On Separability of Unbounded Norm Topology

Home , Band (order theory), Weak order unit

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interior set {un : n ∈ N} whenever (X, τun) is separable. This result enables us

to prove (see Theorem 3.2) that separability and/or second countability of (X, τun) is equivalent to separability of the normed lattice X itself. This result is applied to prove that separability of the unbounded norm topology passes up and down between X and its Banach space completion X. In Section 4 we make one step further.b Instead of considering separability of X

or equivalently separability of principal ideals Iun for each n ∈ N, we rather consider

separability of the sets Cun of components of vectors un equipped with the metric topology inherited from X. For a normed lattice with the projection property we

prove in Theorem 4.2 that it is separable if and only if the set of components Cun is separable for each n ∈ N. We also provide an example that this statement does not hold in general for normed lattices without the principal projection property. In Section 5 we consider separability of order continuous Banach function spaces over semi-ﬁnite measure spaces. In Theorem 5.7 we prove that such a Banach function space is separable if and only if it has a σ-ﬁnite carrier and it is separable with respect to the topology of local convergence in measure.

In Section 6 we are interested in those vector lattices for which the equality Iu = Iv

between principal ideals or the equality Bu = Bv between principal bands yields an

existence of a bijective mapping between Cu and Cv. In Proposition 6.2 we prove

that the equality Iu = Iv for positive vectors in a uniformly complete Archimedean

vector lattice guarantees an existence of a surjective isometry between Cu and Cv. If

we replace the condition Iu = Iv by the weaker condition Bu = Bv, then Example 6.5

shows that it is even possible that there is no homeomorphism between Cu and Cv in the case of Dedekind complete vector lattices. Nevertheless, for order continuous

Banach lattices the condition Bu = Bv is suﬃcient for Bu and Bv to be homeomorphic (see Theorem 6.10). In Section 7 we discuss the separation axioms of a normed lattice equipped with

the unbounded norm topology. Since (X, τun) satisﬁes the Hausdorﬀ axiom, by the

general theory of topological vector spaces the space (X, τun) is completely regular.

We prove that (X, τun) is a paracompact and hence a normal space whenever X is an atomic KB-space.

2. Preliminaries Throughout the paper we will assume that all vector lattices are Archimedean. Let X be a vector lattice. The set of all positive elements is denoted by X+. For ONSEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 3

a subset A we denote by IA and BA the ideal and the band in X generated by A,

respectively. In the case when the set A = {u} is a singleton set we write IA = Iu

and BA = Bu. Ideals and bands generated by a single element are called principal. For a given set A ⊆ X we denote the set {x ∈ X : |x|∧|a| = 0 for all a ∈ A} by Ad and we call it the disjoint complement of A in X. It turns out that Ad is always a band in X. If a band B satisﬁes X = B ⊕ Bd, then B is called a projection

band. To a projection band B we associate the band projection PB which is the identity mapping on B and the zero mapping on Bd. It is well known that band projections always commute. If every principal band in X is a projection band, then X is said to have the principal projection property. A vector lattice X is said to be Dedekind complete or order complete if non-empty bounded from above subsets of X have suprema. If non-empty countable subsets of X have suprema, then X is said to be σ-Dedekind complete or σ-order complete. A vector lattice is said to satisfy the countable sup property whenever every non-empty subset possessing a supremum contains an at most countable subset possessing the same supremum. The countable sup property is equivalent to the following fact: for each

net (xα) in X that satisﬁes 0 ≤ xα ↑ x there is an increasing sequence of indices

(αn)n∈N such that 0 ≤ xαn ↑ x (see e.g. [LZ71, Theorem 23.2(iii)]). A positive vector

a in X is said to be an atom whenever the principal ideal Ia is one-dimensional.

Since X is Archimedean, the principal ideal Ia is a projection band in X. A vector lattice X is called atomic if the linear span of the set of all atoms is order dense in X, i.e., for each non-zero positive vector x ∈ X there exists an atom a in X such that 0 ≤ a ≤ x. A lattice norm k·k on a vector lattice is a norm which satisﬁes kxk≤kyk for all elements x and y ∈ X with 0 ≤|x|≤|y|. A vector lattice equipped with a lattice norm is said to be a normed lattice. A Banach lattice is a normed lattice which is also a Banach space. The Banach space completion of a normed lattice X which is denoted by X is always a Banach lattice. A normed lattice is order continuous -Fatou whenever xα ↓b0 implies xα → 0 in norm. A normed lattice X has the σ + property if for every increasing sequence (xn)n∈N in X with the supremum x it follows that

kxk = sup kxnk. n∈N It turns out that dual Banach lattices and Banach lattice with order continuous norm have the σ-Fatou property (see e.g. [MN91]). A set Q of positive vectors of a normed lattice X is said to be a quasi-interior set if the ideal generated by Q in X is dense in X. A positive vector u ∈ X is 4 M.KANDIC´ AND A. VAVPETICˇ said to be a quasi-interior point whenever the set {u} is a quasi-interior set of X. A positive vector u ∈ X is called a weak unit if |x| ∧ u = 0 implies x = 0. Since norm dense ideals are order dense, every quasi-interior point is a weak unit. Although the converse statement does not hold in general, in order continuous normed lattices these two notions coincide. The remaining unexplained facts about vector and normed lattices can be found in [AA02, AB06, LZ71, MN91, Sch71, Zaa97]. Given a normed lattice X, a positive number ǫ > 0 and a positive vector x ∈ X+ we deﬁne the set

Ux,ǫ = {y : k|y| ∧ xk < ǫ}. Then the collection of all sets of this form is a base of neighborhoods for zero for some locally solid Hausdorﬀ linear topology on X. This topology is called the unbounded norm topology (un-topology for short) on X induced by the norm, and it is denoted by τun. A net (xα) in X is said to un-converge to a vector x ∈ X whenever for + each y ∈ X we have |xα − x| ∧ y → 0 in norm. It should be clear that a net (xα) un-converges to x if and only if it converges to x with respect to τun.

3. Separability of un-topology for general normed lattices In this section we consider the question under which conditions a normed lattice

equipped with the unbounded norm topology τun is a separable or even a second countable topological space. Although in general the unbounded norm topology be- haves diﬀerently than the norm topology, the answer is surprisingly simple. In fact, we prove in Theorem 3.2 that separability or equivalently second countability of (X, τun) is equivalent to separability of X. We start with the following proposition which at ﬁrst glance might appear to be an improvement if [AA02, Exercise 4.2.3] however having in mind Theorem 3.2, it just provides an invaluable tool for proving it.

Proposition 3.1. Let X be a normed lattice such that (X, τun) is separable. (1) Then X has a countable quasi-interior set. (2) If X is a Banach lattice, then X has a quasi-interior point.

Proof. (1) Let {xn : n ∈ N} be a countable τun-dense subset of X. Since the inequality 0 ≤|x+ − y+|∧|z|≤|x − y|∧|z|

+ + holds for all x, y, z ∈ X, the set {xn : n ∈ N} is τun-dense in X . Let F be the set of + all finite subsets of N. For F ∈F we define xF = n∈F xn and we observe that the family D := {xF : F ∈ F} is countable and upwardP directed. The set G of all finite ONSEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 5

sums of elements from D is countable, additive, upward directed and it obviously contains D. Since G is upward directed, it can be considered as an increasing net. Since G is additive, in order to prove that G is a quasi-interior set, by [Drn01, Proposition 2.2] it is enough to show that for every positive vector x the net (x − + x ∧ u)u∈G converges to zero. Pick any positive vector x ∈ X , any set F1 ∈ F and

an arbitrary ǫ > 0. Then x + U|x−xF1 |,ǫ is a τun-neighborhood of x, and since D is + τun-dense in X , there exists F2 ∈ F such that xF2 ∈ x + U|x−xF1 |,ǫ from where it follows that

|x − xF2 |∧|x − xF1 | < ǫ. n n Let us deﬁne F0 = F1 ∪ F2. Then for any u = i=1 xGi ∈ G with F0 ⊆ G := i=1 Gi we have that x − u ≤ x − xF1 and x − u ≤Px − xF2 from where it followsS that + + + + (x − u) ≤ (x − xF1 ) and (x − u) ≤ (x − xF2 ) . Since each vector a ∈ X satisﬁes 0 ≤ a+ ≤|a|, we conclude that

+ + + + + k(x − u) k = k(x − u) ∧ (x − u) k≤k(x − xF1 ) ∧ (x − xF2 ) k

≤ |x − xF1 |∧|x − xF2 | < ǫ.

