Question regarding part of the proof for the typical subspace theorem

Question

Part three (going by N&C page 544) states that $$tr(S(n)\rho^{\otimes n})=tr(S(n)\rho^{\otimes n}P(n,\epsilon))+tr(S(n)\rho^{\otimes n}(I-P(n,\epsilon))).$$ Now I understand how the term on the left of + goes to 0 as n $\to \infty$. However, I am confused how the term on the right does. N&I states that you can set $$0 \le tr(S(n)\rho^{\otimes n}(I-P(n,\epsilon))) \le tr(\rho^{\otimes n}(I-P(n,\epsilon))) \rightarrow 0\,\,\text{ as } n\to \infty.$$

I don't quite understand why this is the case. My only assumption is that the eigenvalues of $\rho^{\otimes n}(I-P(n,\epsilon))$ are bounded in such a way that as $n \to \infty$ it will go to zero. However, I am unsure how to go about calculating this bound, though I assume it is of a similar form to the eigenvalues of $\rho^{\otimes n}P(n,\epsilon)), 2^{-n(S(\rho)-\epsilon)}$

Answer 1

By part 1, we have that for any $\delta > 0$, then for sufficiently large $n$, $tr( \rho ^ {\otimes n} P(n, \epsilon)) \geq 1 - \delta$.

This means that $tr( \rho ^ {\otimes n} P(n, \epsilon)) \rightarrow 1$ as $n \rightarrow \infty$, since it is at most 1.