# Shelah's proof of diamond

This paper is http://www.cs.elte.hu/~kope/ss3.pdf .

In Remark 1 : I want to prove that the set $$D=\{\delta<\lambda^+:\pi\restriction \delta \text{ is a bijection onto }\lambda\times\delta\}$$ is a club in $\lambda^+$.

$D$ is closed : let $\mu<\lambda^+$ and $\delta_1<\delta_2<\dots<\delta_\xi<\dots \in D$ a strictly increasing sequence of elements of $D$ of length $\mu$. Let $\delta=sup(\delta_\xi:\xi<\mu)$. Define $\pi\restriction\delta$ as follows : if $\gamma<\delta$ define $\pi\restriction\delta(\gamma)=\pi\restriction\delta_\xi(\gamma)\in\lambda\times\delta_\xi\subseteq\lambda\times\delta$ for a $\xi<\gamma$ such that $\gamma<\delta_\xi<\delta$. Then $\pi\restriction\delta$ is a bijection onto $\lambda\times\delta$. So $\delta\in D$ and $D$ is closed.

$D$ is cofinal : let $\mu<\lambda^+$. I want to find an element of $D$ between $\mu$ and $\lambda^+$. I don't know how to do that .... construct a strictly increasing sequence of elements greater than $\mu$ so that the sup is in $D$ ? but how ....

In Remark 2 : First, it is needed to see that the cardinality of $[\lambda^+]^{\leq\lambda}$ is $\lambda^+$. My proof is as follows : we have $$[\lambda^+]^{\leq\lambda}=\coprod_{\mu\leq\lambda}\{X\subseteq\lambda^+ : |X|=\mu\}$$ so $$|[\lambda^+]^{\leq\lambda}|=\lambda\times\sup\{|[\lambda^+]^\mu| : \mu\leq\lambda\}= \lambda\times(\lambda^+)^\lambda=(\lambda^+)^\lambda$$ Of course, we know that $2^\lambda\leq(\lambda^+)^\lambda$ and $\lambda^+\leq2^\lambda$ so $(\lambda^+)^\lambda=2^\lambda=\lambda^+$. Hence the enumeration. Next, we want to show that $$C=\{\delta<\lambda^+ : \forall\mu<\delta, \exists\alpha,\beta<\delta \text{ with }\mu<\alpha, Z\cap\alpha=X_\beta\}$$ is a club. It is easy to prove the closeness of $C$. But, I have some problem to prove that $C$ is unbounded .... could somebody you help me ?

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## 1 Answer

In the proof that $D$ is closed in Remark 1 you can’t define $\pi\upharpoonright\delta$: the bijection $\pi$ is already fixed, and therefore so is its restriction to $\delta$. What you have to show is that $\pi\upharpoonright$ maps $\delta$ onto $\lambda\times\delta$. (You already know that it’s one-to-one.) But this is easy: $$\pi\upharpoonright\delta=\bigcup_{\xi<\mu}(\pi\upharpoonright\delta_\xi)\;,$$ so $$\operatorname{ran}(\pi\upharpoonright\delta)=\bigcup_{\xi<\mu}\operatorname{ran}(\pi\upharpoonright\delta_\xi)=\bigcup_{\xi<\mu}(\lambda\times\delta_\xi)=\lambda\times\bigcup_{\xi<\mu}\delta_\xi=\lambda\times\delta\;.$$

To show that $D$ is unbounded in $\lambda^+$, let $\alpha<\lambda^+$ be arbitrary, and let $\alpha_0=\alpha+1$. Let $R_0=\operatorname{ran}(\pi\upharpoonright\alpha_0)\subseteq\lambda\times\lambda^+$. $|R_0|\le\lambda$, so there is a $\beta_0<\lambda^+$ such that $\beta_0>\alpha_0$ and $R_0\subseteq\lambda\times\beta_0$. Since $\pi$ is a bijection from $\lambda^+$ onto $\lambda\times\lambda^+$, and $|\lambda\times\beta_0|\le\lambda$, there must be an $\alpha_1<\lambda^+$ such that $\alpha_1>\beta_0$ and $\operatorname{ran}(\pi\upharpoonright\alpha_1)\supseteq\lambda\times\beta_0$.

