Mathematics Today Vol.29(June-Dec-2013)25-30 ISSN 0976-3228 ESSENTIAL SET AND ANTISYMMETRIC SETS OF CARTESIAN PRODUCT OF FUNCTION ALGEBRAS H. S. MEHTA, R. D. MEHTA AND D. R. PATEL Abstract. Let A and B be function algebras on compact spaces X and Y respectively. Then the Cartesian product A B is also a function algebra on the topological sum X + Y. In this paper, we discuss sets of antisymmetry and essential set of A B. We prove that the collection of maximal sets of antisymmetry for A B is the disjoint union of collection of maximal sets of antisymmetry for A and B. Also we prove that E A B = E A E B, where E A denotes the essential set for A. 1. Introduction A function algebra on a compact Hausdorff space X is a closed subalgebra of C(X), containing constant functions and separating the points of X. If A is a function algebra on X and B is a function algebra on Y, then as we know A B is a commutative Banach algebra with identity (1 X, 1 Y ) and norm (f, g) = max{ f, g } where 1 X and 1 Y are identities of A and B respectively and is supnorm [4]. Definition 1.1. [4] Let (X, τ X ) and (Y, τ Y ) be two topological spaces with X Y =. Then on the disjoint union X Y, the sum topology is defined as follows: τ X Y = {G X Y : G X τ X and G Y τ Y }. We denote this topological space by X + Y. If X and Y are compact and Hausdorff, then X + Y is also compact and Hausdorff [3]. Further, There is an algebra isomorphism from C(X) C(Y ) onto C(X + Y ). In fact, for f C(X) and g C(Y ), h = (f, g) C(X + Y ) is defined as f(z), if z X; h(z) = g(z), if z Y. i.e. C(X + Y ) = C(X) C(Y ). 2010 Mathematics Subject Classification. 46J10. Key words and phrases. Cartesian product of function algebras, Sets of antisymmetry, Essential set.
26 Mathematics Today Vol.29(June-Dec-2013)25-30 If A and B are function algebras on X and Y respectively, then it is easy to check that A B is a subalgebra of C(X + Y ) [3], it contains constants and separates the points of X + Y. Also, as h = max{ f, g } = h for h = (f, g) A B, A B is closed in C(X + Y ). Thus A B is a function algebra on X + Y. In next result, we prove that the converse is also true. Theorem 1.2. Let A and B be subalgebras of C(X) and C(Y ) respectively and A B be a function algebra on X + Y. Then A and B are function algebras on X and Y respectively. Proof. Suppose that A B is a function algebra on X + Y. Note that f (f, 0) and g (0, g) are isometries from A to A B and B to A B respectively. i.e. f = f A = (f, 0) and g = g B = (0, g). Since maximum norm is complete, A and B are complete with respect to norm. Therefore A and B are closed subalgebras of C(X) and C(Y ) respectively. It is clear that 1 X = 1 Y = constant function 1. So A and B contain the constants. To show that A and B separates the points of X and Y respectively, let x 1, x 2 X with x 1 x 2. Then h = (f, g) A B h(x 1 ) h(x 2 ). But h(x 1 ) = f(x 1 ) and h(x 2 ) = f(x 2 ). Therefore f(x 1 ) f(x 2 ). Thus, A separates the points of X. Similarly, B separates the points of Y. Hence A and B are function algebras on X and Y respectively. Now onwards, we shall assume that A and B are function algebras on X and Y respectively. For concepts regarding function algebras we refer to ([1], [2]). 2. Sets of antisymmetry We know that if A and B are function algebras on X and Y respectively, then A B is a function algebra on X + Y. In this section, we discuss the relation between sets of antisymmetry for A and B with that of A B. We recall the definition of an antisymmetric set for A. Definition 2.1. [2] Let A be a function algebra on X. A set K X is said to be a set of antisymmetry for A if f A and f K is real, then f is constant on K. A is said to be antisymmetric algebra if X is a set of antisymmetry for A.
