PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 129, Number 1, Pages 265{270 S 0002-9939(00)05499-X Article electronically published on July 27, 2000 STRONGLY MEAGER SETS AND THEIR UNIFORMLY CONTINUOUS IMAGES ANDRZEJ NOWIK AND TOMASZ WEISS (Communicated by Carl G. Jockusch, Jr.) Abstract. We prove the following theorems: (1) Suppose that f :2! ! 2! is a continuous function and X is a Sierpi´nski set. Then (A) for any strongly measure zero set Y , the image f[X + Y ]isans0-set, (B) f[X] is a perfectly meager set in the transitive sense. (2) Every strongly meager set is completely Ramsey null. This paper is a continuation of earlier works by the authors and by M. Scheepers (see [N], [NSW], [S]) in which properties (mainly, the algebraic sum) of certain singular subsets of the real line R and of the Cantor set 2! were investigated. Throughout the paper, by a set of real numbers we mean a subset of 2! and by \+" we denote the standard modulo 2 coordinatewise addition in 2!.Letus also assume that a \measure zero" (or \negligible") set always denotes a Lebesgue measure zero set. We apply the following definition of sets of real numbers. Definition 1. An uncountable set X is said to be a Luzin (respectively, Sierpi´nski) set iff for each meager (respectively, measure zero) set Y , X\Y is at most countable. We say that a set X is of strong measure zero (respectively, strongly meager)ifffor each meager (respectively, measure zero) set Y , X + Y =26 !. Remark 1. It is well known (see [M] for example) that every Luzin set is strongly measure zero. Quite recently J. Pawlikowski proved that each Sierpi´nski set must be strongly meager as well (see [P]). Let us recall that a set X is called an s0-set (or Marczewski set) iff for each perfect set P one can find a perfect set Q⊆P that is disjoint from X. M. Scheepers showed in [S] that for a Sierpi´nski set X and a strong measure zero set Y , X + Y is an s0-set. Later, in [NSW] it was proven that this also holds when X is strongly meager. We have the following functional version of the M. Scheepers' result. Theorem 1. Let X be a Sierpi´nski set and let Y be a strong measure zero set. Assume also that f :2! ! 2! is a continuous function. Then the image f[X + Y ] is an s0-set. Received by the editors July 16, 1998 and, in revised form, September 9, 1998 and March 10, 1999. 2000 Mathematics Subject Classification. Primary 03E15, 03E20, 28E15. Key words and phrases. Strongly meager set, always first category set. The first author was partially supported by the KBN grant 2 P03A 047 09. c 2000 American Mathematical Society 265 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 266 ANDRZEJ NOWIK AND TOMASZ WEISS Proof. Let P ⊆2! be a perfect set. We can assume that f[2!] \P contains a perfect set. Otherwise, we are done. So, fix fRα : α<!1g, a family of pairwise disjoint ! \ 0 −1 perfect subsets contained in f[2 ] P ,andforeveryα<!1, put Rα = f [Rα]. 0 2N Take α0 <!1 such that Rα0 (negligible sets). We know that Y is of strong 0 0 − measure zero and that Rα0 is closed, so Rα0 Y has measure zero. From this we \ 0 − 0 have that X (Rα0 Y )(letusdenotethissetbyX ) is countable. Thus, f[X + Y ]=f[(X n X0)+Y ] [ f[X0 + Y ]; n 0 and f[(X X )+Y ] is disjoint from Rα0 .Also,sincef is a uniformly continuous function, f[X0 + Y ] is a strong measure zero set. Hence f[X + Y ] is disjoint from some perfect set contained in Rα0 . Definition 2. AsetX is called an AFC' set (perfectly meager in the transitive sense) iff for each perfect set P there is F ,anFσ set containing X, such that for every t 2 2!,(F + t) \ P is meager in the relative topology of P . We will say that X is a wQN-set (weakly Quasinormal set) iff for each sequence of continuous ! ! functions fn : X R,iffn 0 (pointwise),S then there is a subsequence fnk and f g 2 countable family Xn n ! such that X = n2! Xn and fnk converges uniformly on Xn for every n 2 !. It is easy to prove that each Sierpi´nski set is wQN and that for a wQN-set X and every continuous function f :2! ! 