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Quaestiones Mathematicae 28(2005), 1–11.
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°
A NEW COMPACTNESS TYPE TOPOLOGICAL
PROPERTY
Dongsheng Zhao
Mathematics and Mathematics Education, National Institute of Education,
Nanyang Technological University, Singapore 637616.
E-Mail [email protected]
Abstract. By a gauge on a topological space we shall mean a mapping that assigns
each element in the space an open neighbourhood. We investigate some topological
properties which can be characterized using gauges. The main property we will
consider is the gauge compactness. Some problems and possible future work are
listed at the end of the paper.
Mathematics Subject Classification (2000): 54A20, 54C08, 54D30.
Key words: Gauge, M-uniformly continuous mapping, gauge compact space.
1. Introduction. A gauge on a metric space (X, d) is usually understood to
be a positive real valued function δ(·) : X −→ R.
I In defining the HenstockKurzweil integral, gauges play a crucial role [4]. In [3], gauges are used to provide
a Cauchy type characterization for Baire class one functions. Given any positive
real valued function δ(·) on the metric space (X, d), we can define a neighbourhood
assignment η in a natural way: for each x ∈ X, η(x) = B(x, δ(x)) = {y ∈ X :
d(x, y) < δ(x)}. It is easily observed that, in both defining the Henstock-Kurzweil
integral and characterizing Baire class one functions, one can actually use the
induced neighbourhood assignments instead of the positive functions. Since for
an arbitrary topological space, a positive real valued function no longer defines a
neighbourhood assignment, we shall employ neighbourhood assignments directly
and call them gauges.
Given a gauge δ on a topological space X, define the binary relation RδM on X
by
xRδM y iff either x ∈ δ(y) or y ∈ δ(x).
The relation RδM is reflexive and symmetric, but not transitive in general. The
main objective of this paper is to study some topological properties which can
be defined in terms of the relation RδM . In Section 2, we introduce the concept
of M-uniformly continuous mappings and prove a gauge characterization of R 0 –
spaces. In Section 3, we define and study a new compactness type property– gauge
compactness. We prove that gauge compactness implies compactness if the space
satisfies certain conditions. For example, every Tychonoff gauge compact space is
compact. A net characterization of gauge compactness is also proved. In Section 4
we consider a special type of gauges–symmetric gauges. We prove that a positive
function on the real line defines a symmetric gauge if and only if it is a constant
function. In the last section we list some problems and topics for further study.
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2
2.
D. Zhao
M-uniformly continuous mappings and R0 -spaces.
Definition 1. A mapping f : X −→ Y between topological spaces is called Muniformly continuous if for any gauge λ on Y there exists a gauge δ on X such
that for any x1 , x2 ∈ X, x1 RδM x2 implies f (x1 )RλM f (x2 ).
The composition of two M-uniformly continuous mappings is clearly M-uniformly
continuous. Every continuous mapping is M-uniformly continuous. For a space Y ,
the converse conclusion is true for all spaces X if and only if Y is an R0 -space as
is shown in the theorem below.
Recall from [7] that a space X is called an R0 –space if for any two points
x, y ∈ X, x ∈ cl{y} if and only if y ∈ cl{x}, where cl{x} denotes the closure of
{x}. The class of R0 -spaces plays a good role in the theory of nearness spaces. For
example, the category of topological nearness spaces is isomorphic to the category
of R0 -spaces (see Chapter 3 of [7]).
Theorem 2. A space Y is an R0 –space if and only if for any space X, every
M-uniformly continuous mapping f : X −→ Y is continuous.
Proof. Suppose that Y is an R0 –space and f : X −→ Y is M-uniformly continuous.
For any x ∈ X, we show that f is continuous at x. Given any open neighbourhood
U of f (x), define a gauge λ on Y by
(
U
if y ∈ cl{f (x)},
λ(y) =
Y − cl{f (x)} if y 6∈ cl{f (x)}.
