Georgian Math. J. 2019; 26(4): 499ś503 Research Article Jacek Cichoń, Michał Morayne* and Robert Rałowski Images of Bernstein sets via continuous functions https://doi.org/10.1515/gmj-2019-2041 Received October 27, 2018; accepted January 22, 2019 Abstract: We examine images of Bernstein sets via continuous mappings. Among other results, we prove that there exists a continuous function f : ℝ→ℝ that maps every Bernstein subset of ℝ onto the whole real line. This gives the positive answer to a question of Osipov. Keywords: Bernstein set, Vitali set, continuous function MSC 2010: 03E75, 03E15, 54C05, 26A99 || Dedicated to Professor Alexander B. Kharazishvili on the occasion of his 70th birthday 1 Introduction, notations and deőnitions This paper has been inspired by Alexander V. Osipov’s question whether, for every Bernstein set, there are countably many continuous real functions deőned on ℝ such that the union of their images of the Bernstein set is equal to the whole real line ℝ. We answer this question in positive with a stronger conclusion. Namely, we show that there is a continuous function deőned on ℝ which maps every Bernstein set onto ℝ. We adopt the standard set theoretical notation. The őrst inőnite cardinal number is denoted by ω. The Cantor and Bairespaces are deőned, as usual, as 2 ω and ω ω , respectively. For any A, B ⊆ℝ, we deőne A ⋇ B as the algebraic sum of A and B, i.e., A ⋇ B ={x ⋇ y ∈ R : (x, y)∈ A × B}. If A ={a}, then we simply write a ⋇ B instead of {a}⋇ B. Let X be a Polish space. A subset B ⊆ X is called a Bernstein set if, for every nonempty perfect subset P of X, the sets P ∩ B and P \ B are nonempty. It follows from the deőnition that the complement of a Bernstein set is also a Bernstein set. 2 Results Let us start with an easy observation about the size of continuous images of Bernstein sets in the image of the basic space. Namely, suppose that X, Y are Polish spaces, f : X → Y is continuous and that B is a Bernstein subset of X. Then P ∩ f[B] ̸ =0 for each perfect set P ⊆ f[X]. Indeed, f −1 [P]∩ B ̸ =0 since f −1 [P] is closed and uncountable, and f[f −1 [P]∩ B]⊆ P ∩ f[B]. Moreover, since, in every perfect set, there are continuum many disjoint perfect subsets, we have ℘P ∩ f[B]℘ = c. Therefore, if f[X]= Y , then the set f[B] contains a Bernstein subset. *Corresponding author: Michał Morayne, Department of Computer Science, Faculty of Fundamental Problems of Technology, Wrocław University of Science and Technology, 50-370 Wrocław, Poland, e-mail: michal.morayne@pwr.edu.pl Jacek Cichoń, Robert Rałowski, Department of Computer Science, Faculty of Fundamental Problems of Technology, Wrocław University of Science and Technology, 50-370 Wrocław, Poland, e-mail: jacek.cichon@pwr.edu.pl, robert.ralowski@pwr.edu.pl Brought to you by | Tulane University Authenticated Download Date | 1/14/20 4:04 AM