SPACE-OPEN TOPOLOGY: PART II BY FEREYDOON DERAKHSHANI KORI ZANYARI MERIWAN

We continue our study of space-open topology as in [2]. We investigate the behaviour of space-open topology with respect to connectedness, countability and the separation axioms. As usual we use X, Y for topological spaces and U, V for open subsets of topological spaces. K(X, Y ) denotes the set of constant functions from X to Y , C(X, Y ) the set of continuous functions from X to Y and Y X the set of all functions from X to Y .

1. Connectedness Definition 1.1. Let DX ⊆ X be the set of points of X which do not belong to any proper open subset of X. In other words, x ∈ DX ⇐⇒ if x is in an open subset U ⊆ X, then U = X Lemma 1.2. DX has the following properties: It is a closed subset of X, It is connected, It is compact, If ∅ = 6 DX ⊆ Y ⊆ X and Y has the trivial topology, then Y = DX , (5) If DX 6= ∅, then X is connected, (6) If X is Hausdorff and it has more than one element, then DX = ∅. (1) (2) (3) (4)

Proof. (1) In fact, DX is the intersection of all nonempty closed subsets of X and hence it is closed. (2) Obviously, the induced topology on DX is the trivial one, that is, the only open subsets of DX are DX and ∅. Therefore, it is connected. 1

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(3) The only open subset of X intersecting DX is X itself, thus any open covering of DX has a finite subcover. So, it is compact. (4) If there is y ∈ Y which is not in DX , then by definition of DX , there must be a proper open subset U ⊂ X which contains y. Therefore, U ∩ Y is a proper open subset of Y . This is a contradiction because Y has the trivial topology. (5) If X is not connected, then X = U ∪ V such that U, V are nonempty open subsets and U ∩ V = ∅. Then it is easy to see that DX = ∅. (6) Let x, y ∈ X. Then, there are open subsets U, V such that x ∈ U , y ∈ V and U ∩ V = ∅. So, by definition, x, y ∈ / DX . Thus, DX = ∅. £ Though for most topological spaces DX = ∅ but this is not the case for the space-open topology on F (X, Y ). In some cases, this provides a good way to prove that F (X, Y ) is connected. Example 1.3. Suppose that X = Y and F (X, Y ) = Y X . The identity function id : X → Y is in F (X, Y ). Moreover, since id is a surjective function, it does not belong to any proper open subset of F (X, Y ). This means that id ∈ DF (X,Y ) . Therefore, by Lemma 1.2, F (X, Y ) is connected. Theorem 1.4. Suppose that K(X, Y ) ⊆ F (X, Y ). If Y is connected, then F (X, Y ) is also connected. Proof. Suppose that F (X, Y ) = P ∪ Q where P, Q are nonempty open subsets of F (X, Y ) and P ∩ Q = ∅. By definition of space-open topology, there are families of open subsets {Ui }i∈I and {Vj }j∈J such that P = ∪i∈I (X, Ui )F and Q = ∪j∈J (X, Vj )F . Since P ∩ Q = ∅, for any i, j we have (X, Ui ∩ Vj )F = (X, Ui )F ∩ (X, Vj )F = ∅ and so Ui ∩ Vj = ∅ because K(X, Y ) ⊆ F (X, Y ). Therefore, (∪i∈I Ui ) ∩ (∪j∈J Vj ) = ∅. On the other hand, since K(X, Y ) ⊆ F (X, Y ), we have Y = (∪i∈I Ui ) ∪ (∪j∈J Vj ). This is a contradiction because Y is connected. £ Theorem 1.5. Suppose that X is connected but Y is not connected. Then, the set of continuous functions F (X, Y ) = C(X, Y ) is not connected. Proof. Since Y is not connected, there are proper open subsets U, V such that Y = U ∪ V and such that U ∩ V = ∅. Let f ∈ F (X, Y ). Since f is continuous and X is connected, f (X) is also connected.

