arXiv:1701.00075v1 [math.GN] 31 Dec 2016

Extending Baire-one functions on compact spaces
Olena Karlova, Volodymyr Mykhaylyuk
Department of Mathematical Analysis, Faculty of Mathematics and Informatics, Yurii Fedkovych Chernivtsi National University, Kotsyubyns'koho str., 2, Chernivtsi, Ukraine
Abstract We answer a question of O. Kalenda and J. Spurny from [8] and give an example of a completely regular hereditarily Baire space X and a Baire-one function f : X  [0, 1] which can not be extended to a Baire-one function on X.
Keywords: extension; Baire-one function; fragmented function; countably fragmented function 2000 MSC: Primary 54C20, 26A21; Secondary 54C30, 54C50
1. Introduction
The classical Kuratowski's extension theorem [14, 35.VI] states that any map f : E  Y of the first Borel class to a Polish space Y can be extended to a map g : X  Y of the first Borel class if E is a G-subspace of a metrizable space X. Non-separable version of Kuratowski's theorem was proved by Hansell [4, Theorem 9], while abstract topological versions of Kuratowski's theorem were developed in [5, 8, 9].
Recall that a map f : X  Y between topological spaces X and Y is said to be - Baire-one, f  B1(X, Y ), if it is a pointwise limit of a sequence of continuous maps fn : X  Y ; - functionally F-measurable or of the first functional Borel class, f  K1(X, Y ), if the preimage f -1(V ) of
any open set V  Y is a union of a sequence of zero sets in X. Notice that every functionally F-measurable map belongs to the first Borel class for any X and Y ; the converse inclusion is true for perfectly normal X; moreover, for a topological space X and a metrizable separable connected and locally path-connected space Y we have the equality B1(X, Y ) = K1(X, Y ) (see [10]).
Kalenda and Spurny proved the following result [8, Theorem 13].
Theorem A. Let E be a Lindelof hereditarily Baire subset of a completely regular space X and f : E  R be a Baire-one function. Then there exists a Baire-one function g : X  R such that g = f on E.
The simple example shows that the assumption that E is hereditarily Baire cannot be omitted: if A and B are disjoint dense subsets of E = Q  [0, 1] such that E = A  B and X = [0, 1] or X = E, then the characteristic function f = A : E  R can not be extended to a Baire-one function on X. In connection with this the following question was formulated in [8, Question 1].
Question 1. Let X be a hereditarily Baire completely regular space and f a Baire-one function on X. Can f be extended to a Baire-one function on X?
We answer the question of Kalenda and Spurny in negative. We introduce a notion of functionally countably fragmented map (see definitions in Section 2) and prove that for a Baire-one function f : X  R on a completely regular space X the following conditions are equivalent: (i) f is functionally countably fragmented; (ii) f can be extended to a Baire-one function on X. In Section 3 we give an example of a completely regular hereditarily Baire (even scattered) space X and a Baire-one function f : X  [0, 1] which is not functionally countably fragmented and consequently can not be extended to a Baire-one function on X.
2. Extension of countably fragmented functions
Let X be a topological space and (Y, d) be a metric space. A map f : X  Y is called -fragmented for some  > 0 if for every closed nonempty set F  X there exists a nonempty relatively open set U  F such that diamf (U ) < . If f is -fragmented for every  > 0, then it is called fragmented.
Let U = (U :   [0, ]) be a transfinite sequence of subsets of a topological space X. Following [6], we define U to be regular in X, if
Email addresses: maslenizza.ua@gmail.com (Olena Karlova), vmykhalyuk@ukr.net (Volodymyr Mykhaylyuk)

