1974
DOI: 10.1007/978-3-642-65780-1
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The Theory of Ultrafilters

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Cited by 437 publications
(269 citation statements)
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“…A < K. There exist II I-good countably incomplete ultrafilters on I (in fact exp2(III) of them [17]) and this is the maximal degree of goodness possible. The major lemma which will be of use to us here is the following.…”
Section: (Ie S C T => F(t) C F(s)) "Dominates" a "Multiplicative" Gmentioning
confidence: 97%
See 1 more Smart Citation
“…A < K. There exist II I-good countably incomplete ultrafilters on I (in fact exp2(III) of them [17]) and this is the maximal degree of goodness possible. The major lemma which will be of use to us here is the following.…”
Section: (Ie S C T => F(t) C F(s)) "Dominates" a "Multiplicative" Gmentioning
confidence: 97%
“…What we really need to show is that if <Xi: i E I> is any family of compact Hausdorff spaces such that {i: Ind(Xi) < n} E 9, then Ind(Zg, Xi) < n. (1.7) establishes the assertion for n = 0; so we work on the inductive step, and assume the more general statement true for dimensions < n. Assume Ind(X) < (ii) The inequality in (2.2.3), as well as the corresponding one in (2.2.2 Recall that for any space X, the weight w(X) of X is the least infinite cardinal of a basis for X. If X is infinite Boolean, then it is easily shown [17] that w(X) = IB(X)I. We thus immediately get the fact that if X is an infinite Boolean space and 9 is a regular ultrafilter on I then w(ZgX) = w(X)I1.…”
Section: H?g MI Where Mi Is Open (Closed) In Xi It Is Easy To Verifmentioning
confidence: 99%
“…This choice satisfies (1) and (2). Finally, if f n+1 (x n+1 ) lies in the closure of a countable set of weak P -points, fix any such countable set Y n+1 with Y n+1 ∩ {f (x l ) : l ≤ n + 1} = ∅.…”
Section: P -Points and Weak P -Points In C K (X X)mentioning
confidence: 99%
“…For (2), note that by continuity of ev (Lemma 6.4), we have ev { f n , p n : n ∈ ω} ⊆ ev({ f n , p n : n ∈ ω}) = B.…”
Section: Convergent Sequences Inmentioning
confidence: 99%
“…Let h: U(k)-* I be the composition of/ and g. In addition, let h: ßK-* I extend h. Take s G I arbitrarily. Then g"'({¿}) is a nonempty Gs in l/(«) and consequently has nonempty interior, [CN,14.17]. Therefore,/-1g_1({¿}) has nonempty interior (in £/(*)) and consequently we can find a subset E c k so that 0 ¥= Ë n UÍk) G f-xg-xi{s}).…”
mentioning
confidence: 99%