ON FUNCTIONS WHOSE GRAPH IS A HAMEL BASIS II

To the memory of my Mother ON FUNCTIONS WHOSE GRAPH IS A HAMEL BASIS II KRZYSZTOF P LOTKA Abstract. We say that a function h: R R is a Hamel function (h HF) if h, considered as a subset of R 2, is a Hamel basis for R 2. We show that A(HF) ω, i.e., for every finite F R R there exists f R R such that f + F HF. From the previous work of the author it then follows that A(HF) = ω (see [P]). The terminology is standard and follows [C]. The symbols R and Q stand for the sets of all real and all rational numbers, respectively. A basis of R n as a linear space over Q is called Hamel basis. For Y R n, the symbol Lin Q (Y ) stands for the smallest linear subspace of R n over Q that contains Y. The zero element of R n is denoted by 0. All the linear algebra concepts are considered for the field of rational numbers. The cardinality of a set X we denote by X. In particular, c stands for R. Given a cardinal κ, we let cf(κ) denote the cofinality of κ. We say that a cardinal κ is regular if cf(κ) = κ. For any set X, the symbol [X] κ denotes the set {Z X : Z < κ}. For A, B R n, A + B stands for {a + b: a A, b B}. We consider only real-valued functions. No distinction is made between a function and its graph. For any two partial real functions f, g we write f + g, f g for the sum and difference functions defined on dom(f) dom(g). The class of all functions from a set X into a set Y is denoted by Y X. We write f A for the restriction of f Y X to the set A X. For B R n its characteristic function is denoted by χ B. For any function g R X and any family of functions F R X we define g + F = {g + f : f F }. For any planar set P, we denote its x-projection by dom(p ). The cardinal function A(F), for F R X, is defined as the smallest cardinality of a family G R X for which there is no g R X such that g + G F. Recall that f : R n R is a Hamel function (f HF(R n )) if f, considered as a subset of R n+1, is a Hamel basis for R n+1. In [P], it was proved that 3 A(HF(R n )) ω. In the same paper, the author asked whether A(HF(R n )) = ω (Problem 3.5). The following theorem gives a positive answer to this question. 2000 Mathematics Subject Classification. Primary 26A21, 54C40; Secondary 15A03, 54C30. Key words and phrases. Hamel basis, additive and Hamel functions. The work was supported in part by the intersession research grant from the University of Scranton. 1

2 KRZYSZTOF P LOTKA Theorem 1. A(HF(R n )) ω, i.e. for every finite F R Rn, there exists g R Rn such that g + F HF(R n ). Before we prove the theorem we state and prove the following lemmas. Lemma 2. Let b 1,..., b m R be arbitrary numbers. There exists a linear basis C of Lin Q (b 1,..., b m ) such that b i + C is also a basis of Lin Q (b 1,..., b m ), for every i m. Proof. Without loss of generality we may assume that Lin Q (b 1,..., b m ) {0}. Let C = {c 1,..., c k } be any linear basis of Lin Q (b 1,..., b n ). So, for every i m there are p i1,..., p ik Q such that p ij c j = b i. Now, choose q Q \ {0} satisfying the following condition for all i j q j p ij 1. We claim that C = {c 1,..., c k } = 1 q C = { 1 q c 1,..., 1 q c k } is the desired basis. To prove this we need to show that for every i m b i + C is linearly independent. To see this consider a zero linear combination j p ij(b i + c j ) = 0. We have that j p ijc j = b i j p ij. If j p ij = 0 then obviously p i1 = = p ik = 0. So we may assume that j p ij 0. Next we divide both sides of j p ijc j = b i j p ij by j p ij and obtain that p ij j c j pij j = b i. On the other hand j p ij c j = j p ij qc j = b i. So we conclude that p ij j pij = qp ij for all j k and consequently q j p ij = j 1. A contradiction. p ij j pij = Now, since dim(lin Q (b i + C)) = dim(lin Q (C)) and Lin Q (b i + C) Lin Q (C), we conclude that Lin Q (b i + C) = Lin Q (C) = Lin Q (b 1,..., b m ). Let us note here that the above lemma cannot generalized onto infinite case. As a counterexample take {b 1, b 2, b 3,... } = Q and observe that there is no basis C with the required properties. Lemma 3. [PR, Lemma 2] Let H R n be a Hamel basis. Assume that h: R n R is such that h H 0. Then h is a Hamel function iff h (R n \ H) is one-to-one and h[r n \ H] R is a Hamel basis.