SPACE-OPEN TOPOLOGY: PART II

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On the other hand, f (X) = (f (X) ∩ U ) ∪ (f (X) ∩ V ) which implies that f (X) ⊆ U or f (X) ⊆ V . Thus, f ∈ (X, U )F or f ∈ (X, V )F . This implies that F (X, Y ) is not connected because (X, U ∩ V )F = (X, U )F ∩ (X, V )F = ∅. £

2. Countability and separation axioms Definition 2.1. Let y ∈ Y and let By be a set of open subsets of Y containing y. We say that By is a local base at y if for any open subset U which contains y, there is V ∈ By such that V ⊆ U . Lemma 2.2. Suppose that the constant function cy associated to y is in F (X, Y ). If By is a local base at y, then Bcy = {(X, V )F | V ∈ By } is a local base at cy in F (X, Y ). Proof. It is obvious that each member of Bcy contains cy . Now suppose that there is an open subset P of F (X, Y ) containing cy . By definition of space-open topology, there is an open subset U ⊆ Y such that cy ∈ (X, U )F ⊆ P . This implies that y ∈ U . On the other hand, since By is a local base at y, there is V ∈ By such that y ∈ V ⊆ U . Thus, cy ∈ (X, V )F ⊆ (X, U )F ⊆ P . This proves that Bcy is a local base at cy . £ Remember that Y is called first-countable if every point y ∈ Y has a countable local base By . Moreover, Y is called second-countable if it has a countable base, that is if there is a countable set B of open subsets of Y which is a local base at any point y ∈ Y . Theorem 2.3. Suppose that K(X, Y ) ⊆ F (X, Y ). If F (X, Y ) is first-countable then Y is also first-countable and if F (X, Y ) is secondcountable then Y is second-countable. Proof. Suppose that F (X, Y ) is first-countable. Since Y is homeomorphic to K(X, Y ) and since first-countability is inherited by subspaces, then Y is also first-countable. The proof for second-countability is similar. £ Now we turn to separation.

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Theorem 2.4. Suppose that K(X, Y ) ⊆ F (X, Y ). If F (X, Y ) satisfies the separation axiom T1 (or respectively T2 or T3 ) then Y also satisfies the separation axiom T1 (or respectively T2 or T3 ). Proof. Suppose that F (X, Y ) satisfies the axiom T1 . This means that if f 6= g are two elements of F (X, Y ), then there is an open subset P of F (X, Y ) such that f ∈ / P but g ∈ P . Now let y, y 0 ∈ Y . Then, there is an open subset P ⊆ F (X, Y ) which contains the constant function cy0 but it does not contain cy . This implies that there is an open subset U of Y such that cy0 ∈ (X, U )F but cy ∈ / (X, U )F . Therefore, y 0 ∈ U but y ∈ / U . So, Y satisfies T1 . The proof for other axioms is similar. In fact, all the axioms are inherited by subspaces, thus Y naturally inherits these axioms from F (X, Y ) because K(X, Y ) ⊆ F (X, Y ) and Y is homeomorphic to K(X, Y ). £ Example 2.5. Suppose that X = Y = R the set of real numbers and F (X, Y ) = Y X . Now let f 6= g be two surjective elements of F (X, Y ). So, the only open subset of F (X, Y ) which contains f or g is F (X, Y ) itself. Therefore, F (X, Y ) does not satisfy the separation axiom T1 though Y satisfies even the separation axiom T4 . References [1] N. Bourbaki; General topology. Part 2, Elements of Mathematics, AddisonWesley, 1966. [2] F. Derakhshani; Space-open topology: part I. Article of Kori Zanyari Meriwan at www.meriwan.com [3] R. McCoy, I. Ntantu; Topological properties of spaces of continuous functions. Lecture notes in Mathematics, 1315, Springer-Verlag, 1988. [4] S-T Hu; Elements of general topology. Holden-Day, 1964.