Reading Eight Erdős Problems Together

Eliminated ideas

1. If the reachable values form a null set, no rational value is reachable. False. Take the binary series with digit $1$ everywhere except digit $0$ at positions $n_1<n_2<\cdots$, and allow each of those digits to be changed to $1$. The reachable values form a null set when the positions are sparse, by Theorem 6.1(i), yet changing all of them gives $\sum 2^{-n}=1$. So no count of choices against contraction, and no depth depending only on sparsity, excludes a particular rational.

2. Every infinite subset of a host with null subsum set has an irrational sum. A host is a set $B$ of allowed indices, and its subsum set is $\{X_S(2):S\subseteq B\}$. For hosts chosen without reference to the target this is open and is a form of #257 itself. As a universal statement it cannot be a route: the support of any rational subsum with infinite $S$ would be such a host. The subsum set of a host $B$ has positive measure exactly when the complement of $B$ is finite, by Theorem 6.1(i) and the measure at full support.

3. The share of surviving fractions as evidence. Section 9.

4. A wrong locator. A draft of Theorem 6.3(b) cited Theorem 6.1 of [40] as Nishioka’s theorem. That theorem says a value of a Mahler function at an algebraic point is rational or transcendental, which cannot prove irrationality. The proof in Section 6.2 uses Nishioka’s value theorem as quoted in [18], applied there to the two-dimensional system for $(g,1)$ with regular points in $(0,1)$.

5. Algebraic independence for #1049. With $g$ as in Section 6.2, $\sum _{n\ge 1}(t^n-1{)}^{-1}=\sum _{m\mathrm{odd}}g(t^{-m})$, and each $g(t^{-m})$ is transcendental for rational $t>1$. This gives nothing for the infinite sum: limits of transcendental numbers take every value. No applicable value theorem for this decomposition is supplied here.