+ From this it follows that the net ((x−u) )u∈G is norm convergent to zero. Due to the + + equality (x − u) = x − x ∧ u, we conclude that for each x ∈ X the net (x ∧ u)u∈G converges in norm to x.

(2) By (1), there exists a countable quasi-interior set D = {un : n ∈ N} in X. Since the series of positive vectors ∞ un n 2 (kunk + 1) Xn=1 converges absolutely, it converges to some positive vector u ∈ X. Since for each n ∈ N n we have 0 ≤ un ≤ 2 (kunk +1)u, the principal ideal Iu contains the quasi-interior set

D from where it follows that Iu is dense in X. The following theorem which follows from Proposition 3.1 is the main result of this section.

Theorem 3.2. The following assertions about a normed lattice X are equivalent. (1) X is second countable. (2) X is norm separable.

(3) (X, τun) is separable.

(4) (X, τun) is second countable.

Proof. The equivalence (1)⇔(2) holds for general metric spaces. 6 M.KANDIC´ AND A. VAVPETICˇ

(2)⇒(3) If X is norm separable, then X is separable with respect to any topology weaker than the norm topology. In particular, (X, τun) is separable.

(3)⇒(4) Suppose that (X, τun) is separable. By Proposition 3.1 we conclude that X has a countable quasi-interior set, so that (X, τun) is metrizable by [KT18, Theorem

4.3]. Since (X, τun) is metrizable and separable, it is second countable.

(4)⇒(2) Since (X, τun) is second countable, it is separable, so that by Proposi- tion 3.1 we conclude that X has a countable quasi-interior set G. By enumerating the

vectors in G as {u1,u2,...} and after replacing the n-th vector by the sum u1 +···+un,

we may assume without loss of generality that u1 ≤ u2 ≤ ...

Let us consider the order ideal IG generated by G. Since G is a quasi-interior set, the ideal IG is norm dense in X. We claim that the un-topology on IG equals the relative topology induced by the un-topology from X. To see this, note ﬁrst that IG the equality Ux,ǫ := {y ∈ IG : k|y| ∧ xk < ǫ} = Ux,ǫ ∩ IG for all x ∈ IG yields that IG every basis neighborhood Ux,ǫ of zero for the un-topology on IG is open in the relative topology. To prove the converse, pick any x ∈ X and ǫ> 0. Then Ux,ǫ ∩ IG is a basis

open neighborhood for zero in the relative topology on IG induced by the un-topology

of X. Since IG is norm dense in X, there exists a positive vector y ∈ IG such that ǫ IG kx − yk < . Pick any z ∈ U ǫ . From the inequality 2 y, 2 |z| ∧ x ≤ |z|∧|x − y| + |z| ∧ y < ǫ

IG we conclude that U ǫ ⊆ Ux,ǫ ∩ IG which proves the claim. y, 2 Since (X, τun) is second countable, the un-topology on IG is second countable as well. This yields that for each n ∈ N the interval [−nun, nun] is second countable with respect to the relative un-topology induced from IG. Since on order intervals the un-topology agrees with the norm topology, each order interval [−nun, nun] is norm separable. If Fn is a countable norm dense subset of [−nun, nun], then the set ∞ F := n=1 Fn is a countable norm dense subset of IG, so that norm density of IG in X yieldsS norm density of F in X. This proves that X is separable.

Although in metric spaces separability is equivalent to second countability, this is not the case even for locally convex Hausdorﬀ spaces. For example, the space RR equipped with the product topology is separable by [RS64, Theorem 1], yet the space is not second countable since it is an uncountable topological product. The following result tells that second countability and separability of the unbounded norm topology pass up and down between a normed lattice and its norm completion.

Corollary 3.3. For a normed lattice X the following assertions are equivalent. ONSEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 7

(1) (X, τun) is second countable. (2) (X,b τun) is second countable. (3) (X, τun) is separable. (4) (X,b τun) is separable. Proof. Equivalences (1)⇔(3) and (2)⇔(4) follow directly from Theorem 3.2. (1)⇒(2) By Theorem 3.2, the Banach lattice X is separable, so that X is also separable. Another application of Theorem 3.2 yieldsb that (X, τun) is separable. The converse implication can be proved similarly.

The following corollary follows from Theorem 3.2 and from the fact that X is separable whenever its norm dual X∗ is.

∗ Corollary 3.4. Let X be a normed lattice. If (X , τun) is separable, then (X, τun) is separable.

We conclude this section with a characterization of separability of atomic normed lattices with order continuous norm.

Corollary 3.5. For an atomic normed lattice X with order continuous norm the following statements are equivalent. (1) X is separable.

(2) (X, τun) is second countable.

(3) (X, τun) is separable.

(4) (X, τun) is ﬁrst countable.

(5) (X, τun) is metrizable. (6) X has an at most countable quasi-interior set. (7) Every disjoint set in X is at most countable. (8) Every disjoint set of atoms in X is at most countable.

Proof. If dim X < ∞, then all Hausdorﬀ linear topologies on X coincide so that, in this case, all statements between (1) and (5) are equivalent. Furthermore, X has a strong unit and every disjoint set of atoms in X has cardinality at most dim X < ∞. Assume now that dim X = ∞. Then the equivalences (1)⇔(2)⇔(3) follow from

Theorem 3.2 while (5)⇔(6) follows from [KT18, Theorem 4.3]. Since τun is a linear Hausdorﬀ topology on X, (4)⇔(5) follows from [KN76, p. 49]. Furthermore, while Proposition 3.1 yields the implication (3)⇒(6), the implication (7)⇒(8) is obvious.

(6)⇒(7) Suppose that D = {un : n ∈ N} is an at most countable quasi-interior set

in X and pick a disjoint family {xλ : λ ∈ Λ} of positive vectors in X. Then for each 8 M.KANDIC´ AND A. VAVPETICˇ n ∈ N the family {un ∧ xλ : λ ∈ Λ} is an order bounded disjoint family in X. For all n, m ∈ N we consider the set

1 Λn,m := {λ ∈ Λ: kun ∧ xλk ≥ m }.

Fix n ∈ N and suppose that Λn,m0 is inﬁnite for some m0 ∈ N. Take any sequence

(yk)k∈N in {un ∧ xλ : λ ∈ Λn,m0 } of distinct vectors. Since X is order continuous, by [Lux65b, Theorem 64.3] the norm completion X is also order continuous, so that the disjoint sequence (yk)k∈N converges in normb to zero by Theorem (see [AB06, Theorem 4.14]). This contradicts the fact that ky k ≥ 1 for each k ∈ N. Therefore, k m0 for all n, m ∈ N the set Λn,m is at most countable. ∞ Now consider the set Λ0 := n,m=1 Λn,m which is clearly countable. If λ ∈ Λ \ Λ0, then xλ ∧ un = 0 for each n ∈SN. Since D is a quasi-interior set, we conclude that xλ = 0, from where it follows that Λ is at most countable. (8)⇒(1) Pick a maximal set A of atoms in X. Since X is inﬁnite-dimensional, we may enumerate the elements of A as (en)n∈N. For each n ∈ N consider the linear span

Jn of the set {e1,...,en}. Since Jn is spanned by atoms, Jn is a projection band ∞ in X. Since each Jn is ﬁnite-dimensional, the union F := n=1 Jn is a separable subspace of X. Since X is atomic, F is order dense in X, andS since the norm on X is order continuous, F is norm dense in X. This proves separability of X.