Notice that Y is an R0 –space, so if y ∈ cl{f (x)} then f (x) ∈ cl{y}, hence y ∈ U
holds. Thus λ is indeed a gauge. Since f is M-uniformly continuous, there exists
a gauge δ on X such that uRδM v implies f (u)RλM f (v). We claim that if u ∈ δ(x)
then f (u) ∈ U . In fact, as u ∈ δ(x), so uRδM x, which implies f (u)RλM f (x). If
f (u) 6∈ U = λ(f (x)), then by the definition of λ, λ(f (u)) = Y − cl{f (x)}, so
f (x) 6∈ λ(f (u)). But these violate f (u)RλM f (x). Hence we have δ(x) ⊆ f −1 (U ),
therefore f is continuous at x ∈ X. It then follows that f is continuous on X.
Now assume the condition is satisfied. If there are two points a, b ∈ Y such
that a ∈ cl{b} but b 6∈ cl{a}, we let X = {a, b} and assume {a} is the only
non-trivial open set of X. The inclusion mapping f : X −→ Y is obviously Muniformly continuous, but it is not continuous because f −1 (V ) = {b} is not open,
where V = Y − cl{a}. This contradiction proves that Y must be an R0 -space.
2
Given a gauge δ on a space X, one can define another binary relation Rδm on
X by
xRδm y iff x ∈ δ(y) and y ∈ δ(x).
A mapping f : X −→ Y is called m-uniformly continuous if for each gauge λ on Y
there is a gauge δ on X such that x1 Rδm x2 implies f (x1 )Rλm f (x2 ).
Proposition 3. If X is a T1 space and the mapping f : X −→ Y has only finitely
many discontinuity points, then f is m-uniformly continuous.
A new compactness type topological property
3
Proof.
Suppose D = {d1 , d2 , · · · , dn } is the set of all the discontinuity points
of f . Given any gauge λ on Y we define a gauge δ on X as follows: if x ∈ D,
choose an open neighbourhood δ(x) of x that contains no other point in D except
x. If x ∈ X − D, then as X is T1 and f is continuous at x, we can find an open
neighbourhood δ(x) of x such that δ(x) ⊆ (X − D) ∩ f −1 (λ(f (x))). It is then easy
to verify that xRδm y implies f (x)Rλm f (y).
2
In [3], it is proved that a mapping f : X −→ Y between complete separable
metric spaces is Baire class one iff for each ² > 0, there is a positive function ω :
X −→ R
I + such that dY (f (x), f (y)) < ² whenever dX (x, y) < min{ω(x), ω(y)}. It
follows immediately that every m-uniformly continuous mapping between complete
separable metric spaces is Baire class one.
The following example is a function that is Baire class one but not m-uniformly
continuous.
Example 4. Let D = {r1 , r2 , · · · } be the set of all rational numbers in the open
interval (0, 1) arranged in a sequence. Define f : (0, 1) −→ R
I by
(
0
if x 6∈ D,
f (x) = 1
if x = rn .
n
This function is Baire class one. Define a gauge λ on R
I as follows:
(
(x − 21 |x|, x + 12 |x|) if x 6= 0,
λ(x) =
(−1, 1)
if x = 0.
Now let δ be any gauge on (0, 1). For each x ∈ (0, 1), choose a positive number
dx such that (x − dx , x + dx ) ⊆ δ(x). For each integer n > 0, put Fn = {x ∈
(0, 1) − D : dx ≥ n1 }. Then there exists n such that Fn is dense in a subinterval
of (0, 1). So there is a rational r ∈ (0, 1) belonging to cl(Fn ). Choose q ∈ Fn such
that |r − q| < min{dr , n1 }. Hence, r ∈ δ(q) and q ∈ δ(r). But f (q) = 0 6∈ λ(f (r)),
so f (r)Rλm f (q) does not hold. Hence f is not m-uniformly continuous.