ON FUNCTIONS WHOSE GRAPH IS A HAMEL BASIS II 3 Lemma 4. Let X be a set of cardinality c and k 1. TFAE: (a) for all f 1,..., f k : R n R, there exists f R Rn such that f + f i HF(R n ) (i = 1,..., k). (b) for all g 1,..., g k R X, there exists g R X such that g + g i is one-to-one and (g + g i )[X] R is a Hamel basis (i = 1,..., k). Proof. (a) (b) Choose a Hamel basis H R n and a bijection p: R n \ H X. Put f i = (g i p) (0 H). By (a), there exists an f R Rn such that f +f i HF(R n ) (i = 1,..., k). Now, let A Add(R n ) be such that f H = A H and put f = f A. Note that f + f i = (f + f i ) A HF(R n ) Add(R n ) = HF(R n ) (see [P, Fact 3.1]) and also (f + f i ) H 0, (i = 1,..., k). Hence, by Lemma 3 we claim that (f + f i ) (R n \ H) is a bijection onto a Hamel basis. Now define g = f p 1 and note that it is the required function. (b) (a) Let H be like above. Choose A i Add(R n ) such that f i H A i H for every i = 1,..., k. Put X = R n \ H and g i = (f i A i ) X for i = 1,..., k. By (b), there exists a g : X R such that g + g i is a bijection between X and a Hamel basis. Define f = g (0 H) and observe that f + f i = [f + (f i A i )] + A i = [(g + g i ) (0 H)] + A i. Since (g + g i ) (0 H) HF(R n ) by Lemma 3, using [P, Fact 3.1] we conclude that [(g + g i ) (0 H)] + A i HF(R n ) for each i = 1,..., k. Hence f is the required function. Lemma 5. Let X be a set of cardinality c, ω κ < c, and f 1,... f k be functions such that f i [X] = c. R X Then there exist pairwise disjoint subsets A 1,... A n X of cardinality κ + each and satisfying the following property: for every i = 1,..., k and j = 1,..., n the restriction f i A j is one-to-one or constant and f i [ A j ] = κ + (i.e. f i is one-to-one on at least one of the sets). Proof. We prove the lemma by induction on k. If k = 1, then the conclusion is obvious (note that κ + c). Now assume that the lemma holds for k ω and let f 1,... f k+1 R X be functions such that f i [X] = c. Based on the inductive assumption, let A 1,... A n X be sets with the required properties for the functions f 1,... f k. If f k+1 [ A i ] = κ +, then by reducing the original sets A 1,... A n we will obtain sets which work for all the functions f 1,... f k+1. In the case when f k+1 [ A i ] κ, we can find a subset A n+1 X disjoint with n 1 A i such that A n+1 = κ + and f k+1 A n+1 is injective. Now, by appropriately reducing the sets A 1,... A n+1 we will obtain the desired sets.