Continue in this fashion to construct increasing sequences $\langle\alpha_n:n\in\omega\rangle$ and $\langle\beta_n:n\in\omega\rangle$ in $\lambda^+$ such that $\alpha_n<\beta_n<\alpha_{n+1}$ for each $n\in\omega$. Given $\alpha_n<\lambda^+$, let $R_n=\operatorname{ran}(\pi\upharpoonright\alpha_n)$; as before, $|R_n|\le\lambda$, so we can choose $\beta_n<\lambda^+$ so that $\beta_n>\alpha_n$ and $R_n\subseteq\lambda\times\beta_n$, and then find $\alpha_{n+1}<\lambda^+$ such that $\alpha_{n+1}>\beta_n$ and $\operatorname{ran}(\pi\upharpoonright\alpha_{n+1})\supseteq\lambda\times\beta_n$.

Now let $\delta=\sup\{\alpha_n:n\in\omega\}=\sup\{\beta_n:n\in\omega\}$. (The interlacing of the two sequences ensures that these suprema really are equal.) Then $$\pi\upharpoonright\delta=\bigcup_{n\in\omega}(\pi\upharpoonright\alpha_n)\;,$$ so $$\operatorname{ran}(\pi\upharpoonright\delta)=\bigcup_{n\in\omega}\operatorname{ran}(\pi\upharpoonright\alpha_n)\subseteq\bigcup_{n\in\omega}(\lambda\times\beta_n)=\lambda\times\bigcup_{n\in\omega}\beta_n=\lambda\times\delta\;,$$ and $$\operatorname{ran}(\pi\upharpoonright\delta)=\bigcup_{n\in\omega}\operatorname{ran}(\pi\upharpoonright\alpha_n)=\bigcup_{n\in\omega}\operatorname{ran}(\pi\upharpoonright\alpha_{n+1})\supseteq\bigcup_{n\in\omega}(\lambda\times\beta_n)=\lambda\times\bigcup_{n\in\omega}\beta_n=\lambda\times\delta\;,$$ and therefore $\operatorname{ran}(\pi\upharpoonright\delta)=\lambda\times\delta$, and $\alpha<\delta\in D$.

It doesn’t actually matter, but you haven’t quite translated Remark 2 correctly. Here it is:

Remark 2. If $\{X_\alpha:\alpha<\lambda^+\}$ is an enumeration of $[\lambda^+]^{\le\lambda},Z\subseteq\lambda^+$, then for a club set of $\delta<\lambda^+$, the following holds. There are arbitrarily large $\alpha<\delta$ such that for some $\beta<\delta$, $Z\cap\alpha=X_\beta$ holds.

Let $C=\{\delta<\lambda^+:\forall\mu<\delta\,\exists\alpha,\beta<\delta\,(\alpha>\mu\,\land Z\cap\alpha=X_\beta)\}$, as you defined it; the remark doesn’t actually say that $C$ is closed, merely that it contains a club set. However, you’re quite right that $C$ is in fact closed and that this is easy to show.

For $\alpha<\lambda^+$ let $Z_\alpha=Z\cap\alpha$. Suppose first that $|Z|\le\lambda$, so that $Z$ is bounded in $\lambda^+$; then there is some $\eta<\lambda^+$ such that $Z_\alpha=Z$ whenever $\eta\le\alpha<\lambda^+$. There is also some $\beta<\lambda^+$ such that $Z=X_\beta$, and clearly $C\supseteq\{\delta<\lambda^+:\delta>\max\{\eta,\beta\}\}$ and is unbounded in $\lambda^+$.

Now suppose that $|Z|=\lambda^+$. For $\alpha<\lambda^+$ there is a unique $\varphi(\alpha)<\lambda^+$ such that $Z_\alpha=X_{\varphi(\alpha)}$. Let $\eta_0<\lambda^+$ be arbitrary. Given $\eta_n<\lambda^+$, let $\eta_{n+1}=\sup(\{\eta_n+1\}\cup\{\varphi(\alpha)+1:\alpha<\eta_n\})$. Finally, let $\delta=\sup\{\eta_n:n\in\omega\}$. If $\mu<\delta$, there is an $n\in\omega$ such that $\mu<\eta_n<\eta_{n+1}$; let $\alpha=\eta_n$ and $\beta=\varphi(\eta_n)=\varphi(\alpha)$. Then $Z\cap\alpha=Z_\alpha=X_\beta$, $\mu<\alpha<\delta$, and $\beta=\varphi(\eta_n)<\eta_{n+2}<\delta$, so $\eta_0<\delta\in C$, and $C$ is unbounded.

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Thank you for your answer. My native langage is not english so the bad mistakes. – Marc Moretti Apr 30 '12 at 8:42