Separating maps and the nonarchimedean Hewitt theorem

. If X and Y are zerodimensional spaces and T : C(X ) -~-~ C(Y) is a biseparating map then the N-compactifications of X and Y are homeomorphic, provided that K is a commutative nontrivially valued nonarchimedean field whose residue class field has non-measurable cardinality. We deduce the nonarchimedean counterpart to the well-known He-witt theorem, this is, if C(X) and C(Y) are ring-isomorphic, then the N-compactifications of X and Y are homeomorphic. Also, if X and Y are N-compact and T is linear and biseparating, then it is a weighted composition map : for some weight function a in C(Y), T f = a( f o h), being h : Y -~ X a homeomorphism.

proved that X and Y must be homeomorphic when there exists a biseparating map (see below) from C(X) onto C(Y).The aim of this paper is the study of this kind of maps in the nonarchimedean context.We give the general form of such maps when they are linear and the spaces X and Y are N-compact.In particular we prove the continuity of these maps.
Let C(X), C(Y) be the spaces of continuous K-valued functions on the zerodimen- sional spaces X and Y respectively, where K is a nonarchimedean commutative complete nontrivially valued field.For f E C(X), we define the cozero set of f as c( f ) := ~x E X : : f (~) ~ 0~.A map T from C(X) onto C(Y) is said to be separating if it is bijective, additive and if e(f) n c(g) = 0 implies e(Tf) n c(Tg) = 0 for all f , g E C(X ).T is said to be biseparating if both T and T'1 are separating.Important examples of biseparating maps are ring-isomorphisms, but there are many others, as we shall see.
Separating maps were introduced in the nonarchimedean context by E. Beckenstein and L. Narici ([5]) and were first used to prove the non-existence of a nonarchimedean Banach-Stone theorem in [5] and [2].Other papers where nonarchimedean separating maps are studied are [1], [3] and [6].
In this paper we shall denote by ,QoX and voX the Banaschewski compactification and the N-compactification of X respectively ( [13], [4]).If f E C(voX), we denote by its restriction to X.If f E C(X), we denote by voX --~ K its extension to vX (when possible).Being wK any zero dimensional compactification of K, if f E C(X ), we denote by : 03B20X -+ wK the extension of f to /3oX; when K is locally compact we consider wK = K U {~} the one-point compactification of K.For a clopen subset A of X, 03BEA will be the characteristic function of A. If A eX, then cl03B20XA stands for the closure of A in Ix denotes the function identically equal to 1 on X.
Definition 1 : Let T be a separating map from C(X) into C(Y).A point x0 ~ 03B20X is said to be a T-support point of yo e Y if, for every neighborhood of there exists f E C(X) satisfying c{ f ) C U(xo) such that 0.
Lemma 2 : : For every y E Y, there exists a unique T -support point of y in 03B20X.
This result and its proof, for the real and complex case, can be found in [8], p. 260.
A similar proof can be done for the nonarchimedean case.
Lemma 3 : : Suppose that T is separating and xo E 03B20X is the T-support point of yo E Y.
Proof : Suppose that = 0 and 0. Let (an) be a sequence in K such that is strictly increasing and tends to infinity.Then for each n E N consider the clopen subset of X, and let fn = C(X).Then it is not difficult to prove that the functions .00 91 := ~ 92 ~= f -91 n=1 belong to C(X ).Also g2 = 03A3~n=1 hn where hn = for n > 1 and h1 = fgv where V = > 1I2}.
Since f = gi +g2, one of these maps satisfies {Tg=)(ya) = a ~ 0; without loss of generality, suppose this is the case when i == 1.Notice that m, then c( fn) n e( f m) _ 0.
Let us see that g := ~~° ~ T f n belongs to C(Y).From the separating property of T we deduce that g is continuous at every point of Y -bdry U~n=1 c(Tfn).Suppose that g is not continuous at y E bdry c(T f n).Then there exists E > 0 such that for any neighborhood U(y) of y, there is a z E U(y) such that Ig(z)1 > f.Since 0, then there exists nz E N such that z E c(T On the other hand it is not difficult to prove that ~~ 1 an f n belongs to C(X).Therefore oo anfn) which is equal to since T is separating.We deduce that in any neighborhood of y there are arbitrarily big values of a n f n ) and therefore T(03A3~n=1 03B1nfn) cannot be continuous at y, which is absurd.Then we conclude that g belongs to C(Y).
Next let us prove that Tg1 = g.If Tg1 ~ g, then = gi + k where k E C(X) is not equal to zero.Since T is separating, n e(k) ~ 0, that is, there exists no E N such that c( f no ) n c( k) ~ 0. Let U be a nonempty clopen subset of c( f no ) n c( k).Then k03BEU E C(X) and c(k03BEU) n c( fn) = 0 for n ~ no and, by the separating property of T, c(T(k03BEU Let y E Y be such that (T(k~~))(y) ~ 0. Then Since 0 we deduce that there exists nl E N such that yo E c(T f nl ).We can easily see that Xo E cl03B20Xc(fn1), that is, Lemma 4 : : Let T be biseparating and let f , g E C(X) be such that c( f ) C c(g).Then c(T f ) C cl(c(Tg)).
Proof : Suppose that there exists yo E c(T f ) -cl(c(Tg)).Let U be a clopen neighbor- hood of yo such that U n cl(c(Tg)) = 0. Then we have that c(gu) n c(Tg) = 0 and, since T "1 is separating, c(T-103BEU) ~ c(g) = Ø; then c(T-103BEU) n e(f) = Ø and, by the separating property of T, c(~L~) n c(T f ) = 0, which contradicts our assumption on yo.Proposition 5 : : Let T be biseparating.Then the T -support point of every y E Y belongs to voX.