4 KRZYSZTOF P LOTKA Lemma 6. Let X be a set of cardinality c, f 1,... f k R X be functions such that f i [X] = c, B 0, B 1 R be such that B 0 B 1 < c, and y R \ Lin Q (B 0 ). Then there exist y 1,..., y n R and x 1,..., x n X such that (a) n 1 y j = y, (b) {y 1,..., y n }, {y j + f i (x j ): j = 1,..., n} are both linearly independent over Q and Lin Q ({y 1,..., y n }) Lin Q (B 0 ) = Lin Q ({y j + f i (x j ): j = 1,..., n}) Lin Q (B 1 ) = {0} for all i = 1,..., k. Proof. Put κ = B 0 B 1 ω and let A 1,... A n X be the sets from Lemma 5 for functions f 1,... f k. First we will define the values y 1,..., y n. Let {b 1,... b s } be the set of all values such that f i A j b l for some i, j, l. Choose y 2,..., y n to be linearly independent over Q such that Lin Q ({y 2,..., y n }) Lin Q (B 0 B 1 {b 1,..., b s, y}) = {0}. This can be easily done by extending the basis of Lin Q (B 0 B 1 {b 1,..., b s, y}) to a Hamel basis and selecting (n 1) elements from the extension. Next define y 1 = y (y 2 + + y n ). Obviously n 1 y j = y. We claim that {y 1,..., y n } is linearly independent over Q and Lin Q ({y 1,..., y n }) Lin Q (B 0 ) = {0}. Assume that α 1 y 1 + + α n y n = 0 for some rationals α 1,..., α n. From the definition of y 1 we get (α 2 α 1 )y 2 + +(α n α 1 )y n = α 1 y. Based on the way y 2,..., y n were selected we conclude that α 1 = 0 and consequently α 2 = = α n = 0. Next assume that q 1 y 1 + + q n y n = b for some rationals q 1,..., q n and b Lin Q (B 0 ). above, we obtain that (q 2 q 1 )y 2 + + (q n q 1 )y n Then, proceeding similarly like Lin Q (B 0 {y}), which implies that q 1 = = q n. Consequently, if q 1 0, then we could conclude that y Lin Q (B 0 ). That would contradict one of the assumptions of the lemma. Hence q 1 = = q n = 0 and the sequence y 1,..., y n satisfies the required conditions. Before we define the sequence x 1,..., x n, we observe some additional properties of y 1,..., y n. Fix 1 i k. Let A i1,..., A il (i 1 < < i l ) be all the sets on which f i is constant and let b i1,..., b il be the values of f i on these sets, respectively. Note that properties of the sets A 1,..., A n imply that {i 1,..., i l } {1,..., n}. We will show that (1) (y i1 + b i1 ),..., (y il + b il ) are linearly independent, (2) Lin Q ({(y i1 + b i1 ),..., (y il + b il )}) Lin Q (B 1 ) = {0}. To see (1) assume that α 1 (y i1 + b i1 ) + + α l (y il + b il ) = 0 for some rationals α 1,..., α l. This implies α 1 y i1 + + α l y il = (α 1 b i1 + + α l b il ) Lin Q (B 0 B 1 {b 1,..., b s, y}). If i 1 1, then it easily follows that α 1 = = α l = 0. If i 1 = 1, then we can write α 1 y i1 + + α l y il = α 1 y 1 + α 2 y i2 + + α l y il = α 1 [y (y 2 + + y n )] + α 2 y i2 + + α l y il Lin Q (B 0 B 1 {b 1,..., b s, y}).

ON FUNCTIONS WHOSE GRAPH IS A HAMEL BASIS II 5 Consequently, α 1 (y 2 + +y n )+α 2 y i2 + +α l y il Lin Q (B 0 B 1 {b 1,..., b s, y}). Since {i 1,..., i l } {1,..., n}, after simplifying the expression α 1 (y 2 + + y n ) + α 2 y i2 + + α l y il, there will be at least one term y j with the coefficient being exactly α 1. Hence, we conclude that α 1 = 0 and as a consequence of that α 2 = = α l = 0. This finishes the proof of (1). Similar argument shows (2). Next we will define the elements x 1,..., x n X (by induction). Choose x 1 A 1 \ f 1 i [Lin Q (B 1 {b 1, b 2,..., b s, y 1,..., y n })]. i k,f i is 1-1 on A 1 This choice is possible since A 1 = κ + > κ Lin Q (B 1 {b 1, b 2,..., b s, y 1,..., y n }) and together with the condition (2) assures that Lin Q ({y 1 + f i (x 1 ): f i A 1 