Some topologies in the half-plane

Transcription

1 Thai Journal of Mathematics Volume 5(2007) Number 2 : Some topologies in the half-plane M. Grzesiak and L. Jankowski Abstract : In the complex analysis the existence of limits of a function at the boundary of the region is a problem and various limits are considered. We present four topologies in the upper half-plane connected with different approach regions. Their general topology properties are examined. In particular it is proved that all four spaces are sequential but no one of them is normal; some properties connected with compactness are shown. Keywords : topology, sequential space, realcompact space, Čech complete space Mathematics Subject Classification : Primary 54A10, secondary 54D55. 1 Introduction There are some kinds of convergence which are not explicitly connected with any topology. In such a case a question arises whether there is a topology τ such that τ-convergence is equivalent to the original one. There are examples showing that sometimes the convergence cannot be topologized. To see this consider the lin operation introduced by Klee ([2]). Let L be a linear space. For x, y L let [y, x] be the segment including its endpoints, (y, x] = [y, x] \ {y}. Then for X L we define linx = X {y L : (y, x] X for some x X} Now take L = C and define X = {x + iy : x ( 1, 0), y (0, 1) \ Q} {x + iy : x (0, 1), y (0, 1) Q}. Then lin(linx) linx because 0, i lin(lin X) \ linx. It follows that there is no topology for which lin A = A. In the complex function theory we have the following Abel theorem ([5],2.12): If the series F(z) = n a nx n is convergent for z = 1 then F(z) F(1) when the point z tends to 1 from within the circle K(0; 1) in such a way that expression 1 z /(1 z ) remains bounded. Geometrically, this condition means that z should remain in some region cut out from the circle of convergence by a pair of chords with origin at the point 1. This is so-called angular limit; we also say that z 1 non-tangentially.

2 344 Thai J. Math. 5(2007)/ M. Grzesiak and L. Jankowski There is also another problem. Let U(z) be a function analytic in an open region Ω. As a rule, U(z) does not extend continuously on Ω. However, in some special cases (e.g. when Ω is the disk or a half-plane) the limit exists for almost every boundary point z 0 if z z 0 non-tangentially. The limit also exists if z z 0 along a line joining center with z 0 ; this is radial limit. For example, let U(z) be a function which is harmonic in {z : z < 1} and has one of the representations U(re iθ ) = 1 π P r (θ t)f(t)d(t), 2π π F L p ( π, π), U(re iθ ) = 1 π P r (θ t)dµ(t), 2π π where P r (θ t) is the Poisson s kernel and µ a finite signed measure on [ π, π]. Then the study of the pointwise behaviour of U(z) as z tends to points e iθ on the boundary of the unit circle requires a detailed study of Poisson s kernel. Its boundary behaviour depends on the way z tends to the point e iθ. Various problems connected with angular and radial limits are discussed e.g. in [4]. These types of convergence can be considered in slightly different settting. Instead of the unit disk we can take the closed upper half-plane. Then the boundary points form the x-axis. The problems, however, remain essentially the same. It shows up that the above mentioned convergences can be topologized (in the sense mentioned at the begining). We will examine the properties of related topologies. From the above-mentioned examples is clear that if x is a boundary point then the existence of the limit depends on the proper choice of the approach region. So in this paper we will consider four topologies, constructed for various types of approach region and examine their properties from general topology point of view. We believe that knowledge of these properties will help in better understanding of the problems in analytic function theory. Similar problems concerning the boundary behaviour in the upper half-space R n+1 + = {(x, y) : x Rn, y > 0} are considered for example in [6]. We will use the following notation. Let X, X 0 denote the closed and open upper half-plane respectively, i.e. X = {z : Im z 0}, X 0 = {z : Im z > 0}. For u, v C the symbols (u, w), [u, v], [u, v), (u, v] will denote the segments in the plane with endpoints u, v (open, closed, etc. ). Also I = [0, 1], R + = [0, ). For z C, z := Rez + i Imz denotes the complex number symmetric to z with respect to the imaginary axis. K(z 0, r) = {z : z z 0 < r} is an open disk. In general our terminology follows [1]; in particular a neighborhood is always an open set.