3. Gauge compact spaces. Now we introduce a compactness type of topological property using the relation RδM .
Definition 5. A topological space X is called gauge compact if for every gauge
δ on X, there exist finitely many elements x1 , x2 , · · · , xn in X such that for every
x ∈ X there is xi , so that xRδM xi .
Every compact space is obviously gauge compact. The following is an example
of a T1 gauge compact space that is not compact.
Example 6. Let X = R
I be the set of all real numbers equipped with the countable
complement topology. That is, if U ⊆ X and U 6= ∅, then it is open if and only
if X − U is finite or at most countably infinite. It is well known that X is T 1
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and not compact. Now let δ be any gauge on X. If all δ(x) are equal to X then
every x is RδM related to the fixed number 0. If there exists a with δ(a) 6= X and
X−δ(a) = {a1 , a2 , · · · , an , · · · }( when X−δ(a) is finite the T
proof is the same). Since
S
{X − δ(ai ) : i = 1, 2, · · · } is countable, the intersection {δ(ai ) : i = 1, 2, · · · } is
non-empty. Choose an element c ∈ ∩{δ(ai ) : i = 1, 2, · · · }. Now for each x ∈ X,
if x ∈ δ(a) then xRδM a. If x 6∈ δ(a), then x = ai for some i, thus c ∈ δ(ai ) = δ(x)
which implies xRδM c. Hence every element x of X is related by RδM to either a or
c. Therefore X is gauge compact.
Remark 7. (1) Note that the set A = {1, 2, 3, · · · } is a closed subset of X in
Example 6, and A is an infinite discrete subspace which is not gauge compact.
Hence the gauge compactness is not hereditary to closed subsets.
However, the gauge compactness is hereditary to clopen subsets. In fact, if A
is a nonempty open and closed subset of a gauge compact space Y , then for any
gauge δ on A, we extend δ to a gauge λ on Y by letting
(
δ(y)
if y ∈ A,
λ(y) =
Y − A if y 6∈ A.
Since Y is gauge compact, there exist elements a1 , a2 , · · · , an in Y such that for
each y ∈ Y , yRλM ai holds for some i. Obviously if x ∈ A and y 6∈ A then xRλM y
does not hold. Hence among the elements a1 , a2 , · · · , an , some must be from A, let
them be a1 , a2 , · · · , am . Then every x ∈ A is RδM related to some aj (1 ≤ j ≤ m).
Thus A is gauge compact.
(2) If X is gauge compact,
S then for any open cover U of X there exist
U1 , U2 , · · · , Un ∈ U such that {St(Ui , U) : i = 1 · · · n} = X.
Proposition 8. If f : X −→ Y is an M-uniformly continuous mapping onto a
space Y and X is gauge compact, then Y is also gauge compact.
Proof. Given any gauge λ on Y , since f is M-uniformly continuous, there is a
gauge δ on X such that for any u, v ∈ X, uRδM v implies f (u)RλM f (v). Since X is
gauge compact, there exist elements x1 , x2 , · · · , xn in X such that for each x ∈ X,
xRλM xi for some i. Now for every y ∈ Y , since f is surjective, there exists x ∈ X
such that f (x) = y. Suppose xRδM xi , then yRλM f (xi ). Thus every y ∈ Y is RλM
related to some f (xi )(i = 1, 2, · · · , n). Hence Y is gauge compact.
2
Thus, in particular, every continuous image of a gauge compact space is gauge
compact.
A net {xt } in a space X is called gauge clustered if for any gauge δ on X, there
is a subnet {xtk } such that ∩k δ(xtk ) 6= ∅.
Proposition 9. A topological space X is gauge compact if and only if for any net
{xt } in X, either {xt } is gauge clustered or it has a cluster point in X.
