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Monotone insertions of continuous functions - University of ...

Monotone insertions of continuous functions - University of ...

10 CHRIS GOOD AND IAN

10 CHRIS GOOD AND IAN STARES (2) for all κ, there is an operator Φ κ : UL κ (X) → C(X, J(κ)) such that g ≤ Φ κ (g, h) ≤ h and such that if g ≤ g ′ and h ≤ h ′ then Φ κ (g, h) ≤ Φ κ (g ′ , h ′ ), (3) for some κ, there is an operator Φ κ : UL κ (X) → C(X, J(κ)) such that g ≤ Φ κ (g, h) ≤ h and such that if g ≤ g ′ and h ≤ h ′ then Φ κ (g, h) ≤ Φ κ (g ′ , h ′ ). Proof. (2) implies (3) is obvious. To prove (3) implies (1) we use Theorem 1.4. Fix any η < κ. If (g, h) ∈ UL(X) then define g κ , h κ : X → J(κ) by g κ (x) = (g(x), η) and h κ (x) = (h(x), η) (note, without loss of generality, g, h : X → (0, 1)). It is easy to check that these two functions are upper and lower semicontinuous as hedgehog-valued functions and hence (g κ , h κ ) ∈ UL κ (X). Now define Φ(g, h) = π κ ◦ Φ κ (g κ , h κ ) where Φ κ is as in (3). It is clear now that Φ satisfies the conditions in Kubiak’s result and hence X is monotonically normal. (1) implies (2). Assume X is monotonically normal and (g, h) ∈ UL κ (X). First we claim that (π κ ◦ g) ∈ USC(X) (that is, it is an upper semicontinuous real-valued function). This follows since (π κ ◦ g) −1 ((−∞, b)) = {x ∈ X : g(x) = (a, η) for some η < κ and a < b} = g −1 (B(0, b)), which is open since B(0, b) is open in the upper topology and g is upper semicontinuous. Also (π κ ◦ h) ∈ LSC(X) since, (π κ ◦ h) −1 ((b, ∞)) = {x ∈ X : h(x) = (a, η) for some η < κ and a > b} = h −1 ( ⋃ η 0 and ɛ such that 0 < ɛ < a, then it is clear by definition that Φ κ (g, h)(x) ∈ V only if h(x) ∈ (a − ɛ, 1] × {η} and therefore Φ κ (g, h) −1 (V ) = h −1 ((a − ɛ, 1] × {η}) ∩ f(g, h) −1 ((a − ɛ, a + ɛ)). The first of the two sets on the right hand side is open because h is lower semicontinuous and the second because f(g, h) is continuous. We therefore have that Φ κ (g, h) −1 (V ) is open. If V = B(0, ɛ) for some ɛ > 0, then it

MONOTONE INSERTION OF CONTINUOUS FUNCTIONS 11 is easy to check that Φ κ (g, h) −1 (V ) = f(g, h) −1 ([0, ɛ)) which is open by continuity of f(g, h). It remains to check the monotonicity condition, but if g ≤ g ′ and h ≤ h ′ , then π κ ◦g ≤ π κ ◦g ′ and π κ ◦h ≤ π κ ◦h ′ and therefore f(g, h) ≤ f(g ′ , h ′ ). Also, if h(x) ∈ [0, 1] × {η} then h ′ (x) ∈ [0, 1] × {η}. The result now follows. □ The monotone version of Theorem 4.2 is the following. Theorem 4.4. A space X is monotonically normal if and only if for each closed subspace A of X and every (some) κ, there is an operator Φ κ : UL κ (A) → C(X, J(κ)) such that g ≤ Φ κ (g, h)|A ≤ h and such that if g ≤ g ′ and h ≤ h ′ then Φ κ (g, h) ≤ Φ κ (g ′ , h ′ ) Before we prove this result we need two theorems. The following is a monotone version of the Tietze-Urysohn theorem which was proved by the second author in [18]. Theorem 4.5. If X is monotonically normal, then for each closed subspace A of X there is an operator Ψ A : C(A, [0, 1]) → C(X, [0, 1]) such that for each f ∈ C(A, [0, 1]), Ψ A (f) extends f and such that if A 0 ⊆ A 1 are closed subspaces and f i ∈ C(A i , [0, 1]) such that f 1 |A 0 ≤ f 0 and f 1 (x) = 0 for all x ∈ A 1 \ A 0 then Ψ A1 (f 1 ) ≤ Ψ A0 (f 0 ). We now prove that a hedgehog-valued version of the monotone extension property (see [10, Theorem 3.3]) holds in monotonically normal spaces. Recall that if X is monotonically normal then every closed subspace of X is K 1 -embedded in X, that is for each closed subspace A of X there is a function k : τA → τX (where for a space Y , τY denotes the topology on Y ) such that k(U) ∩ A = U for each open U in A and if U ∩ V = ∅ then k(U) ∩ k(V ) = ∅. Without loss of generality we may assume that if U ⊆ V then k(U) ⊆ k(V ). Theorem 4.6. If X is monotonically normal, then for each κ and for each closed subspace A of X there is an operator Ψ κ : C(A, J(κ)) → C(X, J(κ)) such that Ψ κ (f) extends f and if f ≤ f ′ then Ψ κ (f) ≤ Ψ κ (f ′ ). Proof. So assume A is a closed subspace of X and f : A → J(κ) is continuous. For each η < κ define V (f, η) = f −1 ((0, 1] × {η}). Thus (V (f, η)) η

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