Int. J. Contemp. Math. Sciences, Vol. 5, 2010, no. 21, Tree Topology. A. R. Aliabad

Int. J. Contemp. Math. Sciences, Vol. 5, 2010, no. 21, 1045-1054 Tree Topology A. R. Aliabad Department of Mathematics Chamran University, Ahvaz, Iran aliabady r@scu.ac.ir Abstract. In this paper, we introduce topological spaces which apparently look like a tree, and we consider the general properties of these spaces. Applying these spaces, we will reconsider some important results, namely, 1- every P -space is C-embedded in a P -space without any isolated points; 2- there exists a topological space X whose every point is an F -point which is not a P -point; 3- there exists a topological space X whose every point is not almost P -point, but Z = X, for every Z Z(X). Mathematics Subject Classification: primary 54C40, secondary 54C45, 54G05, 54G10 Keywords: C(X), P -space, P -point, almost P -point, F -space, F -point, rank of a point 1. Introduction. 1.1. Throughout this article, C(X) stands for the ring of all real valued continuous functions on a Tychonoff space X. For f C(X), Z(f) denotes the set of zeros of f and Coz(f) denotes X \ Z(f). Z(X) (Coz(X)) denotes all Z(f) (Coz(f)), where f C(X). Suppose p βx where βx is the Čeck compactification of X, bym p (X) and O p (X), we mean the ideals {f C(X) : p cl βx Z(f)} and {f C(X) :p int βx cl βx Z(f)}, respectively. It is wellknown that for every prime ideal P of C(X), there exists a unique p βx such that O p (X) P M p (X). We say X is a P -space if every G δ -set in X is open or equivalently M p (X) =O p (X) for every p βx. Also X is an F -space if O p (X) is a prime ideal, for every p βx. Let p βx,wesayp is a P -point (F -point) with respect to X if O p (X) is a maximal ideal (prime ideal) in C(X). We also call p an almost P -point if for every f C(X) if whenever p cl βx Z(f), then int βx cl βx Z(f). A subset S of X is C-embedded (C -embedded) in X if every real valued (bounded real valued ) continuous

1046 A. R. Aliabad function on S has a continuous extension on X. The reader is referred to [4] for undefined terms and notations. Our main aim in this article is to give a method to construct topological spaces with unusual properties, which turns out to be useful in dealing with some important results. An outline of the article is as follows. In Section 1, we will show that how we can construct a tree topology, where a family of topological spaces have been given. Then we will observe that the new topology inherits most of the important topological properties from the background topological spaces. Clearly, if X is a P -space and x X, then each continuous function at x is constant on a neighborhood of x. P -spaces play a crucial role in the context of C(X). Examples of P -spaces with isolated points are abundant and easy to construct, see 4J, 4K in [4]. However, examples of these spaces without isolated points are very rare and difficult to construct. In Section 2, we present another method for constructing Tychonoff P -spaces without isolated points. We simply show that any P -space can be C-embedded into a P -space without isolated points. Applying this method to certain topological P -spaces, we get a P -space X without isolated points such that each point in X has a local base consisting of open subsets each of which is homeomorphic with X. Also in Section 2, we give an example of topological space whose every point is an F -point, but not a P -point. Finally, in this section, we construct a topological space X in which every point is not an almost P -point while Z = X for every Z Z(X). We construct our structure in two steps, as follows. Step 1: Let D be the set of all isolated points of a topological space X and {Y d } d D be a family of topological spaces which are mutually disjoint. We choose an element in Y d, renaming it as d and pasting Y d and X at the point d. To avoid the trivial case, we assume that d is a non isolated point of Y d. Supposing Y = X {Y d } d D as the quotient space, it is clear that a subset V of Y is open in Y if and only if V X is open in X and V Y d is open in Y d for every d D. Also the neighborhood base at d in Y is the same in Y d. In particular, for every d d in D, Y d and Y d are two separated open sets in Y containing d and d, respectively. It is obvious to see that if U is an open subset of X and W d is an open neighborhood of d in Y d for any d U D, then V = U ( d U D W d ) is an open set in Y such that V X = U. We need the followings for the next step. Proposition 1.2. The following statements are true for the above topological space Y. (a) Y is Hausdorff if and only if X and Y d s are too.