(a) each U is open in X;
(b)  = U0  U1  U2      U = X;
(c) U = < U for every limit ordinal   [0, ). Proposition 1. Let X be a topological space, (Y, d) be a metric space and  > 0. For a map f : X  Y the following conditions are equivalent:
(1) f is -fragmented;
(2) there exists a regular sequence U = (U :   [0, ]) in X such that diamf (U+1 \ U) <  for all   [0, ).
Proof. (1)(2) is proved in [12, Proposition 3.1]. (2)(1). We fix a nonempty closed set F  X. Denote  = min{  [0, ] : F  U = }. Property (c) implies
that  =  + 1 for some  < . Then the set U = U  F is open in F and diamf (U )  diamf (U+1 \ U) < .
If a sequence U satisfies condition (2) of Proposition 1, then it is called -associated with f and is denoted by U(f ).
We say that an -fragmented map f : X  Y is functionally -fragmented if U(f ) can be chosen such that every set U is functionally open in X. Further, f is functionally -countably fragmented if U(f ) can be chosen to be countable and f is functionally countably fragmented if f is functionally -countably fragmented for all  > 0.
Evident connections between kinds of fragmentability and its analogs are gathered in the following diagram.
functional countable fragmentability

functional fragmentability

fragmentability

countable fragmentability

Baire-one

continuity

functional F-measurability

Notice that none of the inverse implications is true.
Remark 1. (a) If X is hereditarily Baire, then every Baire-one map f : X  (Y, d) is barely continuous (i.e., for every nonempty closed set F  X the restriction f |F has a point of continuity) and, hence, is fragmented (see [14, 31.X]).
(b) If X is a paracompact space in which every closed set is G, then every fragmented map f : X  (Y, d) is Baire-one in the case either dimX = 0, or Y is a metric contractible locally path-connected space [11, 12].
(c) Let X = R be endowed with the topology generated by the discrete metric d(x, y) = 1 if x = y, and d(x, y) = 0 if x = y. Then the identical map f : X  X is continuous, but is not countably fragmented.
For a deeper discussion of properties and applications of fragmented maps and their analogs we refer the reader to [1, 2, 7, 13, 15].
Proposition 2. Let X be a topological space, (Y, d) be a metric space,  > 0 and f : X  Y be a map. If one of the following conditions hold
(1) Y is separable and f is continuous,
(2) X is metrizable separable and f is fragmented,
(3) X is compact and f  B1(X, Y ),
then f is functionally countably fragmented.

Proof. Fix  > 0.

(1) Choose a covering (Bn : n  N) of Y by open balls of diameters < . Let U0 = , Un = f -1( kn Bk) for

every n  N and U0 =

 n=0

Un.

Then

the

sequence

(U

:





[0, 0])

is

-associated

with

f.

(2) Notice that any strictly increasing well-ordered chain of open sets in X is at most countable and every open

set in X is functionally open.

(3) By [12, Proposition 7.1] there exist a metrizable compact space Z, a continuous function  : X  Z and a

function g  B1(Z, Y ) such that f = g  . Then g is functionally -countably fragmented by condition (2) of the

theorem. It is easy to see that f is functionally -countably fragmented too.

Lemma 3. Let X be a topological space, E  X and f  B1(E, R). If there exists a sequence of functions fn  B1(X, R) such that (fn) n=1 converges uniformly to f on E, then f can be extended to a function g  B1(X, R).

Proof. Without loss of generality we may assume that f0(x) = 0 for all x  E and

|fn(x)

-

fn-1(x)|



1 2n-1

for all n  N and x  E. Now we put

gn(x) = max{min{(fn(x) - fn-1(x)), 2-n+1}, -2-n+1}

and notice that gn  B1(X, R). Moreover, the series

 n=1

gn

(x)

is

uniformly

convergent

on

X

for

a

function

g  B1(X, R). Then g is the required extension of f .

Recall that a subspace E of a topological space X is z-embedded in X if for any zero set F in E there exists a zero set H in X such that H  E = F ; C-embedded in X if any bounded continuous function f on E can be
extended to a continuous function on X.

Proposition 4. Let E be a z-embedded subspace of a completely regular space X and f : E  R be a functionally countably fragmented function. Then f can be extended to a functionally countably fragmented function g  B1(X, R).

Proof. Let us observe that we may assume the space X to be compact. Indeed, E is z-embedded in X, since X

is C-embedded in X [3, Theorem 3.6.1], and if we can extend f to a functionally countably fragmented function

h  B1(X, R), then the restriction g = h|E is a functionally countably fragmented extension of f on X and

g  B1(X, R).

Fix n  N and consider

1 n

-associated

with

f

sequence U

= (U

:   ).