A new compactness type topological property
5
Proof. Suppose that X is not gauge compact. Then there is a gauge δ on X, such
that for any finite subset A of X there is a point xA , so that xA 6∈ ∪{δ(x) : x ∈ A}
and A ∩ δ(xA ) = ∅. Let Λ be the directed set of all finite subsets of X with the
order of inclusion. Then we have a net {xA }A∈Λ in X. It is easy to see that this
net has no cluster point. For any subnet {xA0 } of {xA } and any point a, there is
A0 such that a ∈ A0 , thus a 6∈ δ(xA0 ). This shows that the intersection ∩A0 δ(xA0 )
is empty.
Now suppose that X is gauge compact and {xt } is a net in X that has no cluster
point. We show that {xt } is gauge clustered.
Let δ be any gauge on X. For every point x ∈ X there is an open neighbourhood Ux of x such that xt 6∈ Ux holds eventually. Put µ(x) = δ(x) ∩ Ux ,
x ∈ X. Thus we have a gauge µ on X. Since X is gauge compact, there exists
a finite set {a1 , a2 , · · · , an } such that for each x ∈ X, xRµM ai for some i. By the
definition of µ, xt 6∈ ∪{µ(ai ) : i = 1, 2, · · · , n} eventually. Hence there is an i0
such that ai0 ∈ µ(xt ) frequently. This implies that there is a subnet {xtk } such
that ai0 ∈ ∩{µ(xtk ) : k} ⊆ ∩{δ(xtk ) : k}. Hence ∩{δ(xtk ) : k} 6= ∅. The proof is
completed.
2
The above theorem can be used to show that the product of two gauge compact
spaces need not be gauge compact.
Example 10. Let X be the space in Example 6 and I = [0, 1] be the unit interval.
Then X and I are gauge compact. Consider the sequence S = {(n, n1 )}n∈N −{1} in
2n+1
2n−1
, 2n(n−1)
)
X ×I. Then S has no cluster point in X ×I. Define δ(n, n1 ) = X ×( 2n(n+1)
1
for n > 1, and δ(x, y) = X × I for other points. Since δ(n, n ) are pair wise disjoint,
∩k δ(nk , n1k ) = ∅ for every subnet. Thus S is not gauge clustered. By Proposition 9,
the cartesian product X × I is not gauge compact. Notice that I is even a compact
space.
Now we prove that gauge compactness implies compactness for Tychonoff spaces.
Consider the following property of a topological space X:
For any open cover U of X, there is a finite U0 ⊆ U such that
|X − ∪U0 | < |X|.
(*)
Lemma 11. Let X be a T1 gauge compact space. Then every nonempty closed Gδ
subset of X satisfies the condition (*).
Proof. Let K be a nonempty closed Gδ subset of X and X − K = ∪{Fi : i ∈ N },
where each Fi is a closed set. For the sake of convenience we assume Fi ⊆ Fi+1 for
all i. Suppose K does not satisfy condition (*). Then there exists an open cover
U = {Ui : i ∈ I} of K consisting of open sets of X such that for each finite U0 ⊆ U,
|K − ∪U0 | = |K|. For each x ∈ K, fix a Ux ∈ U such that x ∈ Ux . Let K0 be the
set of all finite subsets of K and K0 = {Aα : α < λ}, where λ is a limit ordinal such
that |λ| = |K|. Now for each i and α < λ, one can find a point xi,α ∈ K such that
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D. Zhao
xi,α 6∈ Fi ∪ {Ux : x ∈ Aα }. Due to the cardinal reason, we can choose xi,α 6= xj,β
for (i, α) 6= (j, β). Define a gauge δ on X as follows:


if x ∈ X − K,
X − K
δ(x) = Ux − (Fi ∪ Aα ) if x = xi,α ,


Ux
otherwise.
Now for any finite subset A of X, let A = A1 ∪ A2 where A1 ⊆ X − K and
A2 ⊆ K. There is an i such that A1 ⊆ Fi , and α < λ with A2 = Aα . Then for any
x ∈ A, xRδM xi,α does not hold. This violates the gauge compactness of X. The
contradiction completes the proof.
2
Corollary 12. Every T1 gauge compact space satisfies condition (*).