Proof :
Suppose yo E Y and that zo, its T-support point belongs to 03B20X -voX.Then there exists ([4]) a decreasing sequence (Un) of clopen neighborhoods of 2-0 such that For every n EN, let ( f n) _ (~vn ) where Vn is a clopen neighborhood of yo, in such a way that, Vn C Vn-i and also c(T ' 1 In) C Un for every n > 2.
Then, given (an) a sequence in K such that is strictly increasing and tends to infinity, the map oo 9 v._ fn+1] n=l belongs to C(X), so Tg E C(Y) and (Tg)(yo) = a E K. Then we have that given e > 0 there exists a neighborhood U(yo) of yo such that, if y E U(yo), then |(Tg)(y) -a E.
On the other hand, we have that y0 ~ int because, otherwise, there would exist gu E 0, satisfying U C int ~~n=1 Vn and then, according to Lemma 4,

c(T-103BEU) C
Un which obviously cannot be true.
n= ko +2 We have that for n > ko + 2, and since T-1 is separating, then ~~ > E, which is a contradiction.
Remember that if f E C(X) and K is N-compact, then f can be extended to a unique continuous map in C(voX ) ([13], p. 42) and that K is N-compact if and only if and only if its residue class field has nonmeasurable cardinal ( [13], p. 41].Then we can obtain the following result... Proposition 6 : : Suppose that the residue class field of K has nonmeasurable cardinal.If there exists a biseparating map T from C(X) onto C(Y), then voX is homeomorphic to voY.