3 Some topologies in the half-plane Definitions and basic properties We define topologies τ j on X for j = 1, 2, 3, 4. Definition 1. For j = 1, 2, 3, 4 τ j X 0 is the euclidean topology. For z X \ X 0 (that is, z real) the basis of neighborhoods is {z + V }, V B where B is the basis of neighborhoods of 0. The basis B consists of sets V = {0} U, where U X 0 is open in the euclidean topology and fulfills the condition: 1) for j = 1: U contains the segment (0, yi) for some y > 0; 2) for j = 2: for every z X 0 the set U contains the segment (0, t z z) for some t z (0, 1); 3) for j = 3 and any fixed α (0, π/2): U contains an open triangle with vertices 0, re iα, re i(π α) for some r > 0; 4) for j = 4: U = Int(conv{0, z n, z n }) n=1 for some sequence (z n ) with Re z n > 0, arg z n ց 0. Note that τ 3 is defined with the parameter α, so in fact we have the family of topologies, but for any two angles α 1, α 2 the respective topological spaces are homeomorphic. Remark. It is easy to note some similarity of these definitions with the definition of Niemytzki plane (see e.g. [1]). However, as we will see, none of these topologies is homemorphic with the Niemytzki plane. As the set {0} K(ri, r) is τ 4 -open, the topology τ 4 is stronger than the topology of Niemytzki plane. Then from Proposition 2 below it follows that every τ j, j = 1, 2, 3, 4 is essentially stronger than Niemytzki plane. The following proposition is easy to prove. Proposition 1. For j = 1, 2, 3, 4: 1. The topology induced by τ j on the real axis is discrete; 2. Every subset of the real axis is closed in τ j. Proposition 2. We have: τ 4 τ 3 τ 1, τ 4 τ 2 τ 1. However, there is no inclusion between τ 3 and τ 2. Proof. The inclusions are clear. Let V = X 0 \ {e iφ + 1 : φ (0, π)}. Then {0} V is open in τ 2 but not in τ 3. If W = X 0 \ { 1 n eiα : n N}, then {0} W is open in τ 3 but not in τ 2. It follows that τ 3 and τ 2 are incomparable. As τ j X0 is separable we have the following proposition. Proposition 3. τ j is separable for j = 1, 2, 3, 4.

4 346 Thai J. Math. 5(2007)/ M. Grzesiak and L. Jankowski The Lindelöf number (see [1]) of a space Y is the smallest cardinal m such that any open cover of Y has a subcover of cardinality m; The symbol e(y ) denotes the smallest cardinal m such that for any closed set A Y of isolated points is carda m. Proposition 4. For j = 1, 2, 3, 4: 1. The Lindelöf number of (X, τ j ) is c. 2. e(x, τ j ) = c. Proof. By Proposition 1 τ j R is discrete, so there is an open cover P of X such that for every x R there is only one set V P with x V and V R = {x}. Hence cardp c and e(r) = c. Corollary 1. (X, τ j ) is not Lindelöf space (j = 1, 2, 3, 4). Proposition 5. (X, τ j ) is pathwise connected and locally pathwise connected for j = 1, 2, 3, 4. Proof. For the first part it suffices to show that for any z X 0 there is a curve joininig 0 with z (other cases are then easy). Let L = L 1 L 2 be a broken line, where L 1 is the segment [0, i Imz], L 2 = [i Imz, z]. Define f : I X by f(t) = 2ti Imz for t [0, 1 2 ], f(t) = (2t 1)Rez + i Im z for t [1 2, 1]. Then f is continuous and f(i) = L. For the second part we can again restrict ourselves to the case z = 0. Let V be a neighborhood of 0; we have to show that there is U V which is pathwise connected. For j = 1, if r 0 is such that [0, ir 0 ) V, we can take U = {0} r (0,r K(ir, ρ 0) r) where ρ r = dist(ir, X \ V ). For j = 2: for α (0, π) choose r α such that [0, ir α e iα ) V, and let ρ r,α = dist(re iα, X \ V ). Then let U = {0} K(re iα, ρ r,α ). α (0,π) r (0,r α) For j = 3 choose r 0 such that T := Int(conv{r 0 e iα, r 0 e i(π α), 0}) V ; then take U = {0} T. Finally, if j = 4 then there is a sequence (z n ) with T n := Int(conv{z n, z n, 0}) V and z n ց 0, arg z n ց 0 and we can take U = {0} T n. In all cases U is pathwise connected, and the proof is complete. The following theorem gives the criteria for the convergence z n 0. Their modification for the convergence z n r, r R should be clear. Theorem 1. For a sequence (z n ), z n X 0 and j = 1, 2, 3, 4: z n 0 in the topology τ i iff z n 0 in the euclidean topology and: 1. for j = 1: almost all terms of (z n ) are on the imaginary axis; 2. for j = 2: there is a finite number of half-lines beginning at 0 such that almost n=1

