Tomasz Natkaniec

Tomasz Natkaniec: Perfectly everywhere surjective but not Jones functions

Tuesday, November 14, 2017 17:15

Room: D1-215

Speaker:
Tomasz Natkaniec

Title: Perfectly everywhere surjective but not Jones functions

Abstract. Given a function $$f:\mathbb{R}\to\mathbb{R}$$ we say that

1. $$f$$ is perfectly surjective ($$f\in \mathrm{PES}$$) if $$f[P]=\mathbb{R}$$ for every perfect set $$P$$;

2. $$f$$ is a Jones function ($$f\in\mathrm{J}$$) if $$C\cap f\neq\emptyset$$ for every closed $$C\subset\mathbb{R}^2$$ with $$\mathrm{dom}(C)$$ of size $$\mathfrak{c}$$.

M. Fenoy-Munoz, J.L. Gamez-Merino, G.A. Munoz-Fernandez and E. Saez-Maestro in the paper A hierarchy in the family of real surjective functions [Open Math. 15 (2017), 486--501] asked about the lineability of the set $$\mathrm{PES}\setminus\mathrm{J}$$.
Answering this question we show that the class $$\mathrm{PES}\setminus\mathrm{J}$$ is $$\mathfrak{c}^+$$-lineable. Moreover, if
$$2^{<\mathfrak{c}}=\mathfrak{c}$$ then $$\mathrm{PES}\setminus\mathrm{J}$$ is $$2^\mathfrak{c}$$-lineable. We prove also that the additivity number
$$A(\mathrm{PES}\setminus\mathrm{J})$$ is between $$\omega_1$$ and $$\mathfrak{c}$$. Thus $$A(\mathrm{PES}\setminus\mathrm{J})=\mathfrak{c}$$ under CH,
however this equality can't be proved in ZFC, because the Covering Property Axiom CPA implies $$A(\mathrm{PES}\setminus\mathrm{J})=\omega_1<\mathfrak{c}$$.

The talk is based on the joint paper:
K.C.Ciesielski, J.L. Gamez-Merino, T. Natkaniec, and J.B.Seoane-Sepulveda, On functions that are almost continuous and perfectly everywhere surjective but not Jones. Lineability and additivity, submitted.