Let $F\subseteq \mathbb{C}$ be a closed infinite set, and let $\mu (F)$ be the infimum of$$|\{z:|f(z)|<1\}|,$$as $f$ ranges over all polynomials of the shape $\prod (z-z_i)$ with $z_i\in F$.Is $\mu (F)$ determined by the transfinite diameter of $F$? In particular, is $\mu (F)=0$ whenever the transfinite diameter of $F$ is $\ge 1$?

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Write [ \mu(F)=\inf\\{\mathrm{area}\bigl({z\in\mathbb C:\ |p(z)|<1}\bigr):p(z)=\prod_{j=1}^n (z-z_j),\ z_j\in F,\ n\ge 1\\}, ] (where “area” is planar Lebesgue measure). For compact $F$, the transfinite diameter $d(F)$ agrees with the logarithmic capacity (\mathrm{cap}(F)).

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.