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Topology Proceedings
Vol. 19, 1994
MORE ON TOPOLOGICAL COMPLETIONS OF
METRIZABLE SPACES
BEN FITZPATRICK, JR. AI'JD HAOXUAN ZHOU
We continue the investigation of determin­
ing the completions and completion remainders of various
metrizable spaces. We introduce the study of spaces that
are the differences of two completely metrizable spaces
and of spaces that are rim-conlplete: possessing bases
whose elements have complete boundaries. These no­
tions are used to characterize the completion remainders
of lP, the irrationals.
ABSTRACT.
1.
INTRODUCT-ION
All spaces under consideration are assumed to be metriz­
able. A space is called complete provided that it is completely
metrizable (== absolute G 6-set). By analogy with the notion of
rim-compactness, we define a spac~~ to be rim-complete if it has
a basis whose elements have comJ:tlete boundaries. A space X
is called strongly rim-complete provided that if H ~ U ~ X,
with H closed and U open, then there is an open set V in X,
such that H ~ V ~ U and f1x(V) is complete. Here, f1x de­
notes the boundary operator in tlle space X. Our interest in
these notions stems from their usefulness in discussing topolog­
ical completions of metrizable spaces. For example, a separable
space X is a completion remainder of JP>, the irrationals, if, and
only if, X is both rim-complete aIld the difference of two com­
plete spaces. This is shown in Section 2; it provides an answer
to a question raised in [FGO]. Section 3 deals with nonsepara­
97
98
BEN FITZPATRICK, JR. AND HAOXUAN ZHOU
ble spaces. Section 4 provides a sufficient condition fora space
to be rim-complete. Section 5 provides characterizations of
spaces tl1at are the differences of complete (absolute G8-sets)
spaces. Section 6 discusses ambiguous classes and provides two
examples of interest.
2.
COMPLETION REMAINDERS OF THE IRRATIONALS
If X and Yare spaces, Z is a complete space, and there
exists a homeomorphism h : X ~ Z such that h(X) is dense
in Z and Z\h(X) is homeomorphic to Y, then Z is called a
completion of X, and Y is called a completion remainder of
X; in case h (X) can be taken to be open in such a Z then
Y is called a closed completion remainder of X. Denote by
Q or lP the space of rationals or irrationals, respectively, with
the usual topology. It was shown in [FGO] that the completion
remainders of Q are the nowhere locally compact Polish spaces,
and the problem of determining the completion remainders of
JP> was posed.
Theorem 1. In order that the separable space X be a com­
pletion remainder oj' lP, it is necessary and sufficient that 1)
X be the difference oj'two complete spaces, and 2) X be rim­
complete.
Proof: (Necessity) Assume that Z is a complete space, Z ==
A U X, A n X == 0, A is dense in Z, and A ~ P. Then
X == Z\A and is thus the difference of two complete spaces.
Since A is O-dimensional there is a basis for the topology of Z
at points of X whose elements 11ave boundaries missing A. If
{3(g) is such a boundary it is a closed subset of the complete
space Z and is, therefore, complete. So, X is rim-complete.
(Sufficiency) Assume that X is separable, rim-complete, and
the difference of two complete spaces. We may assume that
X == B\C, where Band C are complete and C ~ B.
We first consider the special case in which both C and B\C
are dense in B.
We shall use the following elementary lemma.
MORE ON COMPLETIONS OF METRIZABLE SPACES
99
Lemma 1. If A is dense in X, U is open in A, V is open in
X, and V n A == U, then ,8x(V) n A == ,8A(U).
There is a sequence {G n
covers of X such that
:
111 <
~)}
of countable open (in B)
1) \In < w, G n+1 ~ G n , and if 9 E G n+1 then ClB(g) is a
subset of some element of G~n,
2) \In < w, \lg E G n , diam 9 < 1/111, where the diameter is
taken in a (fixed) complete metric on B,
3) \In < w, \lg E G n , f3x(g n X) == f3B(g) n X is complete.
Let Y == nn<w(UG n ). Since X ~ Y and X is dense in B,
Y is dense in B; also, Y is a G8-set in a complete space and is
complete. We note that X is dense in Y. Each U G n contains
a dense G8-set C n in C, so Y contains nn<w Cn, a dense G8-set
C' in C, and C' is dense in B, and (7nX == 0, so Y\X contains
a dense set C' in Y. That is, both X and Y\X are dense in Y
(and in B).