is 1-1}) Lin Q (B 1 {b 1, b 2,..., b s, y 1,..., y n }) {0} and Lin Q ({y 1 + f i (x 1 )}) Lin Q (B 1 ) = {0} for all i k. Now assume that x 1 A 1,..., x m 1 A m 1 (m < n) have been defined and they satisfy the following property: ( ) {y j + f i (x j ): j = 1,..., m 1} is linearly independent, Lin Q ({y j + f i (x j ): j m 1 and f i A j is 1-1}) Lin Q (B 1 {b 1, b 2,..., b s, y 1,..., y n }) {0}, and Lin Q ({y j + f i (x j ): j = 1,..., m 1}) Lin Q (B 1 ) = {0} for all i = 1,..., k. Choose x m A m such that x m i k,f i is 1-1 on A m The choice of x m implies that f 1 i [Lin Q (B 1 {b 1, b 2,..., b s, y 1,..., y n, f i (x 1 ),..., f i (x m 1 )})]. y m + f i (x m ) Lin Q (B 1 {b 1, b 2,..., b s, y 1,..., y n, f i (x 1 ),..., f i (x m 1 )}) for all i k such that f i is 1-1 on A m. This combined with the inductive assumption ( ) and the conditions (1) and (2) leads to the conclusion that {y j +f i (x j ): j = 1,..., m} is linearly independent, Lin Q ({y j + f i (x j ): j m and f i A j is 1-1}) Lin Q (B 1 {b 1, b 2,..., b s, y 1,..., y n }) {0}, and Lin Q ({y j +f i (x j ): j = 1,..., m}) Lin Q (B 1 ) = {0} for all i = 1,..., k. Based on the induction we claim that the sequence x 1,..., x n X has been constructed and it satisfies the following condition: {y j + f i (x j ): j = 1,..., n} is linearly independent and Lin Q ({y j + f i (x j ): j = 1,..., n}) Lin Q (B 1 ) = {0} for all i = 1,..., k. Summarizing, the sequences x 1,..., x n X and y 1,..., y n R satisfying the conditions (a) and (b) have been constructed. Remark 7. Let A A and f 1, f 2 : A R. If (f 1 f 2 )[A] Lin Q (f 1 [A ]) Lin Q (f 2 [A ]), then Lin Q (f 1 [A]) = Lin Q (f 2 [A]).

6 KRZYSZTOF P LOTKA The remark easily follows from the equality l α i f 1 (x i ) = 1 l α i f 2 (x i ) + 1 l α i [f 1 (x i ) f 2 (x i )]. Proof of Theorem 1. Let X be a set of cardinality c. By Lemma 4, it suffices to show that for arbitrary f 1,..., f k : X R there exists a function g : X R such that g + f i is 1-1 and (g + f i )[X] is a Hamel basis (i = 1,..., k). The proof in the general case will be by transfinite induction with the use of the previously stated auxiliary results. However, in the special case when f i [X] < c for all i, it can be presented without the use of induction. The method is interesting and also used in part of the proof of general case, so we present it here. Assume that f i [X] < c for all i, let V = Lin Q ( f i [X]), and λ < c be the cardinality of a linear basis of V. 1 Choose Z X such that Z = λ and f i Z is a constant function for every i and let {b 1,..., b m } = f i [Z]. Next we define a Hamel basis H. Let C be a basis of Lin Q (b 1,..., b m ) from Lemma 2, H 1 be an extension of C to a basis of V, and finally H be an extension of H 1 to a Hamel basis. Define g : X H as a bijection with the property that g[z] = H 1. We claim that g + f i is 1-1 and (g + f i )[X] is a Hamel basis (i = 1,..., k). To see this recall that b j + C is linearly independent, Lin Q (b j + C) = Lin Q (C) = Lin Q ({b 1,..., b m }) (see Lemma 2), and C H 1. This implies that Lin Q (b j +H 1 ) = Lin Q (H 1 ), b j +(H 1 \C) is linearly independent, and as a consequence, b j + H 1 is linearly independent. Therefore, since f i [Z] = {b j } for some j, we have that (g + f i )[Z] = b j + H 1. Thus (g + f i )[Z] is linearly independent and Lin Q ((g + f i )[Z]) = Lin Q (H 1 ). Finally, since f i [X] Lin Q (H 1 ) = Lin Q ((g + f i )[Z]), we can similarly conclude that (g + f i )[X] is linearly independent and Lin Q ((g + f i )[X]) = Lin Q (g[x]) = Lin