5 Some topologies in the half-plane 347 all terms of (z n ) are on these half-lines, each of them contains an infinite number of terms which form a subsequence converging to 0; 3. for j = 3: if α is the parameter for τ 3 then almost all terms of (z n ) lie in the open angle determined by rays R + e iα, R + e i(π α) ; 4. for j = 4: there is ϕ (0, π/2) such that almost all terms of (z n ) lie in the angle determined by rays R + e iϕ, R + e i(π ϕ). Proof. The if part is clear in all cases, so in the following we will prove the only if part. 1. Suppose z n 0 and there is a subsequence (z nk ) with Re z nk 0 for k N. Then the set U = X 0 \ {z nk : k N} is open in the euclidean topology, so V = {0} U is a τ 1 -neighborhood of 0 hence it should contain almost all z n, a contradiction. 2. To prove Part 2 let z n 0, ϕ n = arg z n and suppose that the set {ϕ n : n N} is infinite. Then there is a strictly monotonic subsequence (ϕ nk ) of (ϕ n ). It follows that (X 0 \ {z nk : k N}) {0} is a neighborhood of 0 (in τ 2 ), so the subsequence (z nk ) cannot converge to 0 a contradiction because z n 0. Hence the set {ϕ n : n N} must be finite, say {ϕ n : n N} = {α 1, α 2,..., α r }. We define subsequences of (z n ) by conditions argz n = α j for j = 1, 2,..., r. At least one of them is infinite. If a subsequence is infinite then it tends to 0. This ends the proof of Part Proof is routine. 4. For the proof of Part 4 assume z n 0 and suppose that for every ϕ (0, π/2) an infinite number of ϕ n = arg z n fulfills ϕ n < ϕ (the case ϕ n > π ϕ is similar). Then for some subsequence (ϕ nj ) there is ϕ nj ց 0. But z nj 0, so there is a subsequence (z nk ) fulfilling also z nk ց 0. Let u k = 1 2 Re z n k + i Imz nk and let U be a neighborhood of 0 defined by the sequence (u k ) (see Definition 1, part 4). Then (z nk ) U for k N, so (z nk ) 0, a contradiction, which ends the proof. Let χ j (z) denote the character of a point z X considered for the topology τ j, j = 1, 2, 3, 4; let χ j denote the character of the whole space. For any real z we have χ j (z) = χ j (0). Also χ j = χ j (0). Theorem ℵ 0 < χ 1 c; 2. χ 2 = c; 3. χ 3 = ℵ 0 ; 4. ℵ 0 < χ 4 c. Proof. Note that χ j c for j = 1, 2, 3, 4 (because the family of all open sets in X 0 has cardinality c). 1. Suppose that χ 1 (0) = ℵ 0, i.e. there is a countable basis of neighborhoods of 0; let {V n : n N} be such a basis. We can assume V n K(0, 1 n ). Take a sequence (z n ), z n V n such that sequences (Re z n ), (Im z n ) are decreasing. Then X 0 \ {z n : n N} {0} is open in τ 1 but it does not contain any V n. It follows that {V n : n N} is not a basis, so χ 1 > ℵ 0.

6 348 Thai J. Math. 5(2007)/ M. Grzesiak and L. Jankowski 2. Let B be a basis of neighborhoods of 0 in τ 2 and V B. Let A V ( π 2, π 2 ) be defined by the condition: α A V iff there exists a component of V \ {0} containing a segment (0, re i(π/2+α) ) for some r and not containing a segment (0, se iβ ) for β π 2 + α and any s. Since V \ {0} has at most a countable number of components, carda V ℵ 0. We claim that for every α ( π 2, π 2 ) there is some V B such that α A V. To prove this, let D α = e iα (C 1 C 1 ) where C k is a circle {k + e it ; t (0, 2π)} for k = ±1. Next, let V α = (X \D α ) {0}. Then A Vα = {α}, which proves the claim. So the cardinality of the sum V B A V is c and it follows that card B = c. 3. Take the basis consisting of triangles T n = Int(conv{0, 1 n eiα, 1 n ei(π α) }) {0}. 4. Suppose that χ 1 (0) = ℵ 0 ; let {V n : n N} be a countable basis of neighborhoods of 0. Take a sequence (z n ), z n V n such that arg z n ց 0 and z n ց 0. Then n=1 Int(conv{0, z n, z n }) {0} is open in τ 4 but it does not contain any V n. It follows that {V n : n N} is not a basis, so χ 1 > ℵ 0. The pseudocharacter ψ(x, X) of a point x in T 1 -space X is defined (see [1]) as a smallest cardinal number U where U is a family of open subsets of X fulfilling the condition {x} = U. Then we define pseudocharacter ψ(x) of T 1 -space X as sup x X ψ(x, X). Let ψ j be the pseudocharacter of (X, τ j ). Proposition 6. ψ j = ℵ 0 for j = 1, 2, 3, 4. Proof. For x X 0 we have ψ(x, X) = ℵ 0, since in X 0 the topology is euclidean. If x R, then we can assume x = 0. Let V be a neighborhood of 0 (as defined in 1) which is bounded in usual sense. Then n=1 1 n V = {0}, so ψ(0, X) = ℵ 0. Remind that a topological space X is called Urysohn space if for any x 1 x 2 X there are neighborhoods U 1, U 2 of x 1, x 2 such that U 1 U 2 =. Clearly X is then a Hausdorff space; however, it need not be regular. Every regular space is Urysohn. A topological space X is called semiregular if it is Hausdorff and there is a basis B consisting of the sets A such that A = Int(A). Every regular space is semiregular. Theorem τ 1 is completely regular, but not normal. 2. τ 2 is Urysohn, but not semiregular. 3. τ 3 is Urysohn and semiregular, but not regular. 4. τ 4 is completely regular, but not normal. Proof. 1. Take any τ 1 -closed set F X with 0 F (other cases are obvious). Let F = F R \ {0}. Then X \ F contains a closed segment I = [0, iε] for some ε > 0. There is a continuous (in euclidean topology) function f : X \ {0} [0, 1] with f I \ {0} = 0, f F = 1. Let f be the extension of f to X defined by f (0) = 0. Then f is τ 1 -continuous, so (X, τ 1 ) is completely regular. We show that it is not normal. Note that the subsets Q and Q := R \Q of the real line are τ 1 -closed. Suppose that there are τ 1 -open and disjoint sets G 1, G 2

