I'm reading my class notes on set theory. There is a proof of Delta system theorem, and there is something unclear about it.
Delta system lemma: Suppose $\{a_{\alpha}: \alpha<\aleph_1\}$ is a collection of finite subsets of $\aleph_1$. Then there exists an uncountable subset $I\subseteq\aleph_1$ such that $\{a_{\alpha}:\alpha\in I\}$ is a $\Delta$-system, i.e there is some set $a$ such that $a_{\alpha}\cap a_{\beta}=a$ for all $\alpha\ne\beta\in I$.
In the proof we use the following result:
Fodor's theorem: Asssume $\kappa$ is a regular uncountable cardinal, $S\subseteq\kappa$ is a stationary set and $f:S\to\kappa$ has the property $f(\alpha)<\alpha$ (which means $f$ is a regressive function) for all $\alpha\in S$. Then there is a stationary set $S^*\subseteq S$ such that $f$ is constant of $S^*$.
Proof of $\Delta$-system theorem: Let $S=\{\alpha<\aleph_1: \omega<\alpha \ is\ a\ limit \ ordinal\}$, this is a stationary subset of $\aleph_1$. We can define $f:S\to \aleph_1$ by $f(\alpha)=|a_{\alpha}|$. Since the sets $a_{\alpha}$ are finite this is a regressive function, so there are $n^*<\omega$ and a stationary subset $S^*\subseteq S$ such that $|a_{\alpha}|=n^*$ for all $\alpha\in S^*$.
Next we define $g:S^*\to \aleph_1$, $g(\alpha)=|a_{\alpha}\cap\alpha|$. Again, $g$ is regressive so there are $n^{**}<\omega$ and a stationary subset $S^{**}\subseteq S^*$ such that $|a_{\alpha}\cap\alpha|=n^{**}$ for all $\alpha\in S^{**}$. Now for each $\alpha\in S^{**}$ we will write the set $a_{\alpha}\cap\alpha$ as an increasing sequence:
$a_{\alpha}\cap\alpha=\{a_{\alpha}^0, a_{\alpha}^1,…,a_{\alpha}^{n^{**}-1}\}$
Now we keep defining regressive functions. Let $g_0: S^{**}\to\aleph_1$, $\alpha\to a_{\alpha}^0<\alpha$. There is a stationary subset $S_0\subseteq S^{**}$ and some $a^0$ such that $a_{\alpha}^0=a^0$ for all $\alpha\in S_0$. Now we define $g_1:S_0\to\aleph_1$, $\alpha\to a_{\alpha}^1<\alpha$. Again, there is a stationary subset $S_1\subseteq S_0$ and some $a^1$ such that $a_{\alpha}^1=a^1$ for all $\alpha\in S_1$. We continue this way and finally get a stationary set $S_{n^{**}-1}$ such that for all $\alpha$ in this set we have:
$a_{\alpha}\cap\alpha=\{a^0, a^1,…,a^{n^{**}-1}\}$
And finally we define the set $a=\{a^0, a^1,…,a^{n^{**}-1}\}$. It can be checked that $C=\{\delta<\aleph_1: a_\alpha\subseteq\delta \ \forall\alpha<\delta\}$ is a Club set, and hence if we define $I=C\cap S_{n^{**}-1}$ then this is a stationary set, hence uncountable. And $\{a_{\alpha}:\alpha\in I\}$ is a $\Delta$-system, because if $\alpha<\beta\in I$ then we have: ($I\subseteq C$, so $\alpha<\beta$ implies $a_{\alpha}\subseteq\beta$)
$a\subseteq a_{\alpha}\cap a_{\beta}\subseteq \beta\cap a_{\beta}=a$
And hence $a_{\alpha}\cap a_{\beta}=a$. So we got what we wanted.
My question is: do we really need the function $f$ and the natural number $n^*$ at the beginning of the proof? I just can't see where did we use it. We never used the size of the sets $a_{\alpha}$ in the proof, only the size of the sets $a_{\alpha}\cap\alpha$. Am I missing something or could we immediately start from the function $g$?
Best Answer
Alright, I'll answer the question. Yes, I found out that we don't need $f$ in this proof. Also, the one thing I didn't explain in the question is why $C$ is a Club set. It follows from the fact that $C$ is the diagonal intersection of the sets $C_{\alpha}=\{\delta<\aleph_1: a_{\alpha}\subseteq\delta\}$ which are all obviously Club sets. This is not the shortest proof the $\Delta$-System lemma, but I think it's still a nice proof.