2!, f[X] is a wQN-set as well (see [BRR]). Thus, using Nowik's theorem which says that any wQN-set is an AFC' set (see [N]), we obtain the following theorem. Theorem 2. If S is a Sierpi´nski set, then for every continuous function f :2! ! 2!, we have that f[S] is an AFC' set. We present an alternative proof of this fact with the hope that it may lead to a positive answer to Question 1 (see below). Lemma 1 (Nowik). For each perfect set P ⊆2!, there exists a continuous function Φ:2! ! 2! such that for any t 2 2!, Φ[P + t]=2!. Proof. See [N]. Corollary 1. Let P ⊆2! be a perfect set. Then there exists an uncountable family H of pairwise disjoint closed subsets of 2! such that for every G 2H, P + G =2!. Proof. Let Φ be as in Lemma 1. Put H = fΦ−1[fhg]:h 2 2!g. Corollary 2. For every perfect set P ⊆2!, there exists an uncountable family H of closed, pairwise disjoint negligible sets such that for each G 2H, P + G =2!. Proof. Obvious, since for H in Corollary 1 we have that jfG 2H: G 62 N gj ≤ !. Proof of Theorem 2. Let P ⊆2! be a perfect set and let f be a continuous function. Without loss of generality we may assume that f maps 2! onto 2!. Suppose that (Pi)i2! is an enumeration of basic clopen sets in the relative topology of P . Assume that for each i 2 !, Hi is an uncountable family of pairwise disjoint, closed sets such that ! 8G2Hi Pi + G =2 : Let H~i = ff −1[G]:G 2Hig: License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use STRONGLY MEAGER SETS AND UNIFORMLY CONTINUOUS IMAGES 267 i We choose from every H~ a negligible set Ai. Suppose that B,aGδ negligible set, is such that [ Ai⊆B i<! and C =2! n B.SinceS is a Sierpi´nski set, S0 = S \ B is at most countable. We have that [ f[C] \ f[Ai]=;; i<! so [ 0 f[S n S ] \ f[Ai]=;: i<! n 0 It is clearS that f[C]isanFσ set; thus f[S S ] is disjoint from some Gδ set A which 2 ! n 0 \ − contains i<! f[Ai]. Finally, for every t 2 , f[S S ] (P t) is disjoint from ! A \ (P − t). From the fact that f[Ai]+Pi =2 (for every i<!)itfollowsthatA is a dense set in P − t. Remark 2. Notice that Corollary 2 is a stronger version of the well-known Erd¨os- Kunen-Mauldin theorem (see [NSW]). Definition 3. For any finite set s 2 [!]<! and infinite A⊆! with max(s) < min(A), let [s; A]=fB 2 [!]! : s⊆B⊆s [ Ag.WesaythatF ⊆[!]! is a completely Ramsey null (CR0) set iff for every so-called Ellentuck basic neighbourhood [s; A], there is B⊆A infinite such that [s; B] \ F = ;. ! Notice that the σ-ideal CR0 is defined on subsets of the set [!] which can be identified with a subset of 2! via characteristic functions. Thus, in the next part we deal with subsets of 2!. Theorem 3. For any [s; A],whereA 2 [!]!, max s<min A,thereexistsanegli- gible set H (even \small" in the sense of T. Bartoszy´nski) such that 8t22! 9B2[A]! [s; B]⊆H + t: Proof. Consider a partition of ! into finite disjoint intervals, say (In)n<!,which satisfies the following conditions: 1. j \ j 8 ln A In +1 ≤ 1 n<! n ; jA \ Inj 2 2. max(s) < min(I0): In By Lorentz's theorem (see for example [NSW]), we can find Hn⊆2 with the properties: 1. ln jA \ I j +1 n jInj jHn|≤ · 2 ; jA \ Inj 2. f n 2 \ g In Hn + ea : a A In =2 ; License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 268 ANDRZEJ NOWIK AND TOMASZ WEISS where en is an element of 2In defined by the following condition: a ( 0ifa =6 b; en(b)= a 1ifa = b: It is clear that the set f 2 ! 91 j 2 g H = x 2 : n x In Hn is negligible; moreover, it is \small" (see [BJ] for the definition of a \small" set). ! Let us fix t 2 2 . For every n 2 !,thereexistsan 2 A such that j n 2 (t In)+ean Hn: ! ! Put B = fangn<!. It is sufficient to show that [s; B]⊆H +t.So,letC 2 [!] ⊆2 satisfy s⊆C⊆s [ B.Wehavethat 91 j n n C In = ean : Thus, 2f 91 j 2 g C x : n (x + t) In Hn = H + t: Theorem 4. Every strongly meager set is a completely Ramsey null set. Proof. Immediately follows from Theorem 3. In the proof of Theorem 1 we used an observation that for a strong measure zero set X⊆2! and for every continuous function f :2! ! 2!, the image f[X]isalso strongly measure zero.