7 Some topologies in the half-plane 349 with Q G 1, Q G 2. For x Q i there is n x N such that x+[0, n x ] G 2. Let Q m = {x Q : n x m}. Then m=1 Q m = Q. It follows that for some m 0 and some (a, b) it is (a, b) Q i m 0. Hence (a, b) (0, m 0 ) G 2. But quotient numbers are dense in (a, b), so for any q Q and some r > 0 we have (q, q + ir) G 1 and there exists an open (in euclidean topology) set U such that (q, q + ir) U G 1. But, on the other side, U G 2. Hence G 1 is not disjoint with G 2, a contradiction. It follows that τ 1 is not normal. 2. Clearly (X, τ 2 ) is an Urysohn space. To show that it is not semiregular let for n N and W n = { 1 n e ikπ 3 n : 1 k 3 n 1, 3 k}, W = W n. n=1 The set X 0 \ W is open in X 0 (for every x X 0 \ W there exists an open disk with center x, disjoint with W), and for every α (0, π) exists r α such that (0, r α e iα ) X 0 \ W. Hence U = {0} (X 0 \ W) is τ 2 -open. Let V U be a τ 2 -open set containing 0. Define A k = {ϕ (0, π) : (0, 1 k eiϕ ) V }. Then k=1 A k = (0, π), so for some k 0 and some segment (ϕ 1, ϕ 2 ) (0, π) we have A k (ϕ 1, ϕ 2 ). But it means that the open circular sector S of radius 1 k 0 between ϕ 1, ϕ 2 is contained in V, hence in Int(V ). Next, V \ {0} X 0 \ W and Int(V ) W S W, so V Int(V ). 3. It is easy to see that τ 3 is Urysohn and semiregular. Let V = {0} T r, where T r is the open triangle with vertices 0, re iα, re i(π α) for some r > 0. Then V = Tr c, where Tr c denotes the closed triangle. Any τ 3 -open set containing 0 must contain some open triangle T r, but T r does not contain any Ts c for s > 0. It follows that the closed set X \ V cannot be separated from Let (z n ) be a sequence with z n ց 0, argz n ց 0 and G = {0} Int(conv{0, z n, z n }), n=1 (as is the definition of τ 4 ). Take y n = 1 2 Re z n + i Imz n. Then F = conv{0, y n, y n} n=1 is τ 4 -closed, F 0 = {0} Int(conv{0, y n, y n}) n=1

