On Cd_0 (K)-spaces

We present, an elementary proof of the (known) fact that a CDo(A')-space is a Banach lattice and is lattice isometrically isomorphic to a particular C(A) for some compact space K.

CDo^Kyspaces were introduced by Yu. A. Abramovich and A. W. Wickstead in [2, 4] and further investigated in [1, 3, 5]. It is known [1, 4] that a. C'£>g(A')-space is a. Banach lattice and a. unital AM-space. In [5] it was shown that CDq^K) is lattice isometrically isomorphic to C^K x {0,1}) with К x {0,1} equipped with a. compact Hausdorff topology. In this note we present elementary proofs of these facts.

Throughout these notes, К stands for a. compact Hausdorff topological space without isolated points. For x E K, let ЛГХ be a. base of open neighborhoods of x in K. As usually, for a. real-valued function f on К and xq E К we write lim /(ж) = r if for every e > 0 there Ж—^Жд exists V E Afx such that /(ж) — r| < e for all ж E V \ {жд}. Note that this notation is not vacuous for every жд E К because К has no isolated points.

We denote by C^K^ the Banach lattice of all continuous functions on K, equipped with sup-norm and point-wise ordering. Denote by cq^K) the set of all real-valued functions / on К such that the set { / >  e} = {ж E К : /(ж) > e} is finite for every e > 0. Clearly, cq (Ai) is a. vector subspace of f^Kf the space of all bounded functions on К equipped with sup-norm.

Lemma 1. f E cgl A") iff lim /(ж) = 0 for every xq E К. ж—^Жо

• <    Suppose that / E c^Kf Fix e > 0. The set {|/ >  e} is finite; since К is Hausdorff there exists V E Af_Xo such that V doesn’t contain any points of this set with the possible exception of жд itself. Thus, /(ж) <  e for all ж E V \ {жд}. Therefore, lim /(ж) = 0.

Suppose now that lim /(ж) = 0 for every xq £ К and assume that the set {|/ > e} is Ж^-Жо                                                                       1         J infinite for some e > 0. Since К is compact, this set must have an accumulation point жд, which contradicts lim /(ж) = 0. >

Ж^-Жо

Lemma 2. c^^K^ is a. closed subspace of f^Kf

• <    Suppose that a. sequence of functions (/„) in cq^K) converges in sup-norm to / E f^Kf Fix e > 0, then ||/„ — f\\ < e/2 for some n. It follows that {|/ >  e} C {|/„ > j}, hence is finite. >

It follows that cg(AT) equipped with the sup-norm is a. Banach space. Define the space C'Dg(AT) as follows: / E CDq(K) if f = g + h for some g E C^K^ and h E cg(A"). Equipped

with the sup-norm, CDq^K) is a normed space, a subspace of L^KY We also equip CDq^K) with the pointwise order. We will see that CDq^K) is a Banach lattice, and, moreover, an AM-spacc.

Lemma 3. If / E CDq^KY namely, f = g + h for some g E C^K^ and h E cq^KY then д(жо) = Um /(ж) f°^r all жд G K.

X^-XQ

• <    By Lemma 1, lim /(ж) = lim g^ + lim Д(ж) = д(жд) for every жд E K. > X^Xo       X^XQ       X^Xq

It follows that every / in CD(YK) has a unique decomposition into a continuous and a discrete part. Indeed, suppose that / = g + h = д' + h' where g, д' E C^K^ and h, h' E cq^KY then for every xq E К Lemma 3 implies д(жд) = lim /(ж) = д'(жд). Hence, g = д' and, X^Xo therefore, h = h'. In the rest of the paper, for / E CDq(K) we will write fc for the continuous component of / and fd for the discrete component. The uniqueness of the decomposition also implies that (/ + g)c = fc + gc and (/ + g^d = fd + 9d for f,g E CD0(K) because / + 9 = fc + fd + 9c + 9d = Ue + 9c) + Ud + 9dY and fc + 9c E C^ while fd + gd E cq^KY

Proposition 1. If lim /(ж) exists for every xq E К then f E CD^KY Iⁿ this ^case X^XQ

/с(ж0) = lim /(ж).