Theorem 13. If X is a T1 gauge compact space that has a basis consisting of Fσ
sets, then X is compact.
Proof.
Let U be an open cover
S of X. If U has no finite subcover, then for
each finite V ⊆ U the set X − V is a nonempty closed Gδ S
set, thus by Lemma
S
11 there is a finite W ⊆ U with V ⊆ W such that |X − W| < |X − V|.
Then we can have a sequence of finite subfamilies Un (n=1,2,...) of U such that
U1 ⊆ U2 ⊆ · · · ⊆ Un ⊆ · · · and |X − ∪Ui+1 | < |X − ∪Ui | holds for all i. However,
the set of cardinals below |X| is well-ordered, so the set {|X − ∪Ui | : i = 1, 2, · · · }
has the smallest element. But this contradicts the property of Ui (i = 1, 2, · · · ).
This contradiction completes the proof.
2
Since every Tychonoff space has a basis consisting of Fσ sets, so we have
Theorem 14. Every Tychonoff gauge compact space is compact.
A space X is called star compact if for any open covering U of X there exists a finite subset A of X such that star(A, U) = X. For Hausdorff spaces star
compactness is equivalent to countable compactness [6]. Obviously every gauge
compact space is star compact. The space W0 given below is countably compact
and Hausdorff, but not gauge compact.
Example 15. Let W0 be the space of all countable ordinal numbers. The topology
on W0 has a basis consisting of all segments (x, y] = {z ∈ W0 : x < z ≤ y} and the
set {0}. Then W0 is countably compact (see Chapter XI, Definition 3.1 of [1]) but
not compact. Also W0 is Tychonoff, so by Theorem 14 the space W0 is not gauge
compact. To see that W0 is not gauge compact more clearly, define a gauge δ on
W0 as follows:
(
{α}
if α is not a limit ordinal,
δ(α) =
(0, α] if α is a limit ordinal.
A new compactness type topological property
7
Now for any finite subset B = {α1 , α2 , · · · , αn } of W0 , let β be the biggest ordinal
in B, then the ordinal β + 1 is not RδM related to any αk in B. Thus W0 is not
gauge compact. Notice that W0 is even a T4 space.
If f : X −→ R
I is a continuous real valued function and X is gauge compact,
then f (X) is a gauge compact subset of R.
I By Theorem 14, f (X) is compact, hence
it is a bounded closed set. Thus f attains its infimum and supremum.
A real valued function f defined on a space X is said to have a bounded local
oscillation if there is an open covering U of X and a positive number B such that
for each U ∈ U, ω(f ; U ) = sup{|f (x) − f (y)| : x, y ∈ U } ≤ B. This is equivalent to
saying that every point x ∈ X has a neighbourhood U such that the oscillation of
f on U is bounded by a fixed number.
Obviously, every continuous function has a bounded local oscillation.
Proposition 16. If X is gauge compact and f : X −→ R
I has a bounded local
oscillation, then f is bounded on X.
Proof. Let f be a real valued function on X that has a local oscillation bound
B. For each x ∈ X, there is an open set δ(x) containing x such that y ∈ δ(x)
implies |f (y) − f (x)| ≤ B. Then δ is a gauge on X. Since X is gauge compact,
there exist points x1 , x2 , · · · , xn such that every y ∈ X is RδM related to some xi .
Put A = max{|f (xi )| : i = 1, 2, · · · , n} + B. Then it follows that |f (y)| ≤ A for all
y ∈ X. So f is bounded by A.
2
Example 17. Let X = N be the set of all positive integers equipped with the
topology consisting of the empty set and all subsets containing the number 1.
Then X is clearly gauge compact. Define f : X −→ R
I by f (n) = n1 , n ∈ X. Then
f has a bounded local oscillation. But f does not attain its infimum 0.