Tree topology 1047 (b) Y is regular if and only if X and Y d s are too. (c) f R Y is continuous if and only if the restrictions of f on X and Y d s are too. So X and Y d s are C embedded in Y. (d) Y is completely regular if and only if X and Y d s are too. Moreover, in this case, if A is a closed subset of Y, c X \ A and f C(X) with f(c) =0 and f(a X) ={1}, then f can be extended to f C(Y ) so that f(c) =0 and f(a) ={1}. (e) Y is normal if and only if X and Y d s are too. (f) Y is a P -space if and only if X and Y d s are too. Proof. (a) and (b) are simple to prove. (c) Suppose f R Y and the restrictions of f on X and Y d s are continuous. Since Y d s are open in Y, it is enough to show that f is continuous at every point x X. Let x X and ɛ>0begiven. Since f X is continuous, there is an open neighborhood U of x in X such that f f(x ) ɛ/2 on U. Also, since the restrictions of f on Y d s are continuous, there exists an open neighborhood W d of d in Y d such that f f(d) ɛ/2 onw d, for every d U D. Taking V = U ( d U D W d ), V is an open neighborhood of x in Y and f f(x ) ɛ on V. Hence f is continuous at x. (d) Suppose X and Y d s are completely regular, A is a closed subset of Y and c/ A. If c/ X, then clearly we are done; otherwise, there exists f C(X) such that f(c) = 0 and f(a X) ={1}. For every d D there is an f d C(Y d ) such that f d (d) =f(d) and f d (A Y d )={1}. Define { f(x) if x X g(x) = f d (x) if x Y d By part (c), g is continuous extension of f, g(c) = 0 and g(a) ={1}. The converse is clear. (e) Let A and B be two closed subsets of Y such that A B =. Since A Y d and B Y d are two disjoint closed sets in Y d, there are two disjoint open sets W da and W db in Y d such that A Y d W da and B Y d W db.in addition, we can choose these two open sets in such away that if d/ A B, then d / W da W db. On the other hand, since A X and B X are two disjoint closed sets in X, there are two disjoint open sets U A and U B in X such that A X U A and B X U B. Now define V A = U A ( d D W da ), V B = U B ( d D W db ). Clearly V A and V B are two disjoint open sets in Y containing A and B. Therefore Y is a normal space. Moreover, note that we can obviously observe that The converse is easy to prove. (f) It is clear, see 4K.5 in [4]. V A X = U A, V B X = U B.

1048 A. R. Aliabad Lemma 1.3. Let X and Y be as in Step 1 and A Y, then X cl Y A =cl X (D cl Y A) cl X (A X). Proof. It is enough to show that X cl Y A cl X (D cl Y A) cl X (A X). Suppose that x/ cl X (D cl Y A) cl X (A X), then there exists an open set V in X such that x V, V D cl Y A =,V A X =. Therefore, for every d V D there exists an open set W d in Y d such that W d cl Y A =. Putting W V =( d V D W d ) V, we clearly see that W V is open in Y, x W V and W V A =. Consequently, x/ cl Y A. Proposition 1.4. Let X, Y d s and Y be as in Step 1, then Y is extremaly disconnected if and only if X and Y d s are too. Proof. ( ) Suppose U is an open subset of X. Put V =( d D U Y d ) U. Clearly V is an open subset of Y and by Lemma 1.3, X cl Y V =cl X U. Since cl Y V is open in Y,cl X U is also open in X. Hence X is extremaly disconnected. Now, suppose U is an open subset of Y d. Since Y d is open-closed in Y, it follows that U is also open Y. Therefore, cl Y U =cl Yd U is open in Y and consequently is open in Y d. Hence Y d is extremaly disconnected, for every d D. ( ) Suppose V is an open subset of Y, then by Lemma 1.3, X cl Y V = cl X (D cl Y V ) cl X (V X). Since V X and D cl Y V are open in X, it follows that X cl Y V is open in X. On the other hand, cl Y V = Y cl Y V =(( d D Y d ) cl Y V ) (X cl Y V ) =( d D (Y d cl Y V )) (X cl Y V )=( d D cl Yd (V Y d )) (X cl Y V ). It follows that cl Y V is open in Y. Hence Y is extremaly disconnected. Similarly, we can show that if Y is basically disconnected, then X and Y d s are too. But the the converse is not true, see example 2.9. Step 2. Let T 1 be a topological space, D 1 be the set of all isolated points of T 1 and {Y d } d D1 be a family of topological spaces which are mutually disjoint. We construct T 2 as it was done for Y in Step (1) and do the same on T 2 to construct T 3 and so on. Now we put T = n N T n, τ T = {U T U T n τ Tn, n N} Next, we present some of the basic properties of the spaces constructed in Step 2. Proposition 1.5. The following statements hold for the above topological space T. (a) T does not contain any isolated point.