Let
A
(Y\X)\(
U U (/3 B (g)))
n<wgEGn
(ync)\(U U (,8B(g)n(YnC))).
n<WgEGn
Since f3B(g) is closed in B, f3B(g) n Y n C is closed in the
complete space Y n C, so Un<w LJgEGn (,8 B (g) n Y n C) is. an
Fa-set in Y n C, so its complement A in that space is a G8-set
in Y n C and is, therefore, complete.
By the Baire Category Theoreln, A is dense in Y\X and,
therefore, in B. Since A has a (cou:ntable) basis whose elements
have empty boundaries, dim A == O. Since both A and its
complement are dense in B, A is llowhere locally compact.
Therefore, A ~ P. We note that A n X == 0 and that A is
dense in A U X .
100
BEN FITZPATRICK, JR. AND HAOXUAN ZHOU
Now,
AUX
Y\(
U U (f3B(g)\f3X(g n X)))
U U (f3B(g) n Y\f3X(g n X) n Y))).
n<wgEGn
Y\
n<WgEGn
Since f3x(gnX) is complete, f3x(gnX)nY is complete, and
f3B(g) n Y is closed in Y, so (f3B(g) n Y)\(f3x(g n X) n Y) is
an Fa-set in Y; so its complement in Y is complete.
This completes the sufficiency proof in the special case.
We now consider the ge11eral case: X == B\O, where Band
o are complete, 0 ~ B, and X is rim-complete. If 0 is empty,
it is known [FGO] that X is a completion remainder of JP>, so
we assume 0 =1= 0 and 0 =1= B.
Let S == {x EX: X is locally complete at x}, and let
T == X\S. Then S is open in X and is complete; so, S
is a closed completion remainder of JP> [FGO]. The space T
is nowhere locally complete. It is straightforward to show
that rim-completeness is inherited by closed subsets; conse­
quently; T is rim-complete. Since S is complete, T == X\S ==
(B\O)\S =='B\(OUS) is the difference of two conlplete spaces.
Let B 1 == OlB(T); let 0 1 == (0 U S) n B 1 ; let T 1 == B 1 \0 1 .
Then B 1 a11d 0 1 are complete, and 0 1 ~ B 1 . Since T is
nowhere locally complete, 0 1 is dense in B 1 . Since T is dense
in B1, B 1 \0 1 is dense in B 1 . By the special case, it follows
that T is a completion remainder of JP>.
We now regard X as a subset of a face F in the Hilbert cube
W
I • ,Since S is open in X, we may assume there is an open
set U in F such that S == U n X and Z == F\U is a compact
space with no isolated points. Applying the construction of the
special case to Z, we obtain a copy K of JP>, K ~ Z, such that K
is dense in K U T, and K U T is complete. Since S is complete,
S ~ F, and S is a closed completion remainder of JP>, [FGO],
it follows that there is a copy J of JP>, J ~ IW\F, J is dense in
JUS, and JUS is complete. We have (JUK)n(SUT) == 0, JUK
is dense in (JUK)UX, and (JUK)UX == (JUS)U(KUT) is
MORE ON COMPLETIONS OF METRIZABLE SPACES
101
the union of two con1plete spaces arld is thus complete. Clearly,
J U K ~ lP'.
This completes the proof of Theorem 1.
Remarks. The O-dimensional case of Theorem 1 was pre­
sented by the second author at the Auburn Spring Topology
Conference, March 1994. It is know'n [FGO] that Q x I is not a
completion ren1ainder of lP'. It is the difference of two complete
spaces, so it is not rim-complete. It can easily be embedded
as a closed subset in JR3 of a connected and locally connected
set, so rim-completeness is not a COJnsequence of these connect­
edness properties. Of course, a s]pace may be rim-con1plete
without being the difference of two complete spaces, for exam­
ple, a O-dimensional subset of 1R that is not a Borel set.
Theorem 2. If X is separable and not compact, then X is a
completion remainder of lP' if, and only if, lP' is a completion
remainder of X .
Proof: Suppose first that lP' is a completion remainder of X.
Then there is a con1plete space Z = A U X, where A ~ lP',
A n X = 0, and X is dense in Z. Then X = P\A, so X is the
difference of two complete spaces. ~Let p E U ~ X, U open in
X. There is an open set W in Z such that W n X = U. Since
A is O-dimensional, there is an ope:n set V in Z, p E V ~ W,
and j3 z (V) n A = 0. Then W n X is open in X, W n X ~ U,
and f'x(W n X) = f'z(V) is complete. Therefore, X is rim­
complete, and by Theorem 1, X is a completion remainder of
lP'.
