GLOBAL SUBDIRECT PRODUCTS
9
Corollary 1.8 Suppose every open subject of I is the union of closed
sets. Then
(i) h(l) is a dense subset of M;
(ii) I is compact if and only if h(l) = M. 1
Lemma 1.9 M is a compact T -space (with the relative topology). 1
If I is a T -space then M is called the Wallman T -compactification
of I.
Next we shall quickly compile some information for the case where "X , is
a Boolean ring. In this case I is a T -space if and only if I is a T.,-
space. Moreover, M = J. If
then 3" is a field of subsets of I and fc = ^ if and only if I 1R, .
Lemma 1.10 If I C $2, then
(i) 7Rs is a prime filter on 3";
(ii) for every prime filter U on R, there exists a unique prime
filter V~ on 3" such that Vi = tfTl£"; moreover, V ± fa ;
(iii) for every prime filter V on 3? , if V ? 1R, then IT^VL
is a prime filter on 'R,. I
Now there are two topologies on I, 3~(%,) and 3"(?). We shall dis-
cuss the space (I, 3*( S")) in case the space (I, T(3£)) is compact.
Lemma 1.11 Suppose (I, 3~(*&)) is compact. Then (I, T(Sr)) is
compact if and only if I 32, or 1Rj does not cover I.
Proof: If (I, 3""(3~)) is compact and 1R, covers I then clearly
I 1R, . Conversely, suppose 1R, does not cover I and ^ c J: has the
finite intersection property. We distinguish two cases. Case 1 "C 0 1R, 4 0.
Pick F t n ^ , Then
Since HZ is Boolean, for every G€£\ F O G E ^ . Since (I, T(lfl)) is
compact, f) t * 0. Case 2 £ 0 « = 0. Then £ c ^ and H C ^
because O 'f t ^ 0. This establishes that (I, S"(Sr)) is compact.
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