Tree topology 1049 (b) A T is closed if and only if A n = A T n is closed in T n for all n N. (c) T n is a subspace and closed subset of T for every n N. (d) f C(T ) if and only if f n = f Tn is continuous for every n N. (e) T n is a C-embedded closed subset in T for every n N. (f) T is Hausdorff if and only if the background topological spaces T 1 and Y d s are too. (g) T is regular if and only if the background topological spaces T 1 and Y d s are too. (h) T is completely regular if and only if the background topological spaces T 1 and Y d s are too. (i) T is a normal space if and only if the background topological spaces T 1 and Y d s are too. Proof. (a) and (b) are easy to prove. (c) Suppose U τ T, then by our definition, U T n τ Tn. Conversely, suppose U n τ Tn. If we put U = U n (T \ T n ), then U n = U T n and it is enough to show that U is open in T. For every m N with m n, T m is a subspace of T n and consequently U T m = U n T m is open in T m ; otherwise, if n<m, then U T m = U n (T m \ T n )isopenint m. Therefore, U τ T. Hence T n is a subspace of T for every n N. Note that T m is closed in T m+1 and by induction T m is closed in T n for every m, n N, where m n. So, supposing m N, then T m T n is closed in T n for every n N and therefore T m is closed in T for every m N. (d) suppose U is an open set in R. Obviously f 1 (U) T n = fn 1 (U) isopen in T n for every n N and so by Step 2, f 1 (U) isopenint. The converse is clear. (e) Let m N and f m C(T m ). By part (c) of Proposition 1.2, f m has a continuous extension f m+1 C(T m+1 ). We continue this process inductively. Now, define f : T R such that f(x) =f n (x) where x T n. Clearly f is well-defined and by part (d), it is continuous. (f)it is obvious. (g) The proof of (g) is similar to part (i). (h) Suppose A is a closed set in T and c/ A. By part (d) of Proposition 1.2, T n is completely regular for every n N. Assuming c T m, also by part (d) of Proposition 1.2, for every n m there exists f n C(T n ) such that f n (c) =0,f n (A T n )={1} and f n+1 Tn = f n for every n m. Now we define f : T R so that f(x) =f n (x) where x T n. Clearly f C(T ), f(c) =0 and f(a) ={1}. The converse is evident. (i) Let A and B be two disjoint closed sets in T. By part (e) of Proposition 1.2, T n is a normal space for every n N. Since A n = A T n, B n = B T n are two disjoint closed sets in T n, then as it is seen in the proof of part (e) of Proposition 1.2, for every n N, there exist two disjoint open sets U n and V n

1050 A. R. Aliabad in T n such that, A n U n,b n V n, U n = U n+1 T n, V n = V n+1 T n. Now, we put U = n N U n and V = n N V n. It is clear that U and V are two disjoint open sets in T containing A and B, respectively. Therefore, T is a normal space. The converse is clear, since T n is a closed subset of T for every n N. As a remark, suppose D n is the set of all isolated points of T n. By T dn we mean the subspace of T which is constructed next over the point d n D n. One can see that T dn is a closed-open subset of T, and if we take the background topological spaces as disjoint copies of one, then T and T dn are homeomorphic for any n N. Proposition 1.6. Let Y d s, T n and T be as in Step 2, then T is extremaly disconnected if and only if the background topological spaces T 1 and Y d s are too. Proof. By Proposition 1.2 and using induction, it is enough to show that T is extremaly disconnected if and only if T n is so, for every n N. To see this, suppose T is extremaly disconnected and U is an open subset of T n. Put V =( dn Dn UT dn ) U, then similar to Proposition 1.4, we see that V is open in T and cl Tn U =cl T V T n and consequently cl Tn U is open in T n. Hence T n is extremaly disconnected. Conversely, suppose T n is extremaly disconnected, for every n N, and V is open in T. Clearly, similar to Lemma 1.3, we see that T n cl T V =cl Tn (D n cl T V ) cl Tn (V T n ). By hypothesis, cl Tn (D n cl T V ) and cl Tn (V T n ) are open in T n, for every n N and so T n cl T V is open in T n, for every n N. Therefore, cl T V is open in T. 2. Some applications. In this Section, we will give some useful applications of the tree topology. Proposition 2.1. In Step (2) of Section 1, if we choose the background topological spaces X and Y d s as P -spaces, then T = n N T n is a P -space with no isolated point. Proof. Suppose G is a G δ -set in T. Clearly G T n is a G δ -set in T n for every n N. Now, by Proposition 1.2 (f), T n s are P -spaces and then G T n is open in T n for every n N. Hence G is open in T. An immediate consequence of Proposition 1.5 and Proposition 2.1 is the following theorem.

