1. In the proof of Pospisil's Theorem (theorem 7.6) that there are $2^{2^\kappa}$ uniform ultrafilters on $\kappa \geq \omega$, the author writes : Let $\mathcal{A}$ be an independent family of subsets of $\kappa$. For every function $f : \mathcal{A} \to \{0,1\}$, consider this family of subsets for $\kappa$ : \begin{align} G_f = \{X : |\kappa - X| < \kappa\} \cup \{X : f(X) = 1\} \cup \{\kappa - X : f(X) = 0\}. \end{align} The family $G_f$ has the f.i.p. so there exists an ultrafilter $D_f$ etc etc... Why do we have to include $\{X : |\kappa - X| < \kappa\}$ in $G_f$?
  2. In the proof of theorem 7.8 that CH implies the existence of a Ramsey ultrafilter, the author writes : Let $\mathcal{A}_\alpha$, $\alpha < \omega_1$ enumerate all partitions of $\omega$. We will construct an $\omega_1$ sequence of infinite subsets of $\omega$ as follows : Given $X_\alpha$, let $X_{\alpha+1} \subset X_\alpha$ be such that either $X_{a+1} \subset A$ for some $A \in \mathcal{A}_\alpha$, or that $|X_{\alpha+1} \cap A| \leq 1$ for all $A \in \mathcal{A}_\alpha$. If $\alpha$ limit, let $X_\alpha$ be such that $X_\alpha - X_\beta$ is finite for all $\beta < \alpha$. Jech notes that such $X_\alpha$ exists because $\alpha$ is countable. Why is that though?
  3. Lastly. At some point (page 79) he mentions that : If $h : B \to C$ is a one-to-one mapping (B, C are Boolean algebras) such that $u \leq v$ if and only if $h(u) \leq h(v)$, then $h$ is an isomorphism. I don't see why h is onto though. It is indeed a homomorphism, but isn't it trivial to find homomorphisms that respect $\leq$ that are not onto. Am I missing something?
  • 2
    $\begingroup$ You copied wrong, in $G_f$ the third part of the union is actually $$\{\kappa\setminus X: f(X)=0\}$$. $\endgroup$
    – Asaf Karagila
    Sep 3, 2012 at 12:56

1 Answer 1

  1. Since we want to generate a uniform ultrafilter (i.e., $|X| = \kappa$ for all $X \in U$), we must ensure that it contains all co-$<\kappa$ sets, which is exactly what this condition guarantees.
  2. As $\{ X_\beta \}_{\beta < \alpha}$ is decreasing, we need only ensure that $X_\alpha \setminus X_\beta$ is finite for a cofinal sequence of $\beta < \alpha$. As $\alpha$ is countable, it in particular has countable cofinality, so let $\{ \beta_n \}_{n \in \omega}$ be an increasing cofinal sequence in $\alpha$. For each $n < \omega$ pick some $i_n \in A_{\beta_n}$, and ensure that $i_{n+1} > i_n$. Then $X_\alpha = \{ i_n : n \in \omega \}$ is as desired. (This is related to the so-called pseudointersection number $\mathfrak{p}$.)
  3. Jech actually says that a one-to-one homomorphism of $B$ onto $C$ is called an isomorphism, so I guess you are missing the onto part of the definition.

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