Remark 3.6. Some implications in Corollary 3.5 do not hold if we relax certain assumptions. (1) The implications (6)⇒(7) and (8)⇒(1) in Corollary 3.5 do not necessarily hold if we replace order continuity of the norm with Dedekind completeness. To see this, for a given index set I consider the Dedekind complete Banach lattice ℓ∞(I) of all bounded sequences over the index set I. Clearly, ℓ∞(I) contains a strong unit. If I is uncountable, then ℓ∞(I) contains uncountably many pairwise disjoint atoms. On the other hand, in ℓ∞ := ℓ∞(N) every disjoint set of atoms is at most countable, yet ℓ∞ is not separable. (2) The implication (7)⇒(1) does not hold for order continuous Banach lattices without atoms. To see this, consider the circle S1 equipped with the Borel σ-algebra and the arc-length measure. Furthermore, consider the uncountable product of cir- cles equipped with the product σ-algebra and the product probability measure µ. Since the coordinate functions are pairwise orthogonal, the Hilbert space L2(µ) is not separable. Since the constant one function is a quasi-interior point in L2(µ) similar arguments as in the proof of implication (6)⇒(7) in Corollary 3.5 show that every pairwise disjoint set in L2(µ) is countable. ONSEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 9

4. Components of positive vectors and separability of un-topology In Section 3 we considered separability of un-topology in comparison with general topological properties of norm topology and un-topology. In this section we study separability of un-topology of normed lattices with the principal projection property in terms of separability of “special” metric spaces which can be considered as their building blocks. These metric spaces are precisely spaces of components of positive vectors equipped with the metric induced by the norm of the underlying normed lattice. The motivation comes from Measure theory.

Given a ﬁnite measure space (Ω, F,µ) one can deﬁne the map dµ : F×F → [0,µ(Ω)] as

dµ(A, B) := kχA△Bk1 = µ(A \ B)+ µ(B \ A).

It is easy to see that dµ is a semi-metric on F. On F we introduce the equivalence relation ∼ with

A ∼ B ⇔ dµ(A, B)=0.

Then the semi-metric dµ induces the well-deﬁned metric d on F/∼ deﬁned as

d([A], [B]) := dµ(A, B).

It should be clear that the semi-metric space (F,dµ) is separable if and only if the metric space (F/∼,d) is separable. The following proposition which immediately follows from Theorem 46.4 and Theo- rem 46.5 proved by Luxemburg in [Lux65a] presents an important connection between separability of the norm topology (or equivalently, separability of the un-topology) and the local structure of normed lattices.

Proposition 4.1. Suppose (Ω, F,µ) is a ﬁnite measure space. Then L1(µ) is separable if and only if the semi-metric space (F,dµ) is separable.

Since each L1-function is a limit of a sequence of step functions, the key role in the proof of Proposition 4.1 actually play characteristic functions in L1(µ) or equivalently their representing sets in F. The concept of a characteristic function can be fruitfully generalized to the realm of vector lattices. Given a positive vector u of a vector lattice X, a positive vector x is a component of u if 0 ≤ x ≤ u and x ⊥ (u − x). Since for each measurable set A ⊆ X we have χA + χΩ\A = 1 and χA ∧ χΩ\A = 0, the notion of a component of a positive vector naturally extends the notion of a characteristic function to vector lattices. The set of all components of a given vector u is denoted by Cu. It should be clear that we always have 0,u ∈ Cu. A component x ∈ Cu is called non-trivial if x is neither 0 nor u. A linear combination of components of 10 M.KANDIC´ AND A. VAVPETICˇ u is called a u-step function. It is easy to see that we can always take pairwise disjoint components when working with a particular u-step function. The following result is clearly a generalization of Proposition 4.1. It will be used in Theorem 5.6 in order to obtain a characterization of separability of order continuous Banach function spaces over semi-ﬁnite measure spaces.

Theorem 4.2. For a normed lattice X with the principal projection property the following statements are equivalent. (1) X is norm separable.

(2) (X, τun) is separable. (3) X contains a countable quasi-interior set and for each countable quasi-interior

set Q and each positive vector u ∈ Q the metric space (Cu, ρ) is separable. (3’) X contains a countable quasi-interior set Q such that for each positive vector

u ∈ Q the metric space (Cu, ρ) is separable. Moreover, if X is a Banach lattice, then (1)-(3’) are equivalent to each of the following statements. (4) X contains a quasi-interior point and for each quasi-interior point u the metric

space (Cu, ρ) is separable.

(4’) X contains a quasi-interior point u such that the metric space (Cu, ρ) is separable.

Proof. While the equivalence (1)⇔(2) follows from Theorem 3.2, implications (3)⇒(3’) and (4)⇒(4’) are obvious.

(2)⇒(3) Suppose that X is τun-separable. By Proposition 3.1, X contains a countable quasi-interior set. Let Q be any quasi-interior set and pick any positive vector

u from Q. Then Cu equipped with the metric d induced by the norm is a subspace

of the metric space (X,d). Since X is separable metric space, its subspace Cu is separable as well. + (3’)⇒(1) Pick any positive vector u ∈ X such that Cu is separable, and let

Q be a countable dense set of Cu. Let G be the set of all linear combinations of

vectors from Q with rational coeﬃcients. We claim that G is dense in Iu. To this

end, pick any positive vector x ∈ Iu and ǫ > 0. By Freudenthal’s spectral theorem

(see e.g. [Zaa97, Theorem 33.2]) there exists an increasing sequence (sn)n∈N of u-step

functions such that 0 ≤ sn ↑ x and that sn → x u-uniformly. In particular, sn → x + in norm, so that the positive part of the linear span of Cu is dense in Iu . Due to the + − equality x = x − x we immediately obtain that the linear span of Cu is dense in ON SEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 11

Iu. Since Cu is separable, by a standard approximation argument one can conclude that Iu is separable as well.

For the general case, let us enumerate the vectors from Q as {u1,u2,...}. For each n ∈ N we denote by wn the vector u1 +···+un. From the identity Iwn = Iu1 +···+Iun and the fact that the deal Iuk is separable for each k ∈ N, we conclude that the ideal

Iwn is separable as well. From this we immediately derive that the increasing union ∞ X0 := n=1 Iwn is separable in X. Since X0 is norm dense in X, we obtain separability of X. S Suppose now that X is also norm complete. For the moreover statement it is enough to see that (1) implies (4). This follows immediately from the fact that every separable Banach lattice has a quasi-interior point and from the implication (1)⇒(3).

The following example shows that Theorem 4.2 does not hold if one does not assume that the underlying normed lattice has the principal projection property.

Example 4.3. Let K be a product of uncountably many copies of the interval [0, 1]. Since K is connected, we have C1 = {0, 1}, so that C1 is separable. On the other hand, C(K) is not separable as K is not metrizable (see [AK06, Theorem 4.1.3]).

5. Applications to Banach function spaces In this section we characterize separable Banach function spaces over semi-ﬁnite measure spaces in terms of measure-theoretical properties of the underlying measure space. Given a measure space (Ω, F,µ), a function space X is an (order) ideal in L0(µ). A function space X is a Banach function space if it is equipped with a complete lattice norm. The following lemma (see [Lux65a, Theorem 46.2]) shows that only order continuous Banach function spaces can be separable.