Q (g[x]) = R. This finishes the proof of the special case. Now we prove the result for arbitrary functions f 1,..., f k : X R. We start by dividing {f 1,..., f k } into abstract classes according to the relation: f i f j iff (f i f j )[X] < c (it is easy to verify that this is an equivalence relation). Put K = i f j f i (f i f j )[X], κ = ω K, and note that κ < c. There exists a set Z X such that Z = κ + and for all i, j the function (f i f j ) Z is one-to-one or constant (the existence of such a set can be shown by using an argument similar to the one from the proof of Lemma 5; obviously, if f i f j, then (f i f j ) Z is constant). Our goal is to define g : Z R for some Z Z such that for every i k g + f i is injective, (g + f i )[Z ] is linearly independent, and K Lin Q ((g + f i )[Z ]). Define V = Lin Q (K) and introduce another equivalence relation among the functions f 1,..., f k : f i = fj iff (f i f j ) Z is constant. Note that =. Let

ON FUNCTIONS WHOSE GRAPH IS A HAMEL BASIS II 7 f i1,..., f il be representatives of the abstract classes of the relation =. Consider l s=1 f j =fis (f j f is )[Z] = {b 1,..., b m }. By Lemma 2, there exists a linear basis C of Lin Q ({b 1,..., b m }) such that b r + C (r m) is also a linear basis for Lin Q ({b 1,..., b m }). Let H 1 be a linear basis of V extending C. Choose a set Z 1 Z such that Z 1 = H 1 and (f j f i1 )[Z 1 ] is linearly independent and Lin Q ((f j f i1 )[Z 1 ]) V = {0} for all f j = fi1. This can be done since Z = κ + > V H 1 and (f j f i1 ) Z is injective for every f j = fi1. Let g 1 : Z 1 H 1 be a bijection and define g : Z 1 R by g = g 1 f i1. Then g+f j is oneto-one for all j, (g + f j )[Z 1 ] is linearly independent for all j, Lin Q ((g + f j )[Z 1 ]) = V for f j = fi1 (see the argument in the special case in the beginning of the proof), and Lin Q ((g + f j )[Z 1 ]) V = {0} for f j = fi1 (the latter follows from the fact that if Y 1 and Y 2 are both linearly independent and Lin Q (Y 1 ) Lin Q (Y 2 ) = {0}, then Y 1 +Y 2 is also linearly independent and Lin Q (Y 1 ) Lin Q (Y 1 +Y 2 ) = Lin Q (Y 1 ) Lin Q (Y 1 +Y 2 ) = {0}). Next choose a set Z 2 Z \ Z 1 such that Z 2 = H 1, (f j f i2 )[Z 2 ] is linearly independent, and Lin Q ((f j f i2 )[Z 2 ]) Lin Q ( k 1 (g + f i)[z 1 ]) = {0} for all f j = fi2 (note that V k 1 (g+f i)[z 1 ] since Lin Q ((g+f i1 )[Z 1 ]) = V ). This choice is possible for similar reasons as in the case of Z 1. Let g 2 : Z 2 H 1 be a bijection and extend g onto Z 1 Z 2 by defining it on Z 2 as g = g 2 f i2. Then g + f j is one-to-one for all j, (g + f j )[Z 1 Z 2 ] is linearly independent for all j, V Lin Q ((g + f j )[Z 1 Z 2 ]) for f j = fi1 or f j = fi2, and Lin Q ((g + f j )[Z 1 Z 2 ]) V = {0} for f j = fi1 and f j = fi2. By continuing this process (or more formally, by using the mathematical induction), we construct a sequence of pairwise disjoint sets Z 1, Z 2,..., Z l Z and a partial real function g : Z R (Z = Z 1 Z l ) such that for each j = 1,..., k, g +f j is one-to-one, (g +f j )[Z ] is linearly independent, and V Lin Q ((g +f j )[Z ]). Observe also that Z κ. Therefore X \ Z = c. In the following part of the proof, we will use the transfinite induction to extend the partial function g onto the whole set X making sure it possesses the desired properties. We will make use of Lemma 6 and Remark 7. First notice that if Z A X and g : A R is any extension of g : Z R then for f j f i we have that ((g + f j ) (g + f i ))[A] = (f j f i )[A] (f j f i )[X] V Lin Q ((g + f i )[Z ]) Lin Q ((g + f j )[Z ]). Hence the remark implies that Lin Q ((g + f i )[A]) = Lin Q ((g + f j )[A]). Thus, when extending the function g it will suffice to consider only the representatives of the abstract classes of the relation. Let f j1,..., f jt be those functions. Let H = {h ξ : ξ < c} be a Hamel basis and {x ξ : ξ < c} be an