4. Symmetric gauges. Let (X, d) be a metric space. Given any positive number a, we can define a gauge δa on X, where δa (x) = {z ∈ X : d(z, x) < a}. The
gauge δa has the following property:
x ∈ δa (y) if and only if y ∈ δa (x).
Definition 18. A gauge δ on a space X is symmetric if for any x, y ∈ X,
x ∈ δ(y) iff y ∈ δ(x).
Example 19. (1) Let f : X −→ Y be a continuous mapping between topological
spaces. If δ is a symmetric gauge on Y , then f −1 (δ) is a symmetric gauge on X,
where for each x ∈ X, f −1 (δ)(x) = f −1 (δ(f (x))). In particular, if f is a continuous
mapping from X to a metric space (Y, d), then for each ² > 0, λ(x) = {z ∈ X :
d(f (x), f (z)) < ²}, x ∈ X, defines a symmetric gauge on X.
(2) If δ and λ are symmetric gauges on X, then δ ∧ λ is also symmetric, where
(δ ∧ λ)(x) = δ(x) ∩ λ(x), x ∈ X.
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D. Zhao
Given a positive function s : R
I −→ R
I on R,
I we obtain a gauge δ on R
I defined
by δ(x) = (x − s(x), x + s(x)), x ∈ R,
I we call this δ the gauge defined by s. Now
we investigate the positive functions that define symmetric gauges on R.
I
Lemma 20. If s is a positive function on the real line R
I that defines a symmetric
gauge on R,
I then s is a continuous function.
Proof.
Suppose that s is not continuous. Then there is a sequence {xn } of
numbers converging to a point x and |s(xn ) − s(x)| > ² > 0, ∀n. We can choose ²
satisfying the additional condition s(x) > ². Take an xn such that |xn − x| < 12 ².
We consider the following cases:
(a) s(x) < s(xn ) . Assume x < xn (when xn < x it can be shown similarly) .
Then xn − s(xn ) < x + 12 ² − s(xn ) = x − (s(xn ) − 12 ²) < x − (s(x) − 12 ²) < x < xn .
Choose y, such that xn − s(xn ) < y < x − s(x). Then y ∈ (xn − s(xn ), xn + s(xn )),
which implies xn ∈ (y − s(y), y + s(y)) because s defines a symmetric gauge. So
x ∈ (y−s(y), y+s(y)) because x lies between y and xn . Again, from the assumption
on s it follows that y ∈ (x − s(x), x + s(x)), which contradicts the assumption on
y.
(b) s(x) > s(xn ). Assume x < xn ( the proof is similar for x > xn ). Choose y
such that x+s(x)− 21 ² < y < x+s(x). It follows from 0 < xn −x < 21 ² and y−x > 21 ²
that x < xn < y. Since y ∈ (x−s(x), x+s(x)), so x ∈ (y −s(y), y +s(y)). Since x <
xn < y, so xn ∈ (y −s(y), y +s(y)), which then implies y ∈ (xn −s(xn ), xn +s(xn )).
But then we have
1
1
y < xn + s(xn ) < x + ² + s(x) − ² = x + s(x) − ²,
2
2
which contradicts the assumption on y. These contradictions show that s must be
continuous.
2
Corollary 21. Let s be a positive function on R
I that defines a symmetric gauge
on R.
I Then for each x, s(x) = s(x − s(x)) = s(x + s(x)).
Proof.
First, if x < y < x + s(x) then, by the assumption on s, x ∈ (y −
s(y), y + s(y)), thus s(y) > y − x. By Lemma 20, s(·) is continuous, it follows that
s(x + s(x)) ≥ s(x). If s(x + s(x)) > s(x), let a = 21 (s(x + s(x)) − s(x)). Then there
is ² > 0 such that s(y) > s(x) + a holds for all y ∈ [x + s(x), x + s(x) + ²). Choose
one y0 with 0 < y0 − x − s(x) < min{², 21 a}. Then for every z ∈ (x − 12 a, x) we
have z ∈ (y0 − s(y0 ), y0 + s(y0 )), thus y0 ∈ (z − s(z), z + s(z)). Hence s(z) ≥ y0 − x.