Lemma 5.1. Every separable Banach function space in L0(µ) is order continuous.

Let X be an order dense order continuous Banach function space in L0(µ) where µ is ﬁnite. Norm separability of X is equivalent to separability of X equipped with un-topology. An application of [KLT18, Theorem 5.2] yields that X is separable if and only if X is separable with respect to the topology τµ of convergence in measure. 0 In fact, separability of (X, τun) depends only on L (µ).

Lemma 5.2. Let (Ω, F,µ) be a ﬁnite measure space and let X be an order dense order continuous Banach function space in L0(µ). Then X is norm separable if and only if L0(µ) equipped with the topology of convergence in measure is separable. 12 M.KANDIC´ AND A. VAVPETICˇ

Proof. Suppose ﬁrst that L0(µ) is separable. Since µ is ﬁnite and X is order continuous, by [KLT18, Theorem 5.2] the topology of convergence in measure is the un-topology on L0(µ) induced by X. Pick a weak unit g in X. Due to order density of X in L0(µ) we have that g is also a weak order unit in L0(µ). Since the norm of X is order continuous, g is a quasi-interior point in X. Hence, by [KLT18, Theorem 3.3] and its proof we conclude that the topology of convergence in measure on L0(µ) is metrizable, and that the associated metric is given by

dg(f1, f2)= k|f1 − f2| ∧ gkX . Since the un-topology on X is just the restriction of un-topology induced by X on L0(µ), the un-topology on X is precisely the topology of convergence in measure restricted to X. Since (X, τµ|X ) is separable as a subspace of a separable metric space by Theorem 3.2 we conclude that X is norm separable. To prove the converse statement, assume that X is norm separable. By Theorem 3.2 it follows that X is τun-separable. To ﬁnish the proof it suﬃces to prove that X is 0 0 τµ-dense in L (µ). Pick any non-negative function f ∈ L (µ). Since X is order dense in L0(µ) and L0(µ) has the countable sup property, there is an increasing sequence

(fn)n∈N in X such that fn ↑ f. This yields that fn → f µ-almost everywhere, and since the measure µ is ﬁnite, by a consequence of the well-known Egorov’s theorem we conclude that fn → f in measure.

Proposition 5.3. Let (Ω, F,µ) be a ﬁnite measure space. For an order continuous order dense Banach function space X in L0(µ) the following assertions are equivalent. (1) X is norm separable. (2) X is separable with respect to the topology of convergence in measure.

(3) The semi-metric space (F,dµ) is separable.

(4) The metric space (Cf , ρ) is separable for each strictly positive function f ∈ X. If F is countably generated, then all the statements (1)-(4) hold.

Proof. Equivalences between (1), (2) and (4) follow from Theorem 4.2. 1 (3)⇔(1) If (F,dµ) is separable, then (C1, ρ) is separable in L (µ). The implication (4)⇒(1) in the case of the Banach lattice L1(µ) yields that L1(µ) is norm separable, and so by an application of Lemma 5.2 we conclude that X is norm separable. The proof of the converse statement can be proved similarly and is therefore omitted. Although the proof that the additional assertion implies (3) is well known, we will

provide its sketch solely for the sake of completeness. Suppose that a countable set F0

generates F. Let F1 be the family of all sets which are obtained after ﬁnitely many ON SEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 13

steps of eﬀecting Boolean operations on the sets of F0. Since F0 is countable, F1 is

countable as well. Let F2 be the closure of F1 in F with respect to the semi-metric dµ. The sets from F2 are precisely the sets which are approximable by sets from

F1. Since the measure µ is ﬁnite, the complement of an approximable set is again an approximable set. Also, countable unions of approximable sets are also approximable.

Hence, F2 is a σ-subalgebra in F. Since F2 contains F0, we conclude F2 = F, and so F is separable as it is the closure of the countable set F1.

In Theorem 5.6 we will extend Proposition 5.3 to order continuous Banach function spaces over semi-ﬁnite measures. First we need a general result about semi-ﬁnite measures.

Lemma 5.4. Let (Ω, F,µ) be a semi-finite measure space which is not σ-finite. Then there exists an uncountable disjoint family of measurable sets in Ω of finite positive measure.

Proof. Let A ⊆ P(Ω) be the family of all families of measurable pairwise disjoint subsets of Ω with ﬁnite positive measure. Since µ is semi-ﬁnite, the family A is non- empty. The family A ordered by set inclusion is a partially ordered set. If C is a chain in A, then it is obvious that the union of C is again in A. Hence, by Zorn’s lemma there exists a maximal element M in A.

Suppose that M is countable. Since E := F ∈M F is a measurable subset of Ω and since µ is not σ-finite, the set Ω \ E has positiveS measure. Due to semi-finiteness of µ there exists a measurable set F ⊆ Ω \ E with a finite positive measure. This is in contradiction with maximality of M.

For a function space X over a measure space (Ω, F,µ) and a measurable set A ∈F 0 we deﬁne the set XA = {χAf : f ∈ X}. Since X is an ideal in L (µ), XA is an ideal 0 in X, so that XA is a function space in L (µ). If X is a Banach function space, then

XA is a Banach function space as well. By F|A we denote the relative σ-algebra on A, i.e., the σ-algebra of all measurable sets contained in A. Whenever µ(A) < ∞, the restriction dµ|A of dµ on F|A induces a semi-metric. If X is an order continuous Banach function space over a σ-finite measure space, then [KLT18, Theorem 5.2] yields that τun agrees with the topology τµ of local convergence in measure on sets of finite measure. An examination of its proof actually reveals that one can replace σ-finite measure spaces with semi-finite ones. Recall that 0 a net (fλ) converges to f in L (µ) with respect to the topology of local convergence in measure if fλχA → fχA in measure for each subset A ∈F with finite measure. 14 M.KANDIC´ AND A. VAVPETICˇ

Lemma 5.5. Let X be an order dense order continuous Banach function space over a semi-ﬁnite measure space. Then a net (fα) in X τun-converges to zero if and only if it converges to zero with respect to the topology of local convergence in measure.

Proof. For the proof of the forward implication, pick a set A of ﬁnite measure. Then fα ∧ χA ∈ XA, and since fαχA τun-converges to zero in X, it also τun-converges to zero in XA. Since µ(A) < ∞, [KLT18, Theorem 5.2] yields that (fαχA) converges to zero in measure on A. This proves the forward implication.

For the proof of the backward implication, assume that (fα) converges to zero with respect to the topology of local convergence in measure. As in the proof of [KLT18, Theorem 5.2] let Z be the set of all functions in X which vanish outside of a set of finite measure. We claim that Z is a non-trivial ideal in X. Pick a non-zero function + 1 f ∈ X . Then there exists n ∈ N such that the set An := {x ∈ Ω: f(x) ≥ n } has a positive measure. Since µ is semi-finite, there exists a measurable set A ⊆ An of 0 finite positive measure. Hence, χA ≤ χAn ≤ nf and since X is an ideal in L (µ), we

have that χA ∈ X. This yields that χA ∈ Z and so Z is non-trivial. The fact that Z is an ideal in X is clear. + Now we claim that |fα| ∧ g → 0 for each g ∈ Z . Since g can be approximated by an increasing sequence of step functions and since X is order continuous, it suﬃces to consider the case when g itself is a step function. Furthermore, an application of the triangle inequality for the norm implies that we need to consider only the case when

g = χA for some set A of positive ﬁnite measure. Since |fα| ∧ χA → 0 in measure on

A, by [KLT18, Theorem 5.2] we have that |fα| ∧ χA → 0 in norm in XA and so in X. + This proves that |fα| ∧ g → 0 for each g ∈ Z . To conclude the proof, due to the standard approximation argument we need to show that Z is norm dense in X. Since X is order continuous, it suffices to prove that Z is order dense in X. Pick a positive vector f ∈ X+ and find a step function s ∈ X+ such that 0 < s ≤ f. Then there exists a measurable set A and a scalar ′ α> 0 such that 0 < αχA ≤ s ≤ f. Since µ is semi-finite, there is a subset A of A of

ﬁnite positive measure, from where it follows 0 < αχA′ ≤ f. To conclude the proof,

note that we have χA′ ∈ Z.