X^XQ

• <    For every xq E K, put д(жд) = lim /(ж), and let h = f — g. Then lim Д(ж) = 0 for X^-Xo                                    X—^Xq

every жо E K, so that h E cq(K) by Lemma 1. It remains to show that g E C^KY Fix xq E К and e > 0, there exists V E М_Жо such that /(ж) — д(жд) <  e for all ж E V \ {жд}. It follows that for every у E V we have

\дЫ - g(®o) = lim /(ж) - д(ж₀) <  e. ▻

Combining Lemma 3 and Proposition 1 we get the following result.

Corollary 1. f E CDq(K) if and only if lim /(ж) exists for every жд E К.

X^XQ

Lemma 4. For every f E CD(fK) we have \\f_c\\ ^ ||/|| ^ ||/_c|| + \\f_d\\.

• <    The first inequality follows from Proposition 1 while the second inequality is just the triangle inequality. >

Corollary 2. CDq(K) is a. Banach space.

• <    Suppose that a sequence (/„) is Cauchy in CDq(KY It follows from Lemma 4 that the sequence of the continuous parts Un)c is Cauchy, and, therefore, the sequence of discrete parts YfnYl is Cauchy. Since СЦ^ and c^K^ are complete, Un)c converges to some g E СЦ^ and (/„) converges to some h E cq^KY Hence (/„) converges to g + h, which belongs to CD^KY ▻

Next, we show that for this topology CD^KY is order isometric to C^K x {0,1}), if the topology on К x {0,1} is defined as follows. We put discrete topology on К x {1}, that is, we put ЛГ(_ж,у = {(ж, 1)} for each x E К. Then all the points of К x {1} are isolated points of К x {0,1}. For a point (ж,0) in К x {0} we take the basic open neighborhoods to be of the form V = (V x {0,1}) \ {(ж, 1)}, where V E Af_x. One can easily verify that these sets indeed form a base of a Hausdorff topology. From now on we consider К x {0,1} equipped with this topology.

One can easily see that К x {0} is a closed subspace of К x {0,1}, and the map ж н (ж, 0) is a homeomorphism between К and К x {0}. Further we will often identify К x {0} and K.

Lemma 5. К x {0,1} is compact.

• <    Consider an open cover of К x {0,1}. By replacing each set in the cover by a union of basic open neighborhoods of all the points in the set, we can assume that the cover is formed

by basic open neighborhoods. Hence, the cover is of the form

{{(ж«’ !)}}qeA u {^7}7er- where ха E К and У7 E Mx^ for some ж7 E K. It is easy to see that {H7}7er is an open cover of K, so that there is a finite sub-cover Vi,... , Vn. But then V} U • • • U Уп only misses finitely many points of К x {0,1}, so that if we add the corresponding open sets from the original cover then we obtain a finite cover of the entire К x {0,1}. >

Theorem 1. CDq(K) is lattice isometrically isomorphic to C^K x {0,1}). In particular, CDq(K) is an AM-space.

• <    Define T: CDo(K) —> C(K x {0,1}) via (Т/)(ж,г) = /_с(ж) + г/^(ж). In other words, TJ agrees with / on К x {1} and with /_c on К x {0}. It follows immediately that T is an isometry. It is obvious that TJ У 0 implies / > 0. On the other hand, if / > 0 then /_c ^ 0 by Proposition 1.

Observe that TJ is indeed a continuous function. Clearly, TJ is continuous on К x {1}, as the later set consists of isolated points. Finally, it is left to show that lim (Т/)(ж,г) = (ж,г)^(ж₀,0)

(Т/)(жд, 0) for every жд E К. Observe that (ж, r) —> (жд, 0) in К x {0,1} implies that ж —> жд in К, so that fc^ —> /с(^жо) and /(ж) —> 0 by Lemma 1. It follows that (Т/)(ж,г) = /с(ж) + г/^ж) -4 /с(жд) = (Т/)(жд,0).

Show that Т is onto. Let F E C^K x {0,1}). For every ж E К define /(ж) = ^(ж,!). Fix жд E К and e > 0, there exists V E ЛГ_Жо such that |Р(ж, г) — Р(жд,0) <  e for all (ж, г) E V. In particular, for every ж E V \ {жд} we have /(ж) — ^(жд, 0) = Р(ж, 1) — ^(жд, 0) < е, so that lim /(ж) = Р(жд,0). It follows from Lemma 1, that / E CDq(K) and J_c^ = Р(ж,0) for all ж E K, so that F = TJ. >