Next, suppose X is a completion remainder of lP'. Then X is
rim-complete and the difference of two complete spaces. If X
is complete, then by Theorem 1 of [I~GO], lP'is a completion re­
mainder of X. Let T = {x EX: X is not locally complete at x}.
Regard X as a subset of a face F of the Hilbert cube. As in
the sufficiency proof of Theorem 1, there is a copy K of JP> in
F such that K n T = 0, K is dense in K U T, and K U T is
complete. But T is also dense in K" U T, or at least it can be
102
BEN FITZPATRICK, JR. AND HAOXUAN ZHOU
so taken, for T is dense in Clp(T), and K can be taken to be
a subset of Clp(T). Since T is closed in X, K n X == 0. So,
X U K == (X\T) U (K U T) is a complete space, X is dense in
it, and the complement of X in it is K, a copy of JID. Therefore,
JID is a completion remainder of X.
Theorem 3. If X is separable, then X is rim-complete if, and
only if, there is a completion Z of X such that dim(Z\X) :::; O.
Proof: If X is itself complete but not compact, then we can add
a single point p to X in such a way that X is dense in X U {p }.
Suppose X is not complete, and let T be the set of all points
of X at which X is not locally complete. Regard X as a dense
subset of a compact space Z. Then Clz(T) is compact, and
T and Clz(T)\T are dense in Clz(T). We proceed as in the
special case of Theorem 1; this time it does not follow that A
is complete. It is true, however, that A is O-dimensional and
that AUT is complete. Then X U A is complete, and X is
dense in X U A.
Remark. The construction can be strengthened to yield that
if X is rim.:complete, separable and not compact, and if Hand
K are disjoint closed sets in X, then X has a completion Z
such that Z\X is O-dimensional and Clz(H) n Clz(K) == 0.
Theorem 4. If X is separable and rim-complete, then X is
strongly rim-complete.
Proof: Suppose H ~ U ~ X, where H is closed and U is
open. Then Hand X\U are disjoint closed sets in X. By the
remark above, there is a completion Z of X such that Z\X is
O-dimensional and Clz(H) n Clz(X\U) == 0. There exist open
set WH and W' in Z such that Clz(H) ~ WH, Clz(X\U) ~
W', and ClZ(WH) n Clz(W') == 0. By a standard theorem
from dimension theory [HW, 16-17], there is an open set V in
Z such that Clz(H) ~ V ~ WH and ,6z(V) n (Z\X) == 0.
Now, j3z(V) is a closed subset of the complete space Z, so it
is complete. Let S == V n X. Then H ~ $ ~ W.Also,
,6x(S) == ,6x(V n X) == ,6zX (V) n X == ,6z(V) and is complete.
MORE ON COMPLETIONS OF METRIZABLE SPACES
103
Remark. We do not have an exanlple of a rim-complete space
that is not strongly rim-complete.
3.
SPACES THAT MAY NC)T BE SEPARABLE
In this section we establish the hereditary character of the
properties under discussion, and we discuss completions of
strongly rim-conlplete spaces.
~'heorem
5. A. If X is rim-com.plete and M is a G8-set in
X, then M is rim-complete. B. I]' X is strongly rim-complete
and M is closed in X, then M is strongly rim-complete.
Proof: (A) Special case 1: M is closed in X. Let p E U, U
open in M. There is an open set 11 in X, V n M == U. There
is an open set W in X, pEW ~: V, with fJx(W) complete.
Then fJM(W n M) ~ fJx(W) n M, and fJx(W) n M is a closed
subset of the complete space fJx(~V), so it is complete. Now,
(3M(W n M) is closed in M, so it is closed in every subset of
M of which it is a subset; i.e., fJA1(W) n M is closed in the
complete space fJx(W) n M, so it is complete.
Special case 2: M is a dense G8-set in X. Then M == nn<w
Un, where Un is open in X and Un ~ Un +1 . Let 9 be open in
X, with fJx(g) complete. Then fJM(g n M) == fJx(g) n M ==
nn<w(fJX(g) n Un). For each n, fJ.x(g)'n Un is open in fJx(g)
and is complete. So, nn<w(fJX(g) r1 Un) is complete.
General case. By special case 1, Clx(M) is rim-complete,
and M is a dense G 8-set in C lx (1\11), so by special case 2, M
is rim-complete.