Proof :
Since the residue class field of K has nonmeasurable cardinal, we can extend T : C(X) -+ C(Y) to a bijective, additive map : C(voX) -+ C(voY) by defining f E C(voX).Next we are going to see that T"° is biseparating : if e(f) ~ e(g) = ~, f, , g E C(voX), then c{g J x ) = ~ and, since T is separating, c(T f Jx) n c(Tg|x) = Ø; if ~ c ( ( T g | x ) 0 3 B D 0 Y ) ~ 0, then its intersection with Y is not empty, this is, there exists yo E Y such that (T f ( x )( yo ) ~ 0 and (T g ( x )( ya ) ~ 0, which cannot be possible.
So we may assume that X and Y are N-compact, and we are going to prove that they are homeomorphic.Define the map h : Y -; X where h(y) is the T-support point of y E Y.For .ro E X, consider the T -1-support point k(xo ) of xa in Y.In this way we can define a map k : X -+ Y. Let us see that h is continuous.
Consider U a clopen subset of X containing h(yo), y0 ~ Y.If y E c(T03BEU), then h(y) E X, by Proposition 5, and ~ 0, by Lemma 3. We deduce that h(y) E U and then h is continuous.In the same way we can prove that k is continuous too.Let us see also that h o k = ix.Let xo E X. Consider a clopen subset U of X containing By Lemma 3, = 0 and applying Lemma 3 to   we have that = 0, this is, a-o E U.This implies xo = h(k(x0)) and then h o k = ix.In the same way we can prove that k o h = i y and then h and k are homeomorphisms.
As a corollary, taking into account that every ring isomorphism is a biseparating map, we obtain a nonarchimedean version of the Hewitt theorem.
Corollary 6 : : Suppose that the residue class field of K has nonmeasurable cardinal.If C(X) and C(Y) are isomorphic as rings, then voX and voY are homeomorphic.
Looking carefully at the proof of Proposition 6, we can see that we make use of the measurability of the cardinal of the residue class field of K just to extend the functions of C(X) to functions defined on voX.But if X and Y are N-compact, then the second part of the proof is valid to show the following result.
Corollary 7 : : Suppose that X and Y are N.compact.If C(X) and C(Y) are isomorphic as rings, then X and Y are homeomorphic.Proposition 8 : : If T is a linear biseparating map from C(X) onto C(Y) and X, Y are N- compact spaces, then there exist a E C(Y), 0 for all y E Y, and a homeomorphism h : : Y --~ X such that (T f)(y) = a(y) f (h(y)) for every f E C(X), y E Y.In particular, if C(X) and C(Y) are endowed with the compact-open topology, then T is continuous.
Let us next show that T is continuous.Take a compact subset K of Y.Given e > 0, if for x E h(h'), I/(x)1 E~ sup{y E K it is easy to check that E, for every y E K. Remarks : 1.In general, if X and Y are not N-compact, a biseparating linear map need not be continuous.As an example of this fact, consider X a pseudocompact not N-compact space and Y its N-compactification, which is compact ([14], Corollary 1.6).Suppose that K is N-compact.It is easy to see that the canonical isomorphism T : : C(X) --~ C(Y), defined as T f := is well defined (since K is N-compact), biseparating, but not continuous : given E > 0, there is no compact subset K of X such that if ~ E for every x E K, then E for every x E voX = 03B20X.
2. If we do not assume T to be linear, then the map a E C(Y) of Proposition 8 may not exist.Suppose that K = Qp.Consider X = Y = ~x~.Let r stand for the map f(x) = r, r E Qp.Take s E Qp -Q.Consider ~1, s~ U {hi: i E I} a basis of C(X) as a vector space over Q. Define Tl := s, T s :=1, Thi := h=, a E I, and extend T by linearity (over Q) to the whole space C(X).
Then T : C(X) -~ C(X) is a biseparating map.If (Tf)(x) = a(x)f(x) for every f E C(X), then taking f = 1, a = s, and taking f = hi, a == 1, which is not possible.
Proposition 9 : : Suppose that K is locally compact.If X is zerodimensional and T is linear and separating, then T is biseparating.
Proof : If T was not separating, there would exist two maps f and g E C(X) such that e( f ) n 0. Take U a nonempty clopen subset of c( f ) n c(g) such that f and g are bounded on U and there exists a E R, a > 0 such that ( f (x)) > a and for every x E U. By putting f = + f ~x-u and g = g~v + it is easy to see that c(T(f03BEU)) ~ c(T(g03BEU)) = 0.
Suppose that yo E Y is such that ~ 0. Let x0 E 03B20X the T-support point of yo.Let 03B3 ~ 0 be such that From Propositions 8 and 9, we deduce the following corollary.Corollary lo : Let K be locally compact.If T is a linear separating map from C(X) onto C(Y) and X, Y are N-compact spaces, then there exist a E C(Y), a(y) ~ 0 for all y E Y, and a homeomorphism h : Y --; X such that (T f)(y) = a(y) f (h(y)) for every f E C(X), y E Y.In particular, if C(X) and C(Y) are endowed with the compact-open topology, then T is continuou3.

Remark :
Note that in Proposition 9 and Corollary 10, if X is compact, we can get the same results, with the same proofs, even if K is not locally compact.