8 350 Thai J. Math. 5(2007)/ M. Grzesiak and L. Jankowski is τ 4 -open and F 0 G. There is a function f on X \ {0} continuous in euclidean topology with f F \ {0} = 0, f X \ ({0} G) = 1. Define f(0) = 0. The extended function f is τ 4 -continuous. This proves that (X, τ 4 ) is completely regular. To show that it is not normal we argue as in 1. Corollary 2. The space (X, τ j ) is not metrizable for j = 1, 2, 3, 4. Note that this Corollary for τ 1, τ 2, τ 3 follows also from Theorem 2. The following notion of c-regularity was introduced by Klee ([3]). Definition 2. A topological space Z which is also a subset of a linear space is called c-regular if for any closed and convex set C Z and any x C there is a open and convex set V with x V such that V C =. Theorem The space (X, τ j ) is c-regular for j = 2, 3, The space (X, τ 1 ) is not c-regular. Proof. 1. Let C be a τ 2 -closed and convex set and x C. The case x X 0 is easy, so it suffices to consider the case x R and we can assume x = 0. Suppose first that there is ε > 0 such that ( ε, ε) C =. We claim that then inf{ z : z C} > 0. If it were 0, there would be a sequence (z n ) C X 0, z n 0 in the euclidean topology but z n 0 in τ 2. Then, since arg(z 1 z n ) arg z 1 : [0, z 1 ] [z 1, z n ] n=2 (the closure in the euclidean topology). Hence (0, z 1 ) C and (since C is τ 2 -closed) 0 C, a contradiction, which proves the claim. So 0 < r := inf{ z : z C}. The set (K(0, r 2 ) X 0) {0} is an open and convex set, disjoint with C. Now suppose that for every ε > 0 we have ( ε, ε) C. Since C is convex and 0 C, we have either (0, ε) C or ( ε, 0) C. We claim that C R. If there were y C X 0 then conv{0, ε, y} \ {0} C; hence 0 C, a contradiction. The set X \ C is an open and convex set with 0 X \ C. For j = 3. As above assume x = 0. If C is closed then there is a neighborhood U of 0 disjoint with C. We can assume U = {0} T r and this set is convex. For j = 4 we argue as for j = Consider the closed semi-disk C with center z 0 = 1 and radius 1, without 0. C is τ 1 -closed and convex. Any convex and open set containing 0 has a nonempty intersection with C. Remind that a space X is called a Fréchet space if for A X and x A there is a sequence x n x with x n A. Theorem The space (X, τ j ) is a Fréchet space for j = 3, 4. The spaces (X, τ 1 ), (X, τ 2 ) are not Fréchet. Proof. 1. τ 3 is first countable, so is Fréchet (see [1], ). Let A denote the τ 4 -closure of a set A and x A \ A. Case x X 0 is clear, so it suffices to show

9 Some topologies in the half-plane 351 that if 0 A \ A then there is a sequence (a n ) A, a n 0. Suppose not. Since 0 A it follows that there is a sequence (a n ) A, a n 0 in euclidean topology. Let X + := {Rez 0, z X}, X := {Re z 0, z X}. Then 0 A X + or 0 A X. Without loosing generality we can assume A X +. Let K ε := {z X : z ε}. Then K ε A X 0 because if K ε A R then 0 A R. Define A r := {argz : z K r A}, α r := supa r. Then α r is a nondecreasing function of r. We claim that lim r 0 α r = 0. If not then inf α r ϕ > 0 and we could find a sequence z n A X 0 such that z n ց 0, ϕ < argz n < π ϕ for almost all n. But then z n 0 in τ 4 and this contradicts our assumption. The claim is then proved and we can find a sequence a n A X 0 such π/2 arg a1 2n. that a n ց 0, arga n ց 0. Define z n = a n e iβn where β n = arga n + Let T n = conv{0, z n, z n} (as before, z n is symmetric to z n with respect to the imaginary axis) and U = n=1 T n. Then U is a neighborhood of 0 disjoint with A, so 0 A. This contradiction shows that τ 4 is a Fréchet space. To show that τ 1 and τ 2 are not Fréchet spaces consider the set A = {z X 0 \ K(1, 1) : Rez > 0}, wher K(1, 1) is a closed disk with center 1 and radius 1. Then 0 A but no sequence in A converges to 0. This ends the proof. Theorem 6. For j = 1, 2, 3, 4 the spaces (X, τ j ) are sequential spaces. Proof. By Theorem 5 (X, τ 3 ) and (X, τ 4 ) are Fréchet spaces, so they are sequential ([1], ). To show that (X, τ 1 ), (X, τ 2 ) are sequential, suppose that a set A fulfills the condition: (a n ) A, a n a 0 a 0 A, and suppose that x A \ A. Clearly x X 0, so x R and we can assume x = 0. For τ 2 we argue as follows: 0 A, so no sequence (a n ) A converges to 0; hence for every α (0, π) there is r α > 0 such that [0, r α e iα ] A =. But also for every α (0, π) there is an open (in X 0 ) set U α, U α A = such that (0, r α e iα ] U α ; then the set U = α (0,π) U α {0} is open in τ 2. This yields 0 A, contrary to our assumption, so τ 2 is sequential. For τ 1 the proof is similar and even easier. Corollary 3. The set A is closed in topology τ j, (j = 1, 2, 3, 4) if A is closed in euclidean topology, or A = A 1 A 2, where A 1 is closed in X 0 (with euclidean topology), A 2 R and for any sequence (a n ) in A 1 if a n a 0 R we have a 0 A 2. We now examine compactness of spaces (X, τ j ). Obviously we cannot expect much; in fact, as (X, τ 2 ), (X, τ 3 ) are not regular, they cannot be even pseudocompact, realcompact, Čech complete or locally compact. So from now on we restrict ourselves to τ 1, τ 4. Remind at a start that the Niemytzki plane is realcompact and Čech complete.