(B) The argument is straightforward and is omitted.
Theorem 6. If X is the difference of two complete spaces and
M is a G 8 -subset of' X, then M is the difference of two complete
spaces.
We omit the proof, as it is straightforward.
We next address the question of completing spaces that are
strongly rim-complete.
104
BEN FITZPATRICK, JR. AND HAOXUAN ZHOU
Lemma 2. If X is strongly rim-complete and dense in Z and
U is a a-discrete open (in Z) cover oj' X, then there exists a
a-discrete open (in Z) refinement V ofU which covers X such
that if V E V then JJz(V) n X is complete.
Proof: For U E U, let {Un(U) : n < w} be a sequence of open
sets such that Clz(Un(U) ~ Un+1 (U) and U = Un<w Un(U).
Let Hn(U) = Clz(Un(U)) n X, Vn(U) = Un+l(U) n X. There
exists Dn(U), open in X, such that Hn(U) ~ Dn(U) ~ Vn(U)
and such that JJx(Dn(U)) is complete. There exists Wn(U),
open in Z, such that Wn(U)nX = Dn(U). Then JJz(Wn(U))n
X is complete. For each n < w, {Wn(U) : U E U is a-discrete,
so {Wn(U) : n < w, U E U} is a-discrete.
Corollary to Lemma 2: If X is strongly rim-complete, then
X has a a-discrete basis whose elements have complete bound­
aries.
Theorem 7. If X is strongly rim-complete and the difference
of two complete spaces, then X is a completion remainder of a
strongly O-dimensional, complete space.
Theorem 8. If X is strongly rim-complete, then there is a
completion Z of X such that dim(Z\X) ~ O.
Indication of Proofs. The proofs are very much like those of
Theorems 1 and 3; Lemma 2 is used to provide a sequence {G n :
n < w} of a-discrete open covers of X with the properties 1),
2), 3) listed before. The set A is now strongly O-dimensional,
because it has a a-discrete basis whose elements have empty
boundaries. Instead of using a face F of the Hilbert cube, we
use a face F of thew-power of a suitable hedgehog.
4. SUFFICIENT CONDITIONS FOR A SPACE TO BE
RIM-COMPLETE
As was noted following Theorem 1, Qx I is not rim-complete.
In this section we discuss spaces X = Un<w X n, where each X n
is closed, X n n X m = 0 for n, =f m" and there is a metric on
MORE ON COMPLETIONS OF METRIZABLE SPACES
105
X such that diam X n --t 0 as n -~ 00. We show that if, in
addition, each X n is rim-complete, then X is rim-complete.
IJemma 3. If, in a metric space ~~, G == {X n : n < w} is a
collection of mutually exclusive closed sets and diam X n --t 0
as n, --t 00, then G is upper semicontinuous.
Proof: Let U be an open set containing Xo. For each x E Xo,
there is an ax > 0 such that B(x, a:J~)' the open ax ball centered
at x, is contained in U. There is a I>ositive integer n x such that
if k ~ n x then diam Xk < a x /3. ~rhere is a 8x > 0 such that
8x < a x /3 and B(x, 8x ) misses.UI<i<n Xi. Let Vx == B(~, 8x ).
If k > 0 and X k n Vx =1= 0, then k- ~x, so diam Xk < Q.x/3,
:>
and dist (x, Xk) < 8x < G- x /3, so Xk ~ B(x, ax) ~ u. Let
== UXEXo Vx . Then Xo ~ V ~ [1, and if X j n V i- 0, then
)(j ~ U. Thus G is upper semicontinuous at Xo, and similarly
(i is upper semicontinuous at each of its elements.
"r
Lemma 4. If G is as in Lemma 3, then the decomposition
space of uG obtained . b y contracting each element of G to a
point is normal; hence, homeomor]Jhic to a subset of Q.
Proof: Normality is given in [K, page 185].
Since the decomposition space is countable and normal, it
is metrizable, and every countable metrizable space is homeo­
nl0rphic to a set of rational numbers.
Theorem 9. If X == Un<wXn, where xnnx m == 0 form, i- n,
diam X n --t 0 as n ~ 00, and each X n is closed in X and
rim-complete, then X is rim-complete.
