A cofinal aleph-subproduct has the cardinality of the aleph-omega product

Enumerate $B$ increasingly as $(e(r){)}_{r<\omega }$. Explicitly, choose its least element first and at each stage its least unused element; a next element exists since otherwise $B$ would be finite. This enumeration covers $B$, since any fixed natural number has only finitely many smaller natural numbers and cannot remain unchosen forever. The map $q\mapsto (r\mapsto q(e(r)))$ injects $Q$ into ${\lambda }^{\omega }$: every value is below ${\mathrm{\aleph }}_{e(r)}<\lambda$ by F2, and agreement on all enumerated coordinates means agreement on $B$. Thus F1 and A1 give $\mathrm{\mid }Q\mathrm{\mid }\le {\lambda }^{{\mathrm{\aleph }}_0}$.

Partition the coordinate set by putting $B_i=\{e(2^i(2j+1)-1):j<\omega \}$ for $i<\omega$. Each positive integer has a unique expression $2^i(2j+1)$: repeatedly divide by two until the quotient is odd; this terminates by strict decrease of positive integers, and uniqueness follows by cancelling the smaller power of two, since an odd integer cannot equal an even integer. Hence the $B_i$ are pairwise disjoint and cover $B$. Each is infinite, so as a subset of the natural numbers it is unbounded. Its increasing enumeration is $b_i(j)=e(2^i(2j+1)-1)$. Because all members of $B$ are at least two, any increasing sequence from $B$ satisfies $b_i(j)\ge j+2$, by induction on $j$.

For $f\in {\lambda }^{\omega }$ and each $i$, let $m_i$ be the least natural number such that $f(i)<{\mathrm{\aleph }}_{m_i}$; it exists by F2. Define $T(f)\in Q$ by giving it the single nonzero value $f(i)+1$ at coordinate $b_i(m_i+1)$ of $B_i$, and zero at every other coordinate of $B_i$. This is a well-defined function because the $B_i$ partition $B$ by step 2.1. Its assigned value is strictly below the required factor: the ordinal successor of an ordinal below the infinite cardinal ${\mathrm{\aleph }}_{m_i}$ is still below that cardinal, which is a limit ordinal; moreover $b_i(m_i+1)\ge m_i+3>m_i$ and F2 gives ${\mathrm{\aleph }}_{m_i}<{\mathrm{\aleph }}_{b_i(m_i+1)}$. All other values are zero, also below their factors. In particular $f(i)=0$ produces the nonzero tag $1$ and is not confused with an unused coordinate.

From $T(f)$ restricted to $B_i$ recover its unique nonzero coordinate and its value $f(i)+1$. An ordinal successor has its original ordinal as unique greatest member, so this value recovers $f(i)$. Hence $T(f)=T(g)$ implies $f(i)=g(i)$ for every $i$, proving that $T$ is injective. F1 and A1 give ${\lambda }^{{\mathrm{\aleph }}_0}\le \mathrm{\mid }Q\mathrm{\mid }$, and the reverse inequality is step 1.1. Antisymmetry of cardinal comparison yields the asserted equality. QED.