Without loss of the generality we

can assume that all sets U+1 \ U are nonempty. Since E is z-embedded in X, one can choose a countable family

V = (V :   ) of functionally open sets in X such that V  V for all     , V  E = U for every   

and V = < V for every limit ordinal   . For every   [0, ) we take an arbitrary point y  f (U+1 \ U). Now for every x  X we put

fn(x) =

y, x  V+1 \ V, y0, x  X \ V.

Observe that fn : X  R is functionally F-measurable, since the preimage f -1(W ) of any open set W  R is an
at most countable union of functionally F-sets from the system {V+1 \ V :   [0, )}  {X \ V}. Therefore,
fn  B1(X, R). It is easy to see that the sequence (fn) n=1 is uniformly convergent to f on E. Now it follows from Lemma 3
that f can be extended to a function g  B1(X, R). According to Proposition 2 (3), g is functionally countably
fragmented.

Corollary 5. Every functionally countably fragmented function f : X  R defined on a topological space X belongs to the first Baire class.

Proof.

For every n  N we choose a

1 n

-associated

with

f

family Un

= (Un,

:   n) of functionally open sets Un,

and corresponding family (n, :   n) of continuous functions n, : X  [0, 1] such that Un, = -n,1((0, 1]).

We consider the at most countable set  =  n=1{n, : 0    n} and the continuous mapping  : X  [0, 1],

(x) = ((x)).

Show that f (x) = f (y) for every x, y  X with (x) = (y). Let x, y  X with (x) = (y). For every n  N

we choose n  n such that x  Un,n+1 \ Un,n . Then y  Un,n+1 \ Un,n and

|f (x)

-

f (y)|



diam(Un,n+1

\

Un,n )



1 n

for every n  N. Thus, f (x) = f (y).

Now we consider the function g : (X)  R, g((x)) = f (x). Clearly, that every set (Un,) is open in the

metrizable

space

(X ).

Therefore,

for

every

nN

the

family

((Un,) : 

 n)

is

1 n

-associated

with

g.

Thus,

g

is functionally countably fragmented. According to Proposition 4, g  B1((X), R). Therefore, f  B1(X, R).

Combining Propositions 2 and 4 we obtain the following result.

Theorem 6. Let X be a completely regular space. For a Baire-one function f : X  R the following conditions are equivalent:

(1) f is functionally countably fragmented;

(2) f can be extended to a Baire-one function on X.

3. A Baire-one bounded function which is not countably fragmented
Theorem 7. There exists a completely regular scattered (and hence hereditarily Baire) space X and a Baire-one function f : X  [0, 1] which can not be extended to a Baire-one function on X.

Proof. Claim 1. Construction of X. Let Q = {rn : n  N}, rn = rm for all distinct n, m  N and

{r2n-1 : n  N} = {r2n : n  N} = Q,





A = {r2n-1}  [0, 1], B = {r2n}  [0, 1].

n=1

n=1

We consider partitions A = (At : t  [0, 1]) and B = (Bt : t  [0, 1]) of the sets A and B into everywhere dense sets At and Bt, respectively, such that |At| = |Bt| = c. Moreover, let [0, 1] = <1 T with |T| = c for every  < 1. For every   [0, 1) we put

Q =

tT At,  is even, tT Bt,  is odd,

Q=

Q, X = Q  {} and X =

X.

<1

<1

Claim 2. Indexing of X. For every   [0, 1) we consider the set

I = {(i)[,1) : |{ : i = 0}|  0}  [0, 1][,1) and notice that |I| = c. Let  : I  T be a bijection and

X =

X,j ,

jI+1

where

X,j =

A+1(j)  {}, B+1(j)  {},

For all ,   [0, 1) with  >  and i  I we put

 is even,  is odd,

Ji, = {j  I : j|[,1) = i}.

In particular, if  = ,  =  + 1 and i  I+1, then we denote the set Ji +1, simply by Ji . Notice that |Ji | = c and we may assume that
X,i = {xj : j  Ji }.

Then

X = {xi : i  I},

since I = iI+1 Ji .