Proof: Let x E unxo, where U is open in X. There is an open
set WI in X such that x E WI and ClX(WI ) ~ u. There is an
open set V{ in X o such that x E V{ ~ WI n X o and f3 x o(V{) is
complete. There is an open set VI iJrl X such that VI n X o == V{
and VI ~ WI. Then Clx(VI ) ~ U an.d f3xo(VlnXo) is complete.
L,et Xb == Xo nVI and let G == {X~} U {Xj; 0 < j < w}. Then
G satisfies the conditions of Lemrrta 3. Let f : U G -+ Y be
the quotient mapping from G onto the decomposition space Y ,
106
BEN FITZPATRICK, JR. AND HAOXUAN ZHOU
and regard Y as a subset of Q. There is an open set V in X
such that V n Xo = X~, V ~ U, and V contains every element
of G that it intersects. Since UG is open in X and j is a
closed mapping, V could be chosen as j-l(E), where E is a
subset of Y without boundary and E contains j(Xb). Clearly,
,8z(V) = ,8xo(V)
5.
=
,8xo(V n X o) = ,8xo(V1 n Xo).
SPACES THAT ARE THE DIFFERENCES OF TWO
ABSOLUTE G6-SETS
It seems appropriate in this section to use the terminology
'absolute G6-set' instead of the equivalent 'complete space,'
because of the constructions which arise.
Theorem 10. The space X is the difference of two absolute
G6-sets if, and only if, X is the union of countably many closed
subsets, each an absolute G6-set.
Proof: Suppose X = B\C, where C ~ Band Band Care,
absolute G6-sets. Now C = nn<w Un, where each Un is open
in B. Then X = B\(nn<w Un) = Un<w(B\Un ). Each B\Un is
closed in the space B and is thus an absolute G6-set; also, each
B\Un is closed in every subset of B of which it is a subset.
Next, suppose X = Un<w G n , where each G n is closed in
X and is an absolute G6-set. Let Z be a completion of X.
Then, for each n, G n = ni<w Un,i, where Un,i is open in Z and
Un,i+l ~ Un,i.
Let VO,i = UO,i' and for 1 ~ n < w, let Vn,i = Un,i\ClZ(Uj<n G j ).
Let Wi = Un<w Vn,i, alld let B = ni<w Wi. Then B is an
absolute G6-set, and X ~ B. L~t D = Ui<w Clz(G i ). It
{ollows that X = B n D. We denote Clz(Gn) by F n , and we
denote the open set Z\Fn by En.
MORE ON COMPLETIONS OF' METRIZABLE SPACES
107
Now,
X
U
B n D == E~ n (
Fn )
n<w
B n ( (L~\En))
n<w
((B n L~)\(B n En))
n<w
U
U
(B n Z)\
fl (B n En),
n<w
so X is the difference of the two absolute G6-sets B
nn<w(B n En).
n Z and
Theorem 11. The space X is the difference of two absolute
G6-sets if, and only if, X == Un<VoJ X n and there is a metric p
on X such that each X n is complete in the restriction of p to
Xn·
Proof: First, let X == B'\C, where Band C are absolute G6­
sets and C ~ B. Let p be acornJ:.lete metric on B. Now, C ==
nn<w Un' where Un is open in B. Then X == B\(nn<w Un) ==
Un<w(B\Un ). Each B\Un is closed in X. Now, fix n, and let
{x m : m < w} be a p-Cauchy seqllence of points of B\Un . The
sequence must converge to a point of B; but B\Un is closed in
B, so it converges to a point of Er\un .
Next, assume X == Un<w X n and there is a metric p on
X such that each X n is complete in the restriction of p to
X n . Let (Z, (5) be a metric cOITlpletion of (X, p). Let C ==
Z\X. Then X == Z\ C It now suffices to show that C is an
absolute G6-set. Each X n is closed in Z; to see this; let Q
be a convergent (in Z) sequence of points of X n . Then Q is
a p-Cauchy sequence; hence, a p-Cauchy sequence; therefore
by hypothesis Q converges to a point of X n . We now have
C == Z\(Un<w X n ) == nn<w(Z\Xn ), so C is an absolute G6- set .
1
•
Theorem 12. The space X is the difference of two absolute
G6-sets if, and only if, X is locally the difference of two absolute
G6-sets .