H. Mehta, R. Mehta & D. Patel - Essential set and antisymmetric sets... 27 Proposition 2.2. Let K 1 X and K 2 Y be sets of antisymmetry for A and B respectively. Then K 1 and K 2 are sets of antisymmetry for A B. Proof. We prove that K 1 is a set of antisymmetry for A B. Let h = (f, g) A B be real valued on K 1. We will show that h is constant on K 1. For any z K 1, h(z) = f(z). So f is real valued on K 1. Since K 1 is a set of antisymmetry for A and f A, f is constant on K 1. But h(z) = f(z), z K 1. Therefore h is constant on K 1. Thus K 1 is a set of antisymmetry for A B. Similarly, K 2 is also a set of antisymmetry for A B. Proposition 2.3. Let K X + Y be a set of antisymmetry for A B. Then K X and K Y are sets of antisymmetry for A and B respectively. Proof. We show that K X is a set of antisymmetry for A. Let f A and f be real valued on K X. Then h = (f, 0) A B. Also h = f on K X and h = 0 on K Y. So h is real valued on K. Since K is a set of antisymmetry for A B, h is constant on K. But h(z) = f(z), z K X and therefore f is constant on K X. Thus K X is a set of antisymmetry for A. Similarly, K Y is a set of antisymmetry for B. Remark 2.4. In general, if A and B are antisymmetric algebras, then A B is not an antisymmetric algebra as the function (1, 0) A B is real valued on X + Y but it is not constant on X + Y. Proposition 2.5. Let K be a set of antisymmetry for A B. Then either K X = or K Y =. Proof. Suppose K X and K Y. Let x K X and y K Y. Consider the function h = (1, 0) A B. Then h is real valued on X + Y and hence on K, h(x) = 1 and h(y) = 0 which contradicts the fact that K is a set of antisymmetry for A B. Therefore we must have either K X = or K Y =. Let K (A) denote the collection of all maximal sets of antisymmetry for A. Theorem 2.6. K (A B) = K (A) K (B). Proof. Let K K (A). Then by Proposition 2.2, K is a set of antisymmetry for A B. Since every set of antisymmetry is contained in some maximal set of antisymmetry, there exits a maximal set of antisymmetry F X + Y for A B such that K F. Clearly K F X and by Proposition 2.3, F X is a set of antisymmetry for A. Since K
28 Mathematics Today Vol.29(June-Dec-2013)25-30 is maximal, we must have K = F X. Therefore by Proposition 2.5, F Y =. So F = F X = K. Hence K is a maximal set of antisymmetry for A B. i.e. K K (A B). Similarly, if K K (B), then also K K (A B). Therefore K (A) K (B) K (A B). Conversely, let K K (A B). Then by Proposition 2.5, either K X = or K Y =. Suppose K Y =. Then K = K X is a set of antisymmetry for A, by Proposition 2.3. So there exits a maximal set of antisymmetry F X for A such that K X F. Then by above discussion F K (A B). Therefore K = F. Thus K K (A). Similarly if K X =, then we get K Y = K K (B). i.e. either K K (A) or K K (B). Therefore K (A B) K (A) K (B). Hence K (A B) = K (A) K (B). 3. Essential set In this section we relate the essential set of A B with that of A and B. Recall that the essential set for A is the hull of the largest closed ideal of C(X) contained in A [2]. So first we relate closed ideals of C(X) C(Y ) and C(X +Y ). We know that any closed ideal I of C(X) is of the form I F, for some closed subset F of X, where I F = {f C(X) : f F = 0}. Also note that hull I F = F [2]. Proposition 3.1. Every closed ideal in C(X +Y ) is the product of closed ideals in C(X) and C(Y ). Proof. Let I F and I K be closed ideals of C(X) and C(Y ) respectively. Then I F I K is a closed ideal of C(X + Y ), where F and K are subsets of X and Y respectively [4]. We shall prove that I F K = I F I K. Let h = (f, g) I F K. Then h F K = 0. For every z F, f(z) = h(z) = 0 and so f I F. Similarly g I K. Thus (f, g) I F I K. Therefore I F K I F I K. Now I F I K = {(f, g) C(X + Y ) : f F = 0, g K = 0}. Let (f, g) I F I K. Then f I F and g I K. i.e. f F = 0 and g K = 0. i.e. (f, g) F K = 0. i.e. (f, g) I F K. Therefore I F I K I F K. Hence I F I K = I F K.
H. Mehta, R. Mehta & D. Patel - Essential set and antisymmetric sets... 29 Conversely, let I H be a closed ideal of C(X+Y ) with H be a closed subset of X+Y. Let H X = F and H Y = K. Then F and K are closed subsets of X and Y respectively. Further, I H = I (H X) (H Y ) = I F K = I F I K. Thus every closed ideal of C(X + Y ) is of the form I F I K including the case of either of F = or K =. i.e. I F = C(X) or I K = C(Y ). We shall denote by E A, the essential set for A; this means I EA = {f C(X) : f EA = 0} is the largest closed ideal of C(X) contained in A. Theorem 3.2. E A B = E A E B. Proof. By Proposition 3.1, I EA I EB clearly contained in A B. So by definition of essential set, = I (EA E B ) is a closed ideal of C(X + Y ) and it is I (EA E B ) I EA B or E A B E A E B Conversely, I EA B = I F I K (by Proposition 3.1) where F = E A B X, K = E A B Y. Further I F A and I K B, as I EA B A B. Therefore again by definition of essential set, I F I EA, and I K I EB. Therefore I EA B I EA I EB = I EA E B. Therefore E A E B E A B. Since E A = X P A, where P A is the union of all singleton maximal sets of antisymmetry for A [2], the above result can also be proved using Theorem 2.6. Remarks 3.3. (i) If A and B are essential algebras, then A B is also an essential algebra, by Theorem 3.2. (ii) Every antisymmetric algebra is essential [1], but if A and B are antisymmetric, then A B is essential but not antisymmetric. Acknowledgement: The research is supported by the SAP programme to the department by UGC.
30 Mathematics Today Vol.29(June-Dec-2013)25-30 References [1] Andrew Browder, Introduction to Function Algebras, W.A.Benjamin, Inc., New York, 1969. [2] G.M.Leibowitz, Lectures on Complex Function Algebras, Scott, Foresman and Co., 1970. [3] R.D.Mehta and Nirav Shah, Chacterazation of maximal ideal space and Šilov boundary for cartesian product of commutative Banach algebras, Prajna (J. of Sardar Patel University), 17 (2009), 149-154. [4] Nirav Shah, Cartesian Product of Banach Algebras, M. Phil. Dissertation, Sardar Patel University, 2007. Department of Mathematics, Sardar Patel University, Vallabh Vidyanagar 388120, India E-mail address: himalimehta63@gmail.com E-mail address: vvnspu@yahoo.co.in E-mail address: satkaival2301@gmail.com