10 352 Thai J. Math. 5(2007)/ M. Grzesiak and L. Jankowski Theorem 7. The spaces (X, τ 1 ) and (X, τ 4 ) are not Čech complete. Proof. By [1], Theorem it suffices to prove that for any countable family {A i } i=1 of open covers of X there is a family F of closed sets which has the finite intersection property and contains sets of diameter less than A i with F =. Claim. For an infinite sequence (V n ), V n A n, 0 V n there is a family F = {F n } n=1 such that all F n are closed and F n+1 F n, F n V n A n (V n a neighborhhod of 0), F n =. For the proof of the claim let (z n ) be a sequence such that z n n k=1 V k, z n ց 0 and for τ 1 : arg z k argz n for k = 1, 2,...,n 1; for τ 4 : arg z n+1 < 1 2 argz n. By Theorem 1 the sequence (z n ) is not convergent, so the sets F n := {z k : k n} n=1 are closed for n N. Also F n+1 F n, n F n V k, k=1 F n =. n=1 Hence F n V n, so the claim (and the theorem) is proved. As every locally compact space is Čech complete, we have the following fact. Corollary 4. The spaces (X, τ 1 ), (X, τ 4 ) are not locally compact. Theorem 8. Let z 0 X. The spaces (X, τ j ) for j = 1, 4 are hereditarily realcompact (that is, every subspace of X is realcompact). Proof. The space Y is hereditarily realcompact iff for any y Y the space Y \ {y} is realcompact ([1], 3.11.B). By [1], Theorem it suffices to prove that every ultrafilter F in the family of functionally closed sets in which has the countable intersection property has nonempty intersection. Claim. For x R the set R \ {x} is functionally closed in X. It suffices to prove this for x = 0. Let V be a neighborhood of 0 (as in Definition 1, in particular V R = {0}). As (X, τ 1 ), (X, τ 4 ) are completely regular there is a continuous function Define f : X I, f X\V = 0, f(0) = 1. g : X I, g(z) = f(z) + inf{imz, 1}. Then R \ {0} = g 1 (0), so R \ {0} is functionally closed in X. Note that R = Im 1 (0) is functionally closed, too.

11 Some topologies in the half-plane 353 We introduce the following notation. For a set A X let U(A) be the family of all ultrafilters in the family of functionally closed sets in A which have the countable intersection property. Now for the main part of the proof let F U(X \{z 0 }). There are two possible cases: (A) R \ {z 0 } F; (B) there exists F F such that F R =. If (A) holds true, we can construct a sequence (F n ), F n F, with nonempty intersection. It goes as follows. Let a 1 R \ {z 0 }; then the sets A 1 = (, a 1 ] (R \ {z 0 }), A 1 = (a 1, ) (R \ {z 0 }) are functionally closed (because intersections of functionally closed sets are functionally closed), and one and only one of them is in F. Denote this set as F 1. Next choose a 2 IntF 1 (IntF 1 in euclidean topology) with a 2 > 2 and consider A 2 = (, a 2 ] F 1, A 2 = (a 2, ) F 1. One of these belongs to F; choose it as F 2. Now suppose that we have defined F 1, F 2,...,F n. If F n is a half-line then choose a n+1 F n, a n+1 > n + 1; if F n is a segment then let a n+1 be its midpoint. Let A n+1 = (, a n+1 ] F n, A n+1 = (a n+1, ) F n, and define F n+1 as this of the two which is in F. As F has the countable intersection property, the family {F n } has nonempty intersection. From the construction of F n it has to be {x 0 } for some x 0 R. Then x 0 F for every F F since the family F, F 1, F 2,... must have nonempty intersection. If (B) holds true, take F F with F R =. Since F X 0 \{z 0 }, the topology in F is euclidean, so F is realcompact. Also F F := {H F : H F} U(F), because every closed subset of X 0 \ {z 0 } is functionally closed. Hence F F ([1], ). Hence F = F F = F F. In both cases F. It follows that X \ {z 0 } is realcompact. This ends the proof. Remind that a space X is pseudocompact, if X is completely regular and every real-valued and continuous function on X is bounded. A Hausdorff space X is sequentially compact, if every sequence of points of X has a convergent subsequence. Sequential compactness always implies countable compactness, and in the case of sequential spaces both notions are equivalent ([1], ). The spaces (X, τ j ) are (obviously) not sequentially compact for any j = 1, 2, 3, 4, and not countably compact ([1], ). They are even not locally sequentially compact (a space X is locally sequentially compact if for every x X and every neighborhood V of x there is a neighborhood U of x such that U is sequentially compact and U V ). To prove the last statement we need the following lemma.