The following two theorems are the main results of this section. They provide a characterization of separability of order continuous Banach function spaces.

Theorem 5.6. For an order continuous order dense Banach function space X over a semi-ﬁnite measure space (Ω, F,µ) the following statements are equivalent. (1) X is norm separable. ON SEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 15

(2) X is separable with respect to the topology of local convergence in measure.

(3) The measure µ is σ-ﬁnite and the semi-metric space (F|A,dµ|A ) is separable for each set A of ﬁnite measure.

(4) The measure µ is σ-ﬁnite and the function space XA is separable for each set A of ﬁnite measure.

(5) X admits strictly positive functions and the metric space (Cf , ρ) is separable for any strictly positive function f ∈ X.

Proof. While the equivalence (1)⇔(2) follows from Theorem 3.2 and Lemma 5.5, the equivalence (3)⇔(4) follows from Proposition 5.3. (1)⇒(4) Assume that µ is not σ-ﬁnite. By Lemma 5.4 there exists an uncountable family M of pairwise disjoint sets of ﬁnite positive measure. Since X is order dense 0 in L (µ), for each E ∈M there exists a function gE ∈ X such that 0

1 Fn := {gE : E ∈M and kgEkX ≥ n }.

Since M is uncountable, there exists m ∈ N such that Fm is uncountable. For each 1 gE ∈ Fn consider the open ball UE with center in gE and radius 2m . We claim that ′ for diﬀerent sets E and E the balls UE and UE′ are disjoint. Indeed, if f is in their 1 ′ ′ intersection, then kgE − gE k≤kgE − fk + kf − gE k < m . However, this is in contradiction with

′ ′ ′ 1 kgE − gE k = k|gE − gE |k = kgE + gE k≥kgEk ≥ m . Since X admits an uncountable family of pairwise disjoint open sets, it cannot be separable. To prove the remaining claim of (4), pick a set A of ﬁnite measure in Ω and note that separability of the normed space X passes down to the Banach function space

XA.

(4)⇒(1) Since µ is σ-finite, there exists an increasing sequence (An)n∈N of sets of ∞ finite positive measure such that n=1 An = Ω. Since each space XAn is separable by ∞ Proposition 5.3, a simple topologicalS argument yields separability of Y := n=1 XAn . We claim that Y is norm dense in X. To this end, it suffices to proveS that Y + + + is norm dense in X . Pick a function f ∈ X . Then the sequence (fχAn )n∈N is

increasing pointwise to f, so that fn → f in order. Since X is order continuous, we

conclude that fn → f in norm which proves the claim. Separability of X follows now from separability of Y and its density in X. 16 M.KANDIC´ AND A. VAVPETICˇ

(1)⇔(5) If X is norm separable, then (1)⇒(3) implies that the measure µ is σ- ﬁnite, so that by [AA02, Corollary 5.22] the function space X contains a function f which is strictly positive almost everywhere on Ω. Such a function is a weak unit, so that from order continuity of the norm we conclude that f is a quasi-interior point in X. Separability is inherited to (Cf , ρ). On the other hand, if X admits a function f which is strictly positive almost everywhere on Ω, then f is a quasi-interior point in X. To ﬁnish the proof we apply Theorem 4.2.

Suppose that µ is σ-ﬁnite. Then for each ideal X in L0(µ) by [AA02, Theorem

1.92] there exists the smallest measurable subset CX ⊆ Ω with respect to µ-almost everywhere set inclusion such that every f ∈ X vanishes µ-almost everywhere on

Ω\CX . The set CX is called the carrier (or the support) of the ideal X. By [AA02, 0 Lemma 1.94], the function space X is always order dense in L (CX ). This immediately yields that a function space is order dense in L0(µ) if and only if its carrier is Ω. If

X is a separable Banach function space, then X has a carrier CX . Indeed, if X is separable, then X contains a quasi-interior point ϕ. Let A := {x ∈ Ω: ϕ(x) > 0}. Since for each g ≥ 0 in X we have g∧nϕ → g, the function g is zero almost everywhere on X \ A and so g ∈ XA. This proves that X = XA. Since ϕ is strictly positive on

A, we have CX = A. For Banach function spaces over semi-ﬁnite measures their separability can be characterized as follows.

Theorem 5.7. For an order continuous Banach function space over a semi-ﬁnite measure the following assertions are equivalent. (1) X is separable. (2) X is separable with respect to the topology of local convergence in measure and

the carrier CX is σ-ﬁnite.

Proof. (1)⇒(2) Since X is separable, it has a quasi-interior point ϕ. By the discussion preceding the theorem, the set A := {x ∈ Ω: ϕ(x) > 0} is the carrier CX of X. We 0 0 claim that X is order dense in L (µ|A). To see this, pick any function 0 < f ∈ L (µ|A). 1 Since f is non-zero, there exists n ∈ N such that the set B := {x ∈ A : f(x) ≥ n } has a positive measure. Then χB ≤ nf. By the deﬁnition of a carrier there exists + 1 a non-zero function g ∈ X such that g ∧ χB =6 0. Hence, 0 ≤ n (g ∧ χB) ≤ f, 0 so that XA is order dense in L (µ|A). Since X = XA is norm separable and µ|A is

semi-ﬁnite, by Theorem 5.6 we conclude that the measure µ|A is σ-ﬁnite and that

X = XA is separable with respect to the topology of local convergence in measure on ON SEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 17

A. We conclude the proof of (2) by observing that each function in X is zero almost everywhere on Ω \ A. (2)⇒(1) The same argument as in the proof of the previous implication shows that 0 X = XA is order dense in L (µ|A), Hence, by Theorem 5.6 we conclude that X = XA is norm separable. Theorem 5.6 and Theorem 5.7 should be compared with [Lux65a, Theorem 46.4 and Theorem 46.5]. The one-dimensional Banach function space L∞(µ) over a singleton set with infinite measure shows that neither Theorem 5.6 nor Theorem 5.7 do not hold whenever the underlying measure space is not semi-finite. The equivalence between (1) and (3) in Theorem 5.6 directly shows that over a semi- finite measure µ the space Lp(µ) is separable if and only if the Lq(µ) is separable for 1 ≤ p,q < ∞. The following corollary shows that this is also the case for general measure spaces.

Corollary 5.8. For every measure space (Ω, F,µ) and 1 ≤ p,q < ∞ the Banach lattice Lp(µ) is separable if and only if Lq(µ) is separable.

Proof. Suppose that Lp(µ) is separable. Then Lp(µ) contains a quasi-interior point f. As before, let us denote the set {x ∈ Ω: f(x) > 0} by A. Since f is a quasi-interior p p p p point in L (µ), we have L (µ)= L (µ|A), and since the function f q is a quasi-interior q q q point in L (µ) we have L (µ)= L (µ|A). To ﬁnish the proof note that Theorem 5.6 p q yields that L (µ|A) (resp., L (µ|A)) is separable if and only if the topological space (B,dµ|B ) is separable for each subset B of A of ﬁnite measure. So far the results in this section provided a connection between separability of a given Banach function space and separability of the underlying measure space. In the following proposition we establish a suitable topological version for L1-spaces.

Proposition 5.9. Let Ω be a second countable topological space. If (Ω, F,µ) is a measure space with an outer regular Borel measure, then Lp(µ) is separable for each 1 ≤ p< ∞.