Tree topology 1051 Theorem 2.2. Every P -space is a C-embedded closed subset of a P -space without isolated points. Now, we are able to construct the following interesting topological space. Corollary 2.3. Let m be an arbitrary uncountable cardinal number. Suppose Y = D {a} with Y = m such that all points of D are isolated and each neighborhood of a is co-countable. Taking copies of Y in Step 2 as the background topological spaces X = T 1 and Y d s, we obtain the topological space T = n N T n with the following properties. (a) T is a Hausdorff and normal P -space without isolated points. (b) Each point of T has a local base consisting of open sets any of which is homeomorphic with T. Proof. (a) It is obvious by Proposition 1.5 and Proposition 2.1. (b) Let x V τ T and k be the smallest number for which x is a limit point in T k. Without loss of generality, we can suppose k = 1 i.e., x = a. Since each neighborhood of a in Y is homeomorphic with Y, Clearly there exists a homeomorphism ϕ 1 : U 1 T 1 where U 1 is an open neighborhood of a in T 1 such that U 1 V. For every a d U 1, consider an open set W d in Y d such that W d V, then U 2 = U 1 ( d U1 W d ) is an open neighborhood of a in T 2, U 2 V and U 2 T 1 = U 1. Similarly, each W d is homeomorphic with Y d and so there exists a homeomorphism ϕ 2 : U 2 T 2 in such a way that ϕ 2 U1 = ϕ 1. Continuing inductively, we will obtain sequences {U n } n N and {ϕ n } n N such that for every n N, U n is open neighborhood of a in T n contained in V, U n+1 T n = U n, ϕ n is a homeomorphism from U n onto T n and ϕ n+1 Un = ϕ n. Now, clearly U = n N U n is an open neighborhood of a in T contained in V and if we define ϕ : U T with ϕ(x) =ϕ n (x) whenever x U n, then clearly ϕ is well-defined and it is a homeomorphism. Remark 2.4. The topological space T that is constructed above is cardinaly connected, in the sense that the cardinal number of each nonempty open set in T is equal to T. Now we are going to construct a topological space X such that rk(x, x) =1 for every x X. First, we need some preliminaries. Definition 2.5. Let X be topological space and p βx. Suppose the number of non maximal minimal prime ideal contained in M p (X) is equal to n N, then we say the rank of p with respect to X is equal to n N, briefly rk(x, p) =n. For further information, one can see [5]. Note that our definition has some differences from [5].

1052 A. R. Aliabad Clearly, if p βx, then rk(x, p) 1 if and only if p is an F -point with respect to X, and rk(x, p) = 0 if and only if p is a P -point with respect to X. Every basically disconnected space is an F -space. In addition, if S is a C -embedded subset of X and p cl βx S = βs, then rk(s, p) rk(x, p). Example 2.6. Suppose X is a discrete topological space with non measurable cardinal and σ βx \ X. Put Σ(X, σ) =X {σ}. Obviously every point of X is isolated and each neighborhood of σ is of the form U {σ}, where U U= Z(M σ (X)) = {Z(f) : f M σ (X)}. In fact, this is the extension of the topological space Σ introduced in [4]4M and it has all the properties that have been listed there. For instance, Σ(X, σ) is an extremaly disconnected space and consequently it is basically disconnected. This implies that Σ(X, σ) is an F -space, see [4]14N. In addition, since X has a non measurable cardinal, by [4]12.2, X is real compact and so M σ (X) is hyper real. Therefore, by [4]5.14, U has not the countable intersection property. Hence, there exists {U n : n N} Usuch that n N U n =. Clearly, Z n = U n {σ} is a closed-open subset of Σ(X, σ) and so is a zero set, for every n N. Thus, {σ} = n N Z n is a zero set in Σ(X, σ). Therefore, σ is an F -point which is not P -point; i.e., rk(σ) =1. Theorem 2.7. Let m be an infinite cardinal number. There exists a topological space X such that X = m and rk(x, x) = 1 for every x X, i.e., every point of X is an F -point which is a non P -point. Proof. Suppose D is a discrete space with cardinal number m. Letting X and Y d s be the copies of (D, σ), we claim that the topological space T constructed as in Step 2 is the desired space. Since (D, σ) is extremally disconnected, by Proposition 1.6, T is also extremally disconnected and consequently T is an F -space. Therefore, rk(t,t) 1 for every t T. On the other hand, suppose that t T, then clearly there exists a Y d such that t is the unique non isolated point of Y d. Since Y d is a copy of Σ(D, σ), it follows that rk(y d,t) = 1. Clearly Y d is C-embedded in T and consequently rk(t,t) 1. Therefore, rk(t,t) = 1 for every t T. Certainly, by using the tree topological space, we can find some other examples with different purposes. For instance, if m and n are two cardinal numbers such that n is an infinite cardinal and m n, then we can construct a topological space X with cardinal number n such that rk(x, x) m, for every x X (note that the concept of rank can be defined for any cardinal number, see Definition 4.3 of [2] ). To see this, let A be a set with cardinal number n. By the above discussion, the rank of point σ in Σ(A, σ) is equal to one. Suppose Λ = m and {X λ } λ Λ is a family of disjoint copies of Σ(A, σ). By pasting X λ s at σ, we will obtain a topological space X such that by Theorem 4.4 of

