Separation exponent at least 2.05

The theorem $\mathrm{SensitivityCampaign.reach\_205}$ asserts that for every real number $\varepsilon>0$ and every natural number $S\ge 2$, there exist a natural number $n$ and a function $f:\{0,1\}^{n}\to\{0,1\}$ such that $S\le s(f)$ and $s(f)^{41/20-\varepsilon}\le b(f)$, where the latter inequality is in the real numbers and the natural numbers $s(f)$ and $b(f)$ are converted to real numbers before taking the real power or comparing. The decimal $2.05$ denotes exactly $41/20$. For an input $x$ and a subset $B$ of the coordinate set $\{0,\ldots,n-1\}$, let $x^{B}$ be the input obtained by negating each Boolean coordinate in $B$ and leaving every other coordinate unchanged. Then $s(f)$ is the maximum, over all inputs $x$, of the number of coordinates $i$ for which $f(x^{\{i\}})\ne f(x)$. The number $b(f)$ is the maximum, over all inputs $x$, of the maximum number of members of a finite set $\mathcal B$ of coordinate subsets satisfying both that distinct members of $\mathcal B$ are disjoint and that every member $B$ is nonempty and obeys $f(x^{B})\ne f(x)$; equivalently, in the literal maximum over all such finite sets of subsets, a set failing either condition contributes zero. The empty set $\mathcal B$ is permitted and contributes zero. The witnesses $n$ and $f$ may depend on both $\varepsilon$ and $S$; no single pair is asserted to work for all such parameters, and neither witness is required to be unique. The existential quantifier permits $n=0$, but then $s(f)=0$ and the requirement $S\le s(f)$ cannot hold for $S\ge2$. There is no upper bound on $\varepsilon$, so the exponent $41/20-\varepsilon$ may be positive, zero, or negative; for any witnesses satisfying the assertion, its base is at least $2$. The assertion imposes no requirement for $\varepsilon\le0$ or $S<2$.