Proof. By Corollary 5.8 it is enough to prove that L1(µ) is separable. If there are no sets of positive finite measure in Ω, then Lp(µ) = {0}. Otherwise, let W be a countable basis for X and let W′ be the family of all finite unions of elements of W of finite measure. We will show that the non-empty countable set

S = q χ : K ⊂ W′ is a ﬁnite set and q ∈ Q W W W WX∈K 18 M.KANDIC´ AND A. VAVPETICˇ is dense in L1(µ). Let f ∈ L1(µ) and ε > 0. Then there exists a step function n 1 n ε Q k=1 qkχAk ∈ L (µ) such that kf − k=1 qkχAk k1 < 2 where qk ∈ . Since µ is Pouter regular for every k there existsP an open set Vk ⊇ Ak such that µ(Vk \ Ak) < ε k+2 . Since Vk is a (countable) union of elements from W, there exists 2 max{1,|qk|} ′ ε Wk ∈ W , such that Wk ⊆ Vk and µ(Vk \ Wk) < k+2 . Then 2 max{1,|qk|}

ε kχA − χW k1 = µ(Ak \ Wk)+ µ(Wk \ Ak) ≤ µ(Vk \ Wk)+ µ(Vk \ Ak) < k+1 , k k 2 max{1,|qk|} and therefore n n n n f − q χ ≤ f − q χ + q χ − q χ < ε. k Ak k Wk k Wk k Ak Xk=1 1 Xk=1 1 Xk=1 Xk=1 1

The following example shows that there exist a topological space that is not second countable, yet the corresponding L1-space is separable.

Example 5.10. Consider the S¨orgenfrey line RS, i.e., the real line R equipped with the topology deﬁned by the basis {[a, b): a, b ∈ R}. It is well known that RS is separable but not second countable. The corresponding Borel σ-algebra BS is the 1 standard Borel σ-algebra on R, and hence, L (RS, BS, m) is separable.

In comparison with the Sorgenfrey line one can also ﬁnd a non-separable topological space Ω for which the function space L1(µ) is still separable.

Example 5.11. Let τE be the Euclidean topology on R and let

B = {U \ C : U ∈ τE and C is countable}

be a basis of a topology τ on R. Let Bτ and BE be the Borel σ-algebras generated

by τ and τE, respectively. Because τ is stronger than τE, we have BE ⊆ Bτ . On the

other hand, every set U \ C is in BE, so that B ⊆ BE, and therefore, Bτ ⊆ BE. As 1 in the previous example we have that L (R, Bτ , m) is separable. However, the space (R, τ) is not separable, since for every countable set Q we have Q ∩ (R \ Q)= ∅ and R \ Q ∈ τ.

6. The metric space of components In this section we are interested in ﬁnding conditions on the underlying lattice

under which either Iu = Iv or Bu = Bv implies that the corresponding spaces of

components Cu and Cv are homeomorphic or even isometric with respect to their natural metrics induced by the norm. First we recall the classical representation theory due to Kakutani. ON SEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 19

Given a positive vector x in a vector lattice X, consider the principal ideal Ix generated by the vector x. It is known that

Ix = {y ∈ X : |y| ≤ λx for some λ ≥ 0} and that the mapping k·kx on Ix deﬁned as

kykx := inf{λ> 0 : |y| ≤ λx} is a lattice seminorm on Ix. If X is Archimedean, then k·kx is a lattice norm on

Ix such that for every y ∈ Ix we have |y|≤kykx x. If X is uniformly complete,

then (Ix, k·kx) is a Banach lattice with a strong unit x so that by an application of the classical Kakutani representation theorem of AM-spaces with strong units

(see [AB06, Theorem 4.29]) (Ix, k·kx) is lattice isometric to a Banach lattice C(K) for some compact Hausdorﬀ space K. Furthermore, one can require that the isomorphism maps x to the constant function 1K on K. We start by a simple lemma whose proof is provided only for the sake of completeness.

Lemma 6.1. Let u and v be positive vectors of an Archimedean vector lattice X. If

Iu = Iv, then the identity mapping I : (Iu, k·ku) → (Iv, k·kv) is an isomorphism of normed lattices.

Proof. Since Iu = Iv, the identity mapping between I : (Iu, k·ku) → (Iv, k·kv) is clearly a lattice isomorphism. To prove that it is also an isomorphism of normed lattices,

we first find λ ≥ 0 such that u ≤ λv. Then for x ∈ Iu the inequality |x|≤kxkuu immediately yields that |x| ≤ λkxkuv, from where it follows that kxkv ≤ λkxku. This proves that I : (Iu, k·ku) → (Iv, k·kv) is continuous. Similarly one can prove that its inverse is continuous as well. Consider the vector lattice C(K) of all real-valued continuous functions ordered pointwise on a non-empty compact Hausdorff space K. Pick non-negative functions u, v ∈ C(K) such that Iu = Iv. Then their zero sets Zu = {t ∈ K : u(t)=0} and

Zv = {t ∈ K : v(t)=0} coincide. Since Iu = Iv, there exists λ ≥ 0 such that u ≤ λv from where it follows that for each x ∈ Iu = Iv the function Sx deﬁned by

v(t) x(t), t/∈ Zu, (Sx)(t)=  u(t) 0, t ∈ Zu,

is continuous. This yields that S : Cu→ Cv is an isomorphism of lattices which in general is not isometric. Nevertheless, the Kakutani representation theorem allows us to modify S in a way that it becomes also isometric. 20 M.KANDIC´ AND A. VAVPETICˇ

Proposition 6.2. Let X be a uniformly complete Archimedean vector lattice. Suppose

that for positive vectors u and v we have Iu = Iv. Then there exists a surjective

isometry between metric spaces Cu and Cv.

Proof. By Lemma 6.1, (Iu, k·ku) and (Iv, k·kv) are isomorphic as Banach lattices. By the Kakutani representation theorem there exist compact Hausdorﬀ spaces K

and L such that (Iu, k·ku) and (Iv, k·kv) are lattice isometric to (C(K), k·k∞) and

(C(L), k·k∞), respectively. From this we conclude that C(K) and C(L) are isomorphic as Banach lattices. Let T : C(K) → C(L) be a Banach lattice isomorphism between C(K) and C(L). By [AB06, Theorem 2.34] there exists a unique non-negative function g ∈ C(L) and a mapping ζ : L → K which is continuous on the set {y ∈ L : g(y) > 0} satisfying (T f)(y)= g(y)f(ζ(y)). If g(y) = 0 would hold for some y ∈ L, then (T f)(y) = 0 for all f ∈ C(K) which would imply that T is not surjective. Hence, g is strictly positive, so that ζ is continuous on L. Since g is strictly positive and since T is a lattice isomorphism, the mapping S : C(K) → C(L) deﬁned as (Sf)(y)= f(ζ(y)) is a lattice and algebra isomorphism

which satisﬁes S(1K)= 1L. An application of [EFHN15, Lemma 4.14] yields that the mapping ζ is a homeomorphism, so that the mapping S is isometric.

Hence, there exist a Banach lattice isometric isomorphism S : (Iu, k·ku) → (Iv, k·kv) which maps u to v. To conclude the proof observe that its restriction S| is a e Cu surjective isometry between metric spaces Cu and Cv. e Question 6.3. Does Proposition 6.2 still hold if the assumption of uniform completeness is removed?

Since Banach lattices are Archimedean and uniformly complete, the following corollary immediately follows Proposition 6.2.

Corollary 6.4. Let u and v be positive elements of a Banach lattice X. If Iu = Iv, then there exists a surjective isometry between metric spaces Cu and Cv.

The natural question whether the equality between principal ideals could be re- placed by the equality between principal bands has unfortunately a negative answer even when the underlying Banach lattice is Dedekind complete as the following example shows.