Tree topology 1053 [2], rk(x, σ) = m. Now, we construct a tree topology T whose background topological spaces are disjoint copies of X. Suppose t T, then there exists a C-embedded copy of X in T such that t X and t has the role of σ in X. Therefore, rk(t,t) rk(x, t) =m. Suppose X, Y d s and Y are as in Step 1. If f 1 C(X), Z(f 1 )=A Z(X) and Z d Z(Y d ) for every d A D, then clearly there exists f 2 C(Y ) such that Z(f 2 )=( d D Z d ) A and f 2 X = f 1 Theorem 2.8. Let m be a cardinal number c. There exists a topological space X such that X = m and every point of X is not almost P -point, but Z = X for every Z X. Proof. Suppose D is an uncountable discrete space with cardinal number m and X = D {x} is the one point compactification of D. Clearly x is not a P -point in X and if x Z Z(X), then Z = m. Let T be the tree topology whose background topological spaces are disjoint copies of X. We claim that T is the desired topology. If Z Z(T ), then clearly Z = m = T. Now, suppose t T, and k is the smallest number for which t is a non isolated point of T k. We must show that t is not an almost P -point. Clearly t is a non P -point in T k, hence there exists a Z(f k )=Z k Z(T k ) such that t / int Tk Z k. Let D k be the set of isolated points of T k, for every d k D k Z k, there exists a Z dk Z(Y dk ) such that d k / int Ydk Z dk. So, we can find f k+1 C(T k+1 ) such that f k+1 Tk = f k, Z(f k+1 )=( dk D k Z k Z dk ) Z k and T k int Tk+1 Z(f k+1 )=. We can continue this process, by induction, and we will have a sequence of functions like f k,,f n, such that f n C(T n ), f n+1 Tn = f n and T n int Tn+1 Z(f n+1 )= for every n k. By part (d) of Proposition 1.5, there exists f C(T ) such that f Tn = f n for every n k. It is enough to show that int T (Z(f)) =. Clearly int T (Z(f)) = n k (int T (Z(f)) T n ) n k (int Tn+1 Z(f n+1 ) T n =. Hence t Z(f) and int T (Z(f)) = ; consequently, t is not an almost P - point. We conclude the paper with an example by which we show that if the background spaces X and Y d s in Step 1 and Step 2 are basically disconnected, then the constructed topological space need not be basically disconnected. Theorem 2.9. Suppose D is an uncountable discrete space and X = D {a} is the one point compactification of D and Y d s are disjoint copies of (A, σ) where A is a discrete space with nonmeasurable cardinal. Clearly X and Y d s are basically disconnected. We claim that the topological space Y constructed in Step 1 is not basically disconnected. To see this, suppose D 1 is an uncountable subset of D such that X \D 1 is also uncountable. We can easily

1054 A. R. Aliabad define a function f C(Y ) in such away that V = Coz(f) =( d D1 Y d ) \ D 1. Obviously, cl Y V X = D 1 {a} is not open in X. Therefore, cl Y V is not open in Y. We can easily see that the same is true for topological spaces occurred in Step 2. References [1] A. R. Aliabad, z -ideals in C(X), Ph.D. thesis, 1996. [2] A. R. Aliabad, Pasting topological spaces at one point, zechoslovak Mathematical Journal, 56 (131) (2006), 1193-1206 [3] R. Engelking, General topology, PWN-Polish Scientific Publishing, 1977. [4] L. Gillman and M. jerison, Rings of continuous functions, Van. Nostrand Reinhold, New York, 1960. [5] S. Larson, f-rings in which every maximal ideal contains finitely many minimal prime ideals, Communications in algebra, 25(12) (1997), 3859-3888. [6] S. Willard, General Topology, Addison Wesly, Reading, Mass., 1970. Received: October, 2009