Since s(·) is continuous we deduce that s(x) ≥ y0 − x > s(x), a contradiction. So
s(x + s(x)) = s(x). Similarly one can show s(x) = s(x − s(x)).
2
Proposition 22. If s is a positive function on R
I that defines a symmetric gauge,
then s is a constant function.
A new compactness type topological property
9
Proof. Suppose that s is not a constant function. By Lemma 20 and Corollary 21
we can choose a point a such that s(a) ≥ s(x) for all x ∈ [a, a + s(a)] and s(·) is not
constant on [a, a + s(a)]. Since s(·) is continuous, there exists c, a < c < a + s(a)
such that s(c) is the minimum value of s(·) on [a, a + s(a)]. Take d, c < d < a + s(a)
such that s(c) < s(d) < s(a) and s(x) ≤ s(d) for all x ∈ [c, d]. Consider the interval
[c + s(c), d + s(d)]. By the continuity of s, for each y ∈ [c + s(c), d + s(d)], there is
x ∈ [c, d] with x + s(x) = y, thus s(y) = s(x) ≤ s(d). Similarly one can show that
for any positive integer k and y ∈ [c + ks(c), d + ks(d)], s(y) ≤ s(d). Choose k such
that k(s(d) − s(c)) > s(a). Then for some m, a + ms(a) ∈ [c + ks(c), d + ks(d)].
But then we have s(a) = s(a + ms(a)) ≤ s(d). This contradiction completes the
proof.
2
Corollary 23. Let s(·) be a positive function on R
I 2 ( or R
I n ) that defines a
symmetric gauge. Then s(·) is a constant function.
Proof.
For each straight line l through the origin, the restriction of s(·) on l
defines a symmetry gauge on l, which is a constant function by proposition 22.
Thus s(u) = s(0, 0) for all u ∈ l. It follows that s(·) must be a constant function.
2
If a subspace X of the real line is not connected, then there exists a non constant
positive function s(·) on X that defines a symmetric gauge on X.
Recall that a sequence {xn } in a metric space (X, d) is called Cauchy if for any
² > 0 there is n0 such that d(xn , xm ) < ² holds for all m, n ≥ n0 .
For a general space we can define a similar class of sequences using symmetric
gauges, we call them gauge-fine sequences.
Definition 24. A sequence {xn } in a space X is called gauge-fine if for any symmetric gauge δ(·) there exist k such that for all m, n ≥ k,
xn RδM xm .
Remark 25. (1) In a metric space, every gauge-fine sequence is Cauchy. However,
even for the real line we still do not know whether the converse conclusion is true.
(2) If h : X −→ Y is a continuous mapping and {xn } is a gauge-fine sequence
of X, then {h(xn )} is a gauge-fine sequence in Y .
Definition 26. A topological space X is called G-complete if every gauge-fine
sequence converges.
Proposition 27. Every normal and sequentially compact space is G-complete.
Proof. Let {xn } be a gauge-fine sequence in X. Since X is sequentially compact,
the sequence has a convergent subsequence, say {xnk }. Suppose limk→∞ xnk = a.
10
D. Zhao
For each open neighbourhood U of a, choose an open set V with a ∈ V ⊆ cl{V } ⊆
U . There exists a continuous function f : X −→ R
I such that f (U c ) = {1} and
f (V ) = 0 because X is normal. Let δ be the symmetric gauge on X defined by
δ(x) = {y ∈ X : |f (x) − f (y)| <
1
}.
2
Then there is n0 such that xn ∈ δ(xm ) whenever m, n ≥ n0 . Choose nk ≥ n0
with xnk ∈ V . Then for all n ≥ n0 , xn ∈ δ(xnk ), which implies xn ∈ U . So {xn }
converges to a.