108
BEN FITZPATRICK, JR. AND HAOXUAN ZHOU
Proof: We first observe that a union of a locally finite collection
of complete spaces is complete. Now, suppose X is locally the
difference of two absolute Go-sets. Since X is paracompact and
since being the difference of two complete spaces is inherited
by G 8-subsets, it follows that 1) there exists a locally finite
collection U == {U a : a < A} of open sets covering X such that
each UO/. is the difference of two absolute Go-sets, and 2) there
is a closed collection :F == {Fa: a < A} such that Fex ~ Ua,
Fa is closed, and :F covers X. Each Fa is the difference of two
absolute Go-sets, so by'Theorem 10, Fa == Un<w Fa,n, where·
Fa ,n is an absolute Go-set and Fa ,n is closed in Fa and therefore
in X. For each n, let H n == Uex<-x Fa,n. Then H n , as a union
of a locally finite collection of closed sets, is closed, and by the
initial observation it is an absolute Go-set. Also, X == Un<w
H n , so by Theorem 10, X is the difference of two absolute
Go-sets.
6. AMBIGUOUS CLASSES; TWO EXAMPLES
Definition. A set A is of ambiguous class 2 if A E GoanFao .
We note that if A == B\C, for B, C E Go, then A == B n C',
where C' is an Fa; so that A E Goa n Fao . The complement of
A is a union of an Fa and a Go, so it is also of ambiguous class
2. Example 1 below shows that a set may be of ambiguous
class 2, even the union of an absolute Go and a a-compact set,
without being the difference of two absolute Go-sets. Example
2 provides an instance of a Goa-set which is not of ambiguous
class 2. These examples show that Theorem 10 is the best
along those lines that can be expected.
In I x I, let Xl == Q x I, X2 == IP x lP,
Q, X == Xl U X 2 . Then X is
the union of a Go and countably many compact sets, so it is
of ambiguous class 2. It is not, however, the difference of two
,Go-sets. We suppose that it is. Then X3 is a union of an Fa-set
Un<w F n and a Go-set G. For each r E Q, let L r == JP> x {r}.
Example 1.
X3 == 1 2\(X I U X 2 ) == lP x
MORE ON COMPLETIONS OF METRIZABLE SPACES
109
Since each F n is first category in L r (because each compact set
in L r is nowhere dense), then Gr,n == L r\Fn is a dense G6-set.
'rhe first projection Projl(Gr,n) is G~,n' We have countably
luany dense G6-sets in I, namely, (i~n, r E Q, n < w. Let t be
in their intersection. Then {t} x ~~ is a closed subset of G, so
it is a G6-set, which is a contradiction.
Example 2. In I W, let B == (~, A == IW\B == Un<wAn
where An == Ili<w Xi, Xi == I for i =I- n, and X n == P. Then
. An is a G6-set, and A is a· G6a-set, but A is not an Fa6, so
. A is not ambiguous of class 2. If A E F a 6 then B E G6a.
Assume B == Un<w G n , where eac:h G n is an absolute G6-set.
'We claim that each G n is nowhere dense in B. If not, then
there is a basic open set, for examI>le, C == U x ijW ~ ClB(GO),
I
where U == (a, b) ,n Q. Then Go (I C is a dense G6-set in C.
Since each C i == {ri} x Q is a closed nowhere dense set in
C for a < ri < b, ni<w(C\Ci ) n (io is a dense G6-set in Go;
llence, ni<w (C\C i ) n Go =I- 0, because Go is complete. But
rli<w(C\Ci ) == 0, a contradiction.
Since Go is nowhere dense, there is a basic open set Vo ==
. Do x Q, where Do == Do,o X Do,l x·· · x Do,no is an open set in
Qno such that VonG o == 0. In particular, there is to E Qno such
that to x ijW-non Go == 0. Repeating the argument, we obtain a
tl E Qno x ... x Qno+n 1 , such that 1:0 x tl x ijW-(no+n 1 ) n G 1 == 0.
Finally, there is a sequence t == {ti : i < w}, t E ijW t/:. Un<w G n .
'Therefore ijW =1= Un<w Gn'
Acknowledgement.
We are indebted to the referee for
llelpful suggestions.
REFEREN,CES
[FGO] B. Fitzpatrick, Jr., G. F. Gruenhage, and J. W. Ott, Topological
completions of metrizable spaces, F-roc. A.M.S. 117 (1993),259-267.
[HW] W. Hurewicz and H. Wallman, Dimension Theory, 1941, Princeton
University Press.
[K] K. Kuratowski, Topology, Acadenlic Press, New York and London,
1976.
110
BEN FITZPATRICK, JR. AND HAOXUAN ZHOU
Incarnate Word College
4301 Broadway
San Antonio, Texas 78209