there exists a separating element E in the 6*-complete Boolean algebra *ßt . But /rn^CE )->-^гСЕ), which is a tradiction. Hence (АСплг)^ 1' M(nit) . Put M^=-f'W,: i ^ ^? for /rt € N . Clearly, M^Our) ^ l-9\^Cnv) j hence ^(плг)^ ^ ik^Cnir)^ 2*M^0uj')«l'A>4jujfi ryrij (плг ) for all m,e N

A (^w) é: 1 • Zum /rn^ (itr ) .

( 4 . 3 ) Lemma* Let ^ be a ^-complete Boolean algebra, let M be a sequentially compact nonvoid subset of clC^) and let oir be a filter of % with a countable basis» Then

A

Proof» If there is a number d such that I'M(тлг) <d< '^fACvr) and if iAji^}^^^ is a monotone basis for the ter oir then the set M-«{лп.еМ ^/m.CA^) >' cL 1 is nite for each jfe e N . Indeed, if Mj,^ is finite set then pLCnàr)я b\. Our) ^ iMj^C^ur) , hence there Is a set A e vr, A с A^. such that /vofv M^CA ) ^ ci . Then M Cnjtr) ^ ébafiMCA>^ci jWhich is a contradiction»

Now choose nru e M- . If M^ =? '^^^^^^ -^^^ i ^^ ^ sub- sequence of the sequence ^^Щк}!^»^ then yUcfz.M^(A^> ^ ci föT each M. e H . НепсеЙ^С'Ш');^с1>2-/ЙГ'и^) ai l-îÀ^C'u^) )

yi ( %r ) >l*f^X'ur) ."Ртош Theorem (1.3Î it follows that there о о

is a separating element in ^ for the set M^ . Hence the subsequence ^Щ^^^.^^^а does not converge and the set M is not sequentially compact, which is a contradiction» ^^•4) Theorem» Let *^ be a в'-complete Boolean algebra. Let U Ы| a nonvoid subset of сиСШ) such thatlc9LI=^o and let M be a sequentially compact subset of ciC^) (in

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