Is this a correct way of understanding the third law of thermodynamics? The third law of thermodynamics can be stated as:
$$\lim_{T \to 0 \ \mathrm{K}}S = 0 $$
Now, from statistical mechanics: $S = k_B \log W $ where $W$ is the number of microstates. Notice that if $W=1$, then $S=0$. In other words, is the third law a consequence of the fact that systems tend to a single microstate as temperature drops to $0$?
 A: Nernst's law  is usually expressed as stating that as $T\to 0$ the entropy of the system approaches a constant. This is more like saying that the heat capacity goes to zero and is more general than just saying that $S(T=0)=0$ because the former also includes the ground state degeneracies, see the wikipedia article for such examples.
A: The most general statement of the third law is that as $T \rightarrow 0$, system's entropy approaches a unique value, independently of the value of any other thermodynamic variable. Since thermodynamic entropy is defined within an additive constant, it is always possible to put
$$
S(T=0)=0.
$$
Apparently, the microcanonical formula from Statistical Mechanics:
$$
S= k_B \log W,\tag{1}
$$
where $W$ is the number of microstates compatible with a given macrostate at fixed energy, volume, and number of moles would imply the non-degeneracy of the ground state once we translate the limit for vanishing temperature into the limit for energy approaching the ground state.
However, such a conclusion is not consistent with the fact that formula $(1)$, as all the similar formulae of statistical mechanics, does correspond to the thermodynamic entropy only at the thermodynamic limit, i.e., in the limit of a large system. Indeed, a degenerate ground state may be compatible with equation $(1)$ and with the third law, provided the degeneracy of the ground state does not grow with the system size so fast to give a contribution to the entropy per particle (or per volume).
Therefore, the correct statement should be the third law is a consequence of the fact that systems tend to a ground state whose degeneracy $g$ is sub-extensive as temperature drops to zero. Formally, we have to require that
$$
\lim_{T \rightarrow 0^{+}} g(E,V,N)/V=0
$$
whenever $E,V,N$ go to infinity by keeping constant their ratios. That is a milder condition than the non-degeneracy of the ground state.