2
One can easily check that every complete metric space is G-complete; a closed
set
Q of a G-complete space is G-complete. If a sequence {xn } in a cartesian product
λ∈Λ Xλ of topological spaces is gauge-fine then for each λ ∈ Λ, {p λ (xn )} is
a gauge-fine sequence in Xλ . It follows then that the cartesian product of any
collection of G-complete spaces is G-complete.
5. Remarks on some further work. We shall end the paper by listing some
possible future work. We leave the problems for interested readers to explore.
1). It is still unclear whether there is a Hausdorff gauge compact space which
is not compact. Note that every gauge compact Hausdorff space is countably
compact. To find a counter example, we may think about a space Y = βω (X),
where as a set Y = β(X) is the Cech-Stone compactification of X, and F ⊆ Y is
closed in Y iff for every countable set A ⊆ F , cl(A) ⊆ F , where cl(A) is the closure
of A in β(X). One easily shows that Y is countably compact and not compact.
Consider X = W0 ,then W1 = β(X)(see [5]). Now {ω1 } is clopen in βω (W0 ) and so
the subspace W0 is also clopen and is not gauge compact, as is shown in Example
15. Hence βω (W0 ) is not gauge compact. However, we still do not know whether
the space βω (N ) is gauge compact.
2). We can use symmetric gauges to define the corresponding M-uniformly
continuous mappings and gauge compactness. It might be fruitful to explore such
structures. Also more can be done for studying G-complete spaces.
3). Given a gauge δ on a space X, an element a is called RδM linked to element b if there exist elements c1 , c2 , · · · , cm in X such that c1 = a, cm = b and
ci RδM ci+1 (i = 1, · · · , m − 1). We write xEδM y if x is RδM linked to y. Clearly EδM
is an equivalence relation on X.
A space X is called gauge connected if for each gauge δ on X, there exist finitely
many elements x1 , x2 , · · · , xn such that for every x ∈ X there is xi satisfying
xEδM xi . Thus every gauge compact space is gauge connected. One can also show
that every connected space is gauge connected.
Consider the space W0 in Example 15. From Theorem 1.9.2 of [1] one can deduce
that if δ is a gauge on W0 , then there must be a β0 ∈ W0 such that ∀β, ∃α ≥ β
with β0 ∈ δ(α). From the fact that [0, β0 ] is a compact subspace, we know that W0
is gauge connected. However W0 is neither connected, nor gauge compact. Gauge
connectedness might be another interesting topic for further study.
A new compactness type topological property
11
4). It is well known that a topological space is compact iff every net in the space
has a cluster point. Suppose X is a space in which every net is gauge clustered.
Then by Proposition 9, X is gauge compact. We call a space X s-gauge compact
if every net in X is gauge clustered. One can also explore such spaces.
Acknowledgement. I would like to express my hearty thanks to the referees for
their many valuable comments and suggestions for the improvement of the paper. Theorem 13 and its proof are essentially from them. Their suggestions also
encouraged me to work out more on symmetric gauges.
References
1. J. Dugundji, Topology, Allyn and Bacon, Inc., Boston, 1966.
2. D. Lecomte, How can we recover Baire class one functions, 2003, preprint.
3. P.Y. Lee, W.K. Tang and D. Zhao, An equivalent definition of functions of the
first Baire class, Proc. Amer. Math. Soc. 129 (2001), 2273–2275.
4. P.Y. Lee and R. Vyborny, The Integral: An Easy Approach after Kurzweil and
Henstock, Cambridge Press, Cambridge, 2000.
5. R. Engelking, General Topology, Panstwowe Wydawinictwo Naukowe, City?, 1975.
6. M.V. Matveev, A survey on star covering properties, 1998, preprint.
7. G. Preuss, Theory of Topological Structures – An Approach to Categorical Topology,
D. Reidel Publishing Company, City?, 1988.
8. L.A. Steen and J.A. Seebach, Counterexamples in Topology, Dover Publications,
Inc., New York, 1978.
Received 1 July, 2003 and in revised form 24 May, 2005.