed in the original BCS paper. See also Bogoliubov's treatment of superconductivity (far more complicated though).

I do not detail more the answer, since you can find more precise answers in any textbook on superconductivity: Tinkham is a really popular one. The variational exactness of the BCS Ansatz is treated in great details in deGennes. Both books called

superconductivitysomething...