Claim 3. Topologization of X. For all   [1, 1), i  I and x = xi  X we put

L<x = {xj  X : j  Ji ,}, Lx = {x}  L<x.
<

Notice that for all x  X and y  X with    either Lx  Ly, or Lx  Ly = . Now we are ready to define a topology  on X. Each point of X0 is isolated. For any   [1, 1) and a point
x  X we construct a base Ux of  -open neighborhoods of x in the following way. Take i  I and q  Q such that x = xi = (q, )  X. Let Vq be a base of clopen neighborhoods of q in the space Q equipped with the topology induced from R2. Then we put
Ux = {(V  [0, 1))  (Lx \ Ly) : V  Vq and Y  L<x is finite}.
yY
Claim 4. Complete regularity and scatteredness of X. We show that the space (X,  ) is completely regular. We prove firstly that every set Lx is clopen. Since the inclusion v  Lu implies Lv  Lu, every set Lx is open. Now let y  Lx. Then Ly  Lx = . Therefore, Ly  Lx or Lx  Ly. Assume that y  Lx. Then x  L<y and for a neighborhood W = Ly \ Lx of y we have W  Lx = , which implies a contradiction. Thus, Lx = Lx and the set Lx is closed.
Notice that for every clopen in Q set V the set (V  [0, 1))  X is clopen in X. Therefore, every U  Ux is clopen in X for every x  X. In particular, (X,  ) is completely regular.
In order to show that (X,  ) is scattered we take an arbitrary nonempty set E  X and denote  = min{  [0, 1) : E  X = }. Then any point x from E  X is isolated in E.
Claim 5. X =  X for every   [0, 1). It is sufficient to prove that

X  X
<
for every   [1, 1). Let   [1, 1), x = (q, )  X, V be an open neighborhood of q in Q, Y  L<x be a finite set and
U = (V  [0, 1))  (Lx \ Ly).
yY
We show that U  ( < X) = . For every y  Y we choose y <  such that y  Xy . We put  = max{y : y  Y } and  =  + 1. Since  < ,   . We choose i  I such that x = xi and choose j  J such that j|[,1) = i. We consider the set X,j. Recall that X,j = At  {} or X,j = Bt  {}, where t =  (j). Therefore, the set
P = {p  Q : (p, )  X,j}
is dense in Q. Thus, the set P  V is infinite. Moreover, |Ly  X|  1 for every y  Y . Hence, the set
S = {p  P : (p, )  Ly}
yY
is finite. Therefore, the set (P  V ) \ S is infinite, in particular, it is nonempty. We choose a point p  (P  V ) \ S. Then z = (p, )  X,j. Thus, z  Lx. Moreover, z  V  [0, 1) and z  yY Ly. Thus, z  U . Since X,j  X, z  < X. Therefore, Ux  < X =  for every Ux  Ux and x  < X.
Claim 6. Construction of a Baire-one function f . We put

C=

X

<1,  is even

and show that the function f : X  [0, 1],

f = C,

belongs to the first Baire class. Consider a mapping  : X  Q, (x) = r if x = (q, ) and q = (r, t) for some t  [0, 1] and  < 1. The
mapping  is continuous, because for every open in Q set V the set

-1(V ) = (V  [0, 1]  [0, 1))  X

is open in X. Clearly, the function g : Q  R,

g(t) =

1, t = r2n-1, 0, t = r2n,

belongs to the first Baire class. Therefore, the function f (x) = g((x)) belongs to the first Baire class too.

Claim 7. The function f is not countably fragmented. Finally, we prove that f is not countably frag-

mented. Assume the contrary and take a countable sequence U = (U :  < ) of functionally open sets such that

U

is

1 2

-associated

with

f.

We show that U  < X for every   . We will argue by induction on . For  = 0 the assertion is

obvious. Assume that the inclusion is valid for all  <   . If  is a limit ordinal, then

U = U 

X = X.

<

< <

<

Now let  =  + 1. Suppose that there exists x  U \ (  X). Notice that x  X  X according to Claim 6. Therefore, there exist z1  U  X and z2  U  X. According to the inductive assumption, we have U  < X. Therefore, z1, z2  U \ U = U+1 \ U. Thus, we have

1

=

|f (z1)

-

f (z2)|



diam(U+1

\

U )

<

1 2

,

a contradiction. Theorem 6 implies that f can not be extended to a Baire-one function on X.

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