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I'm reading, from Kunen's set theory, the proof that $\diamond ^+$ implies that there is a family $\mathfrak F\subset P(\omega_1)$ such that

  1. $\forall \beta <\omega_1 (|\{X\cap \beta: X\in \mathfrak F\}|\leq \omega)$
  2. $\forall A <\omega_1 (|A|=\omega_1\rightarrow \exists X\in\mathfrak F(|X|=\omega_1\wedge X\subseteq A))$.

Then for given $C\subseteq \omega_1$ and $\xi<\omega_1$ the author defines $s(C,\xi)=\sup(C\cup \{0\})\cap (\xi+1)$, and for given $A,C\subseteq \omega_1$ $$X(A,C)=\{\xi\in A:¬\exists\eta\in A(s(C,\xi)\leq \eta <\xi)\}.$$

Now let $C$ be a c.u.b. of $\omega_1$ and $A\subseteq \omega_1$ of size $\omega_1$. Fix $\beta<\omega_1$. Let $\alpha=s(C,\beta)$, let $\xi$ be the least element of $A-\alpha$, and put $B=A\cap \alpha$ and $D=C\cap \alpha$. Then the author states: $X(A,C)\cap \beta=X(B,D)$ if $\xi\geq \beta$ and $X(A,C)=X(B,D)\cup \{\xi\}$ if $\xi<\beta$, I don't understand why this holds, so I would like to ask why this is so.

Thanks

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To set the background:

  • $\alpha$ is the largest element of $C$ that is $\le\beta$;
  • $\xi$ is the smallest element of $A$ that is $\ge\alpha$;
  • $B$ is the set of elements of $A$ that are less than $\alpha$, so $B=A\cap\xi$; and
  • $D=C\cap\alpha$.

Suppose that $\gamma\in X(B,D)$; then $\gamma\in B$, and there is no $\eta\in B$ that is both less than $\gamma$ and at least as large as $s(D,\gamma)$, the largest element of $D$ that is $\le\gamma$. In other words, $B\cap\big[s(D,\gamma),\gamma\big)=\varnothing$. $B\subseteq A$, so $\gamma\in A$. Moreover, $\gamma<\alpha$, and $D$ is closed in $\alpha$, so $s(D,\gamma)$ is the largest element of $D$ that is $\le\gamma$ and hence the largest element of $C$ that is $\le\gamma$, since $D=C\cap\alpha$. Thus, $s(C,\gamma)=s(D,\gamma)$, and

$$A\cap\big[s(C,\gamma),\gamma\big)=B\cap\big[s(D,\gamma),\gamma\big)=\varnothing\;.$$

It follows that $\gamma\in X(A,C)$ and hence that $X(B,D)\subseteq X(A,C)\cap\beta$. (Recall that $\gamma<\alpha\le\beta$.)

Now suppose that $\gamma\in X(A,C)\cap\beta$. Then $\gamma\in A\cap\beta$, and $A\cap\big[s(C,\gamma),\gamma\big)=\varnothing$. If $\xi\ge\beta$, then $A\cap\beta\subseteq A\cap\xi=A\cap\alpha$, so $\gamma<\alpha$. We’ve already seen that if $\gamma<\alpha$ we have $s(C,\gamma)=s(D,\gamma)$ and hence

$$B\cap\big[s(D,\gamma),\gamma\big)=A\cap\big[s(C,\gamma),\gamma\big)=\varnothing\;,$$

so that $\gamma\in X(B,D)$. Thus, in this case we have $X(A,C)\cap\beta=X(B,D)$.

Suppose now that $\xi<\beta$ and that $\gamma\in X(A,C)\cap\beta$. It’s still true that if $\gamma<\alpha$, then $\gamma\in X(B,D)$ and hence that $X(A,C)\cap\alpha=X(B,D)$. Suppose that $\alpha\le\gamma\in X(A,C)\cap\beta$; $\gamma\in A$, and $\xi$ is the smallest element of $A$ that is $\ge\alpha$, so $\gamma\ge\xi$. Thus, $\alpha\le\xi\le\gamma<\beta$. $C\cap(\alpha,\beta]=\varnothing$, so $s(C,\gamma)=\alpha$, and hence $A\cap[\alpha,\gamma)=\varnothing$. But $\xi\in A$, so this implies that $\gamma\le\xi$ and hence that $\gamma=\xi$. Finally, it’s clear from the definition of $\xi$ that $A\cap[\alpha,\xi)$ actually is empty, so $\xi\in X(A,C)$, and in this case we have $X(A,C)\cap\beta=X(B,D)\cup\{\xi\}$.

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  • $\begingroup$ @Camilo: You’re welcome. $\endgroup$ – Brian M. Scott Mar 21 '13 at 19:16

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