12 354 Thai J. Math. 5(2007)/ M. Grzesiak and L. Jankowski Lemma 1. Let V be a neighborhood of 0 in the topology τ j (j = 1, 2, 4) and let (z n ), z n V be a sequence such that z n ց 0, argz n converges monotonously to α 0 (where α 0 = π 2 for τ 1, α 0 [0, π] for τ 2, α 0 = 0 or π for τ 4 ). Then there is a continuous (in τ j ) function f : X R such that f(z n ) = n and f (X\V ) {0} = 0. Proof. Because argz n converges monotonously to α 0 there is a sequence (r n ) such that 0 < r n < Im z n, r n < dist(z n, Re iα0 ), r n ց 0 with K(z n, r n ) K(z m, r m ) = for m n. Define f : X R + by conditions: 1. f is continuous on K(z n, r n ); 2. f(z n ) = n; 3. f X \ n=1 K(z n, r n ) = 0. We claim that f is τ j -continuous on X. Clearly it is continuous on X 0. Next, f R = 0 and for x R \ {0} there is d > 0 such that f K(x,d) X = 0, so f is continuous on R \ {0}. Let G = f 1 (0, ε) for ε > 0; it suffices to show that G is a neighborhood of 0 in τ j. For j = 1: G contains the imaginary half-axis {yi : y > 0} and G X 0 is open in the euclidean topology, hence G is a τ 1 -neighborhood of 0. For j = 2: from the definition of sets K(z n, r n ) it follows that for every α (0, π) there is r α > 0 such that [0, r α e iα ] G. As above, G X 0 is open in euclidean topology and we infer that G is τ 2 -open. For j = 4 assume (as we can) that arg z n 0. Define a k = sup n k { z n r n }, φ k = sup n k {argz n + arcsin rn z n } and let w k = a k e iφ k, T k = Intconv{0, w k, w k }. Then for k, n N we have T k K(z n, r n ) =. If we define W = T k, k=1 then W G and W is a τ 4 -neighborhood of 0. From this lemma we have the following corollary. Corollary 5. Let V be as above. Then V is not pseudocompact. Theorem 9. The spaces (X, τ j ) are not locally sequentially compact for any j = 1, 2, 3, 4. Proof. From Theorem 3 we know that τ 1, τ 4 are completely regular. Let V be a neighborhood of 0. Were V sequentially compact, then it would be countably compact, hence pseudocompact ([1], ) and it is not. So X is not locally sequentially compact for j = 1, 4.

13 Some topologies in the half-plane 355 Consider now (X, τ 2 ). Let V be a neighborhood of 0. There is a sequence (z n ) such that z n V, z n ց 0, arg z n α 0 (monotonously), where α 0 [0, π]. From Theorem 1 this sequence is not τ 2 -convergent. So V is not sequentially compact. Finally, let V = {0} T 0 be a τ 3 -neighborhood of 0 (as in the Definition 1) and U V be any τ 3 -neighborhood of 0. Since U must contain some triangle T with the same angle as T 0, so we cannot have U V. Theorem 10. The set A X is τ j -compact (j = 1, 2, 3, 4) iff 1. A is compact in the euclidean topology; 2. A R is finite or empty; 3. there is r > 0 such that for x A R the set K(x, r) A is contained in x + ir + when j = 1; some finite set of half-lines beginning at x when j = 2; {x} Int(conv{0, u, u }), where u = re iα when j = 3; {x} Int(conv{0, u, u }), where u = ae iβ for some a > 0, β (0, π 2 ) when j = 4. Proof. Assume A is τ j -compact. Condition 1 is obvious, condition 2 follows from Proposition 1. Further, condition 3 is clear if A R consists of isolated point only. If there is x A R which is not isolated suppose that condition 3 does not hold. Now, if j = 1, this means that for every n N the set (K(x, 1 n ) A)\(x+iR +) is nonempty. Hence there is a sequence (w n ), w n A, w n x < 1 n, Re w n x. Denote W = {w n : n N}. Then X 0 \ W is open in euclidean topology and X 0 \ W (x + ir + ) \ {0}, so {x} (X 0 \ W) is open in τ 1. Also X \ W is open in τ 1. Hence W is τ 1 -closed. But, as an infinite set of isolated points, W is not compact. On the other side W is a closed subset of a compact set A, so it is compact, a contradiction. For j = 2 the reasoning is similar. If condition 3 does not hold then we can find a sequence (w n ) such that w n A, w n x < 1 n and arg w n is monotonously convergent to some ϕ [0, π]. Denote W = {w n : n N}. Then {x} (X 0 \ W) is open in τ 2 because each half-line starting at x can have only one point common with W. The rest of the proof is as above. For j = 3, if condition 3 does not hold then there is (w n ) such that w n A, w n x < 1 n, and (arg w n α or arg w n π α).