Example 6.5. Consider the Banach lattice ℓ∞ of all bounded sequences. Since the 1 ∞ 1 1 vector := (1, 1,...) is a strong unit in ℓ and the vector u = (1, 2 , 3 ,...) is a weak ON SEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 21

∞ ∞ unit in ℓ we clearly have that ℓ = B1 = Bu. Since we have

Cu = {uχA : A ⊆ N} and C1 = {χA : A ⊆ N},

it follows that the mapping uχA 7→ χA is a bijection between Cu and C1. Since the

metric space C1 is discrete, any mapping φ: C1 → Cu is continuous.

We claim that there is no continuous injective mapping φ: Cu → C1. To this end,

suppose that there exists a continuous injective mapping φ: Cu → C1. For each n ∈ N 1 ∞ let us consider the component un := (1,..., n , 0,...) of u. Since un → u in ℓ , we would have φ(un) → φ(u). Since both φ(un) and φ(u) are distinct components of 1 in ∞ ℓ , the sequence φ(un) − φ(u) would need to have at least one non-zero term. Since the components of the vector 1 in ℓ∞ are sequences consisting only of zeros and ones,

we would conclude that for each n ∈ N we would have kφ(un) − φ(u)k ≥ 1. This

contradiction shows that Cu and C1 are not homeomorphic.

It turns out that in Corollary 6.4 we can replace principal ideals by principal bands in the case when the underlying Banach lattice is order continuous (see Theorem 6.10).

We will prove this result by an explicit construction of a homeomorphism between Cu

and Cv. The deﬁnition of the aforementioned homeomorphism will rely on a very easy yet useful description of components of a given vector in the case when the lattice has the principal projection property.

Lemma 6.6. Let u be a positive vector of a vector lattice X with the principal projection property. Then

Cu = {Pu : P is a principal band projection }.

Proof. Let us denote the right-hand side set appearing in the lemma by S. If P is a principal band projection, then for every positive vector u ∈ X we have Pu ⊥ (u−Pu) and u = Pu +(u − Pu). Hence, Pu is a component of u which proves that S ⊆ Cu.

To prove that Cu ⊆ S, pick x ∈ Cu and consider the principal band projection P onto the principal band generated by x. Since x ⊥ (u−x), we have that P (u−x) = 0, and since P x = x we conclude that Pu = P (u − x)+ P x = x.

Now it should be quite obvious that the required homeomorphism ϕ: Cu → Cv should be deﬁned as ϕ: Pu 7→ P v for all principal band projections P .

Lemma 6.7. Let X be a vector lattice with the principal projection property. Suppose

that Bu = Bv for some positive vectors u, v ∈ X.

(1) Then the mapping ϕ: Cu → Cv is a well-deﬁned bijection.

(2) If X is a normed lattice and u ∈ Iv, then ϕ: Cv → Cu is a continuous bijection. 22 M.KANDIC´ AND A. VAVPETICˇ

Proof. It suﬃces to prove the statement only for the special case when u and v are weak units of X. Indeed, suppose that the statement holds for the case of weak units and consider the principal band X0 := Bu = Bv of X. By deﬁnition, u and v are weak units of X0, and since X0 is also an ideal in X, by [LZ71, Theorem 25.2] the vector lattice X0 has the principal projection property. Since Cu and Cv are subsets

of X0, the general result follows from the special case.

(1) We need to check first that ϕ is well-defined. Since by Lemma 6.6 P v ∈ Cv for each principal band projection P , it remains to prove that Pu = Qu yields P v = Qv. Assume, therefore, that Pu = Qu for some principal band projections P and Q. Since Pu = Qu trivially implies (P − Q)2u = 0 and since (P − Q)2 = P (I − Q)+ Q(I − P ) ≥ 0 is an order continuous operator, the fact that u is a weak unit in X yields that (P −Q)2 = 0. On the other hand, from positivity of operators P (I −Q) and Q(I −P ) and from the identity P (I −Q)+Q(I −P )=(P −Q)2 = 0 we conclude that P = P Q and Q = QP . Since band projections always commute, we finally obtain that P = Q which proves that ϕ is well-defined.

The same argument shows that the mapping ψ : Cv → Cu deﬁned by ψ(P v)= Pu is well-deﬁned which is clearly the inverse of ϕ.

(2) Suppose that a sequence (Pnv)n∈N converges to some component P v in Cv.

Since u ∈ Iv, there exists λ ≥ 0 such that u ≤ λv. Hence, from the inequality

|Pnu − Pu| ≤ λ|Pnv − P v| we conclude that Pnu → Pu, so that ϕ is continuous. The following corollary should be compared with Proposition 6.2 where an existence of a surjective isometry between Cu and Cv is proved for uniformly complete Archimedean vector lattices. If we replace the assumption on uniform completeness with the principal projection property, the mapping ϕ deﬁned above is only a homeomorphism. Moreover, ϕ is isometric if and only if u = v. This will follow from Proposition 6.11.

Corollary 6.8. Let X be a vector lattice with the principal projection property. If

Iu = Iv for some positive vectors u and v, then the mapping ϕ: Cu → Cv is a homeomorphism.

If we replace the condition Iu = Iv in the corollary above by a weaker condition

Bu = Bv, in general, the mapping ϕ may not be continuous even when X is Dedekind complete as it was shown in Example 6.5. Continuity of ϕ follows automatically ON SEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 23

when the underlying Banach lattice is order continuous. Before we state and prove Theorem 6.10 which is the main result of this section, we need the following result which is of independent interest.

Proposition 6.9. Let X be an Dedekind complete normed lattice which has the σ-

Fatou property or is a Banach lattice. Suppose that (Pn)n∈N is a sequence of band + projections such that Pnu → Pu for some band projection P and some u ∈ X . Then

there exists a subsequence (Pnk )k∈N which converges in order to P on Bu and strongly on Iu .

Proof. Let us ﬁrst consider the case when P = 0. Then Pnu → 0, so that there is a 1 subsequence (Pnk )k∈N such that for each k ∈ N we have kPnk uk ≤ k2 . For each k ∈ N we consider the band projection ∞

Qk := Pnj . j_=k

It is clear that the sequence (Qk)k∈N is decreasing.

We claim that Qku → 0. Consider ﬁrst the case when X is a Banach lattice. Then ∞ for each k the series j=k Pnj u converges in X to some positive vector uk which satisﬁes P ∞ ∞ 1 ku k ≤ kP uk ≤ → 0 k nj j2 Xj=k Xj=k l l as k → ∞. Since for each l ≥ k we have j=k Pnj u ≤ j=k Pnj u ≤ uk, due to order closedness of the positive cone we conclude W P ∞ P u ≤ u . nj k j_=k ∞ Since uk → 0, we have Qku = j=k Pnj u → 0 as k →∞. Let us now consider the case W that X has the σ-Fatou property. Pick any ǫ > 0. ∞ ∞ 1 Since j=k kPnj uk ≤ j=k k2 → 0 as k →∞, there exists k0 ∈ N such that P Pl l ∞ P u ≤ kP uk ≤ kP uk < ǫ nj nj nj j_=k Xj=k Xj=k

l for all l ≥ k ≥ k0. Since j=k Pnj u ↑ Qku as l →∞ and since X has the σ-Fatou property, for each k ≥ k0 Wwe obtain kQkuk ≤ ǫ. This proves that Qku → 0 also in the second case.

