What Assumptions Govern the Applicability of the Boltzmann Distribution

Question

In the book "Concepts in Thermal Physics" the Boltzmann distribution is derived with the following assumptions:

1. There are two systems, one enormous heat reservoir and one comparatively miniscule system.
2. The two systems are in thermal equilibrium.
3. The heat reservoir is so large that any energy the smaller system can remove makes no change to its overall temperature.
4. The large system has an incredibly large number of possibly microstates.
5. By contrast the small system is assumed to have 1 microstate for every possible energy.

Therefore the energies of each system are $\epsilon$ for the small system and $(E-\epsilon)$ for the large system, with the total energy being $E$. This allows you to formulate the probability of the small system have energy $\epsilon$ as:

$$P(\epsilon) \propto \Omega(E-\epsilon)\times1$$

Where $\Omega(E-\epsilon)$ is the total number of microstates for the reservoir, and $1$ represents the total number of microstates for the system. This allows you to formulate the Boltzmann distribution:

$$P(\epsilon) \propto e^{-\epsilon/k_BT}$$

My question is, when does assumption 5 apply? Is it allowed because the reservoir is assumed to be so much larger than the small system? Or would you need to derive the Boltzmann distribution differently for different systems?

Edit: The answer to this question seems to be given here Derivation of Boltzmann distribution – two questions

Answer 1

Assumption 4 isn't needed to talk about the Boltzmann distribution. Each state has a probability which is proportional to $\exp(-E/k_B T)$, and if there are $g(E)$ different states with the same energy (degeneracy), then the probability to be at any state of energy E is just proportional to $g(E) \cdot \exp(-E/k_B T)$.