14 356 Thai J. Math. 5(2007)/ M. Grzesiak and L. Jankowski Hence there is r > 0 such that so (as above, W = {w n : n N}) w n {x} Int(conv{0, re iα, re i(π α }), W ({x} + Int(conv{0, re iα, re i(π α )})) =. Hence {x} (X \ W) is τ 3 -open and next we reason as above. For j = 4, if condition 3 is not fulfilled then we can find a sequence (w n ): w n A, w n x ց 0, and arg(w n x) 0 or arg(w n x) π monotonously, and further as above (that is, one has to show that the set W = {w n : n N} is closed). We have proved that for every x A R there is r x fulfilling Condition 3. As A R is finite, so for r = inf{r x : x A R} Condition 3 is true. Assume now conditions 1 3. For every y A and any its neighborhood V there is (by condition 3) r y such that K(y, r y ) A V. (2.1) Let P be an τ j -open cover of A. We know that R A is finite; let R A = {x 1,..., x n } and take V k such that x k V k P. Then A \ n k=1 V k is τ j -closed, and as A \ n k=1 V k X 0 it is closed in euclidean topology. For y A \ n k=1 V k there is (by 2.1) r y > 0 and U y P such that K(y, r y ) (A \ n k=1 V k) U y. So there are y 1,..., y m such that A \ n k=1 V k m k=1 U y k and hence A n m V k U yk. k=1 k=1 As V k, U k P this proves that A is τ j -compact. Corollary 6. The set A X is τ j -compact (j = 1, 2, 3, 4) iff A is τ j -sequentially compact. Proof. Assume A is τ j -compact. Then from condition 1 of Theorem 10 it is closed in euclidean topology. Hence every sequence (w n ) has a subsequence (w nk ) convergent to some u A. If u X 0 then w nk u in τ j, so we have to consider only the case u R. If u is an isolated point of A then w nk = u for almost all k. If u is not an isolated point of A then by Condition 3 of Theorem 10: if j = 1: almost all w nk are on the line x + ir, so by Theorem 1 (w nk ) is convergent; if j = 2: by condition 3 of Theorem 1 w nk u; if j = 3: by condition 3 we have condition 3 of Theorem 1 and hence w nk u;

15 Some topologies in the half-plane 357 if j = 4: by condition 4 we have condition 4 of Theorem 1 and hence w nk u. In every case the sequence (w n ) has a convergent subsequence (w nk ), so A is sequentially compact. Conversely, if A is τ j -sequentially compact then conditions 1 and 2 of Theorem 10 are obvious. Condition 3 is also obvious if A R has only isolated points. So, assuming that Condition 3 is not true, there is x A R which is not isolated and for any n N the set K(x, 1 n ) A is not contained in x + ir + when j = 1; some finite set of half-lines beginning at x when j = 2; {x} Int(conv{0, u, u }), where u = re iα for some r > 0, when j = 3; {x} Int(conv{0, u, u }), where u = ae iβ for some a > 0, β (0, π 2 ) when j = 4 (as listed in Theorem 10). This means that there is a sequence (w n ) K(x, 1 n ) A X 0 such that in the case j = 1 : w n x + ir + ; in the case j = 2 : the sequence arg(w n x) is monotonously convergent to some ω [0, π]; in the case j = 3 : n N (arg w n α or arg w n π α; in the case j = 4 : arg(w n x) ց 0 or arg(w n x) ր π. As w n converges to x in the euclidean topology and the topologies τ j, j = 1, 2, 3, 4, are stronger than the euclidean topology, so the only possible τ j -limit of w n is x. But from the way w n were chosen we infer that no subsequence of (w n ) cannot tend to x in τ j, so, by Theorem 1, (w n ) cannot have a convergent subsequence. It follows that A is not τ j -sequentially compact, a contradiction. References [1] R.Engelking, General Topology, PWN, Warszawa 1977; [2] V.L.Klee, Convex sets in linear spaces, Duke Math. J. 18(1951), ; [3] V.L.Klee, Convex sets in linear spaces III, Duke Math. J. 20(1953), ; [4] P. Koosis, Introduction to H p Spaces, Cambridge Univ. Press, 1980; [5] S.Saks, A.Zymund, Analytic functions, PWN Warszawa-Wroc law 1952; [6] J.B.Twomey, Tangential boundary behaviour of harmonic and holomorphic functions, J.London Math. Soc.(2) 65(2002), (Received 30 May 2007) Maciej Grzesiak Institute of Mathematics,

16 358 Thai J. Math. 5(2007)/ M. Grzesiak and L. Jankowski Poznań University of Technology, Poznań,Poland Leszek Jankowski Institute of Mathematics, Poznań University of Technology, Poznań,Poland