To prove that Pnk → 0 strongly on Iu, pick a positive vector x ∈ Iu and an ǫ> 0. ǫ Since u is a quasi-interior in Iu, one can ﬁnd n ∈ N such that kx − x ∧ nuk < 2 . Since 24 M.KANDIC´ AND A. VAVPETICˇ

ǫ Qku → 0, there exists k0 ∈ N such that for all k ≥ k0 we have kQkuk < 2n . Since for each k ≥ k0 we have

ǫ kPnk xk≤kQkxk≤kQk(x − x ∧ nu)k + kQk(x ∧ nu)k ≤ 2 + kQk(nu)k < ǫ,

we have Pnk x → 0.

Now we prove that Pnk → 0 in order on Bu. Since 0 ≤ Pnk ≤ Qk on Bu, it suffices to prove that Qk ↓ 0 on Bu. As Bu is Dedekind complete, by [AB06, Theorem 1.19] it + + is enough to prove that Qkx ↓ 0 for each x ∈ Bu . Suppose that some vector y ∈ Bu satisfies 0 ≤ y ≤ Qkx for each k ∈ N and observe that for a fixed positive integer n

and for all positive integers k ≥ k0 we have

0 ≤ y ≤ Qkx = Qk(x − x ∧ nu)+ Qk(x ∧ nu) ≤ Qk0 (x − x ∧ nu)+ nQku.

Since Qku ↓ and Qku → 0, we have that Qku ↓ 0 and so by letting k go to inﬁnity, for each n ∈ N we obtain

0 ≤ y ≤ Qk0 (x − x ∧ nu).

Since u is a weak unit in Bu, we have x∧nu ↑ x as n →∞, so that order continuity of

Qk0 yields Qk0 (x − x ∧ nu) ↓ 0. This proves that y = 0 from where it follows Qkx ↓ 0.

For the proof of the general case, let us denote by Rn the operator Pn − P . Since

Rn = Pn − PnP + PnP − P = Pn(I − P ) − P (I − Pn) and since band projections

Pn(I − P )=(I − P )Pn and P (I − Pn) have disjoint ranges, they are disjoint in the

vector lattice of all regular operators. This yields that |Rn| = Pn(I − P )+ P (I − Pn)

which implies that |Rn| is a band projection. Since Rnu → 0, by the special case

above one can ﬁnd a subsequence (Rnk )k∈N which converges to 0 strongly on Iu and in order on Bu. Proposition 6.9 can be applied in two particular cases. If u is a weak unit in X,

then there is a subsequence (Pnk )k∈N which converges in order to P on Bu = X, and if u is a quasi-interior point in X, then there is a subsequence (Pnk )k∈N which converges in order and strongly to P on Iu = X.

Theorem 6.10. Let X be an order continuous Banach lattice. If two positive vectors u and v satisfy Bu = Bv then ϕ: Cu → Cv is a homeomorphism of metric spaces.

Proof. We ﬁrst consider the case when u and v are weak units of X. It follows from

Lemma 6.7 that ϕ: Cu → Cv deﬁned as ϕ(Pu)= P v is a bijection where P runs over all principal band projections.

We claim that ϕ is also continuous. Suppose that a sequence (xn)n∈N of vectors in Cu converges to a vector x0 ∈ Cu. For each n ∈ N0 there exists a principal band ON SEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 25 projection Pn such that xn = Pnu. Suppose that the sequence (Pnv)n∈N does not converge to zero. Hence, by passing to a subsequence (if necessary) there exists ǫ> 0 such that kPnv−P vk ≥ ǫ for all n ∈ N. By Proposition 6.9 there exists a subsequence

(Pnk )k∈N which converges in order to P . Again, by passing to a further subsequence

(if necessary), we may also assume that (Pn)n∈N converges to P in order. Hence, there exists a sequence of positive operators (An)n∈N such that 0 ≤ |Pn − P | ≤ An ↓ 0. Since X is order continuous, from the inequality

0 ≤|Pnv − P v| = |Pn − P |v ≤ Anv → 0 we conclude that Pnv → P v in norm which contradicts the fact that kPnv − P vk ≥ ǫ for all n ∈ N. Similarly, one can prove that the inverse mapping ϕ−1 : P v → Pu is also continuous.

For the proof of the general case, consider the closed sublattice F := Bu = Bv and note that F in its own right is an order continuous Banach lattice with quasi-interior points u and v.

The following proposition shows that the mapping ϕ: Cu → Cv is never an isometry unless u = v.

Proposition 6.11. Let X be a normed lattice with the principal projection property.

Suppose that for positive vectors u and v we have Bu = Bv. If the natural mapping

ϕ: Cu → Cv is an isometry, then u = v.

Proof. As in the proof of Lemma 6.7 and Theorem 6.10 it is enough to prove the statement for the case when u (and so also v) is a weak unit. + Pick any α > 1 and consider the vector w := (u − αv) . Let Pw be the band projection onto the principal band generated by the vector w. Since

+ − + Pw(u − αv)= Pw(u − αv) − Pw(u − αv) = Pw(u − αv) ≥ 0, we conclude that Pwu ≥ αPw(v). Since ϕ is an isometry, we have

kPwvk = kϕ(Pwu)k = kPwuk ≥ αkPwvk, so that Pwv = 0. Since v is a weak unit and since Pw is positive and order continuous, + it follows that Pw = 0. This yields that (u − αv) = 0 from where it follows that u ≤ αv for each α> 1. Letting α ↓ 1 we obtain u ≤ v. Similarly one can show that v ≤ u. 26 M.KANDIC´ AND A. VAVPETICˇ

7. Some remarks on the normality of the unbounded topology We conclude this paper with this short section where we consider the question under which conditions a given normed lattice X equipped with its unbounded norm topology is a normal topological space. Since metrizable spaces are normal, (X, τun)

is certainly normal in this case. By [KT18, Theorem 4.3], τun is metrizable if and only if X contains an at most countable set which generates a norm dense ideal in

X. In particular, if X is a Banach lattice with a quasi-interior point, then (X, τun) is metrizable. The following theorem whose proof is a combination of results from general topology, topological vector spaces and the unbounded norm topology provides a more

general situation when (X, τun) is a normal space. In order to do that we ﬁrst recall the following notions. A Banach lattice X is a KB-space whenever every increasing norm bounded sequence in X+ is norm convergent. A Hausdorﬀ topological space is paracompact whenever it admits a partition of unity subordinate to any open cover.

Theorem 7.1. If X is an atomic KB-space, then (X, τun) is a paracompact topological space.

Proof. Since X is an atomic KB-space, the closed unit ball BX is un-compact by

[KMT17, Theorem 7.5], so that (X, τun) is a σ-compact topological space. By [DOT17], the unbounded norm topology on X is a locally solid Hausdorﬀ topology, so that

(X, τun) is regular (see e.g. [Sch71]). Since (X, τun) is σ-compact, it is clearly a Lin- del¨of space, so that paracompactness of the unbounded norm topology ﬁnally follows from Morita’s theorem [Dug66, Theorem VIII.6.5]. The following corollary is an easy consequence of [Dug66, Theorem VIII.2.2] and Theorem 7.1.

Corollary 7.2. If X is an atomic KB-space, then (X, τun) is a normal topological space.

The proof of Theorem 7.1 heavily depends on Morita’s theorem from general topology. It would be of interest to ﬁnd a functional analytical proof. We conclude the paper with the following open question.

Problem 7.3. Under which conditions on a Banach lattice X the topological space

(X, τun) is normal?

The answer remains unknown even for order continuous atomic Banach lattices. ON SEPARABILITYOFUNBOUNDEDNORMTOPOLOGY 27

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Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19, SI- 1000 Ljubljana, Slovenija

Institute of Mathematics, Physics and Mechanics, Jadranska 19, SI-1000 Ljubl- jana, Slovenija Email address: [email protected]

Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19, SI- 1000 Ljubljana, Slovenija

Institute of Mathematics, Physics and Mechanics, Jadranska 19, SI-1000 Ljubl- jana, Slovenija Email address: [email protected]