Fermion and dilute grand-canonical formulas

[Single orbital] The vector $u_k$ is an eigenvector of ${\sum }_{m=0}^{1}m{T}_{u_m,u_m}$ with eigenvalue $k$, so $e^{-\beta ({\epsilon }_1-\mu )N_{-}}={\sum }_{k=0}^1{e}^{-\beta ({\epsilon }_1-\mu )k}{T}_{u_k,u_k}=F$ on a complete orthonormal family, hence everywhere; $F$ is a positive diagonal finite-rank operator with eigenvalues $1$ and $e^{-\beta ({\epsilon }_1-\mu )}>0$. By [F2] its trace is $1+e^{-\beta ({\epsilon }_1-\mu )}$ and the weighted average of the eigenvalues $0,1$ is $\frac{0\cdot 1+1\cdot e^{-\beta ({\epsilon }_1-\mu )}}{1+e^{-\beta ({\epsilon }_1-\mu )}}=\frac{1}{e^{\beta ({\epsilon }_1-\mu )}+1}$. This proves (a).

[Dilute remainders] For $0<q<1$, the bosonic signed deviation is $\frac{q}{1-q}-q=\frac{q^2}{1-q}>0$. For every $q>0$, the fermionic signed deviation is $\frac{q}{1+q}-q=-\frac{q^2}{1+q}<0$. Their absolute values are $q^2\mathrm{/}(1-q)$ and $q^2\mathrm{/}(1+q)$ on these respective domains; when $0<q\le 1\mathrm{/}2$, each is bounded by $2q^2$ because $1-q\ge 1\mathrm{/}2$ and $1+q\ge 1$. The bosonic geometric sums are [F1], and the fermionic finite sums are $1+q$ and $q$. This proves (c).

[Fermionic products] If infinitely many $j$ have $q_j>1\mathrm{/}2$, the product over any $k$ of these indices is at least $(3\mathrm{/}2{)}^k$, so $Z_{\mathrm{F}}=+\mathrm{\infty }$ and ${\sum }_j{q}_j=+\mathrm{\infty }$. Otherwise their finitely many factors are finite positive constants, and [F3] gives $q_j\mathrm{/}2\le q_j-q_j^2\le \log (1+q_j)\le q_j$ on the remaining tail. Thus the nonnegative logarithmic partial sums are bounded exactly when ${\sum }_j{q}_j<\mathrm{\infty }$. Since ${\prod }_{j\le n}(1+q_j)=\exp ({\sum }_{j\le n}\log (1+q_j))$, the increasing partial products have a finite limit exactly in that case. Every finite subset is contained in an initial segment and every factor is at least one, so their supremum equals that limit. Finally ${\sum }_j{q}_j=e^{\beta \mu }{\sum }_j{e}^{-\beta {\epsilon }_j}$, giving precisely the fermionic equivalence in (b).

[Bosonic products] If some ${\epsilon }_j\le \mu$, then $q_j\ge 1$ and $Z_{\mathrm{B}}=+\mathrm{\infty }$ by the stated convention; also $\mu <{\inf }_j{\epsilon }_j$ fails. Suppose all $q_j<1$. By [F4], $q_j\le -\log (1-q_j)$ for every index. If infinitely many $q_j>1\mathrm{/}2$, both corresponding sums diverge; otherwise discard only that finite set and use $q_j\le -\log (1-q_j)\le 2q_j$ on its complement. Consequently ${\sum }_j-\log (1-q_j)<\mathrm{\infty }$ exactly when ${\sum }_j{q}_j<\mathrm{\infty }$, and the reciprocal partial products, whose logarithms are these partial sums, have a finite limit exactly then. Their limit equals the finite-subset supremum because the factors are at least one. If ${\sum }_j{q}_j<\mathrm{\infty }$, then $q_j\to 0$, so ${\epsilon }_j\to +\mathrm{\infty }$ since $q_j=e^{-\beta ({\epsilon }_j-\mu )}$ and $\beta >0$. In particular, all but finitely many ${\epsilon }_j$ exceed ${\epsilon }_1$; hence the infimum is attained in a nonempty finite initial set. Since every ${\epsilon }_j>\mu$, this gives $\mu <{\inf }_j{\epsilon }_j$. Conversely, $\mu <{\inf }_j{\epsilon }_j$ and ${\sum }_j{e}^{-\beta {\epsilon }_j}<\mathrm{\infty }$ imply all $q_j<1$ and ${\sum }_j{q}_j=e^{\beta \mu }{\sum }_j{e}^{-\beta {\epsilon }_j}<\mathrm{\infty }$, so the scalar criterion just proved gives finite $Z_{\mathrm{B}}$. This proves the full bosonic equivalence without assuming attainment in the hypotheses and without invoking an Axiom-of-Choice spectral theorem.

[Conclusion] Steps 1.1, 1.2, 1.3 and 1.4 establish (a), (c), the fermionic half of (b) and the bosonic half of (b) respectively; in the finite cases the displayed products are the limits, and the mode occupancies in (c) are the normalized geometric sums of [F1]. The closing paragraph of the statement records that no physical identification of the parameters or of the products with thermodynamic quantities is asserted.