8 KRZYSZTOF P LOTKA enumeration of X \ Z. We will define a sequence of pairwise disjoint finite sets {X ξ : ξ < c} such that ξ<c X ξ = X \ Z, x ξ β ξ X β and an extension of g onto X such that for each ξ < c the following condition holds (P ξ ) g + f jr is one-to-one, (g + f jr )[Z β ξ X β] is linearly independent, and h ξ Lin Q ((g + f jr )[Z β ξ X β]) for all r = 1,..., t. Notice that this will finish the proof of our main theorem. To perform the inductive construction, fix α < c and assume that the sets X ξ have been defined for all ξ < α and the function g extended onto Z ξ<α X ξ in such a way that (P ξ ) is satisfied for each ξ < α. If x α Z ξ<α X ξ, then define g(x α ) t r=1 Lin Q((g + f jr )[Z ξ<α X ξ] {f jr (x α )}). This assures that g+f jr is one-to-one and (g+f jr )[Z ξ<α X ξ {x α }] is linearly independent (r = 1,..., t). Next, if h α (g + f j1 )[Z ξ<α X ξ {x α }], then put X α1 =. Otherwise, we apply Lemma 6 to functions f jr f j1 : X \ (Z ξ<α X ξ {x α }) R (r = 2,..., t), B 0 = Lin Q ((g + f j1 )[Z ξ<α X ξ {x α }]), B 1 = Lin Q ( t r=2 (g + f j r )[Z ξ<α X ξ {x α }]), and y = h α. Hence there exist y 1j1,..., y n1j 1 R and x 1j1,..., x n1j 1 X\(Z ξ<α X ξ {x α }) such that the conditions (a) and (b) from the lemma are satisfied. We define X α1 = {x 1j1,..., x n1j 1 } and g(x ij1 ) = y ijr f j1 (x ij1 ) (i = 1..., n 1 ). By repeating the above steps for the other functions f j2,..., f jt (the sets B 0 and B 1 need to be appropriately extended in each step) we obtain pairwise disjoint sets X α1,..., X αt X \(Z ξ<α X ξ {x α } and an extension of g onto Z ξ α X ξ (where X α = X α1 X αt {x α }). Observe that the conditions (a) and (b) from Lemma 6 imply that (P α ) holds. This completes the inductive construction and also the proof of Theorem 1. References [CA] A.L Cauchy, Cours d analyse de l Ecole Polytechnique, 1. Analyse algébrique, V., Paris, 1821 [Oeuvres (2) 3, Paris, 1897]. [C] K. Ciesielski, Set Theory for the Working Mathematician, London Math. Soc. Student Texts 39, Cambridge Univ. Press 1997. [CM] K. Ciesielski, A.W. Miller Cardinal invariants concerning functions whose sum is almost continuous, Real Anal. Exchange 20 (1994-95), 657 672. [CN] K. Ciesielski, T. Natkaniec, Algebraic properties of the class of Sierpiński-Zygmund functions, Topology Appl. 79 (1997), 75 99. [CR] K. Ciesielski, I. Rec law, Cardinal invariants concerning extendable and peripherally continuous functions, Real Anal. Exchange 21 (1995-96), 459 472. [H] G. Hamel, Eine Basis aller Zahlen und die unstetigen Lösungen der Funktionalgleichung f(x + y) = f(x) + f(y), Math. Ann. 60 (1905), 459-462. [HW] W. Hurewicz and H. Wallman, Dimension Theory, Princeton University Press, 1948. [MK] M. Kuczma, An Introduction to the Theory of Functional Equations and Inequalities, Polish Scientific Publishers, PWN, Warszawa, 1985. [P] K. P lotka, On functions whose graph is a Hamel basis, Proc. Amer. Math. Soc. 131 (2003), 1031 1041.

ON FUNCTIONS WHOSE GRAPH IS A HAMEL BASIS II 9 [PR] K. P lotka, I. Rec law, Finitely Continuous Hamel Functions, Real Anal. Exchange, 30(2) (2004-05), 1 4. Department of Mathematics, University of Scranton, Scranton, PA 18510, USA E-mail address: Krzysztof.Plotka@scranton.edu