# Closed subset of stationary set and AC questions

Hope I'm not spamming too much by asking questions on separate threads.

I have 2 more questions, not connected one to another, in any way:
1. Show that every stationary set in $\aleph_1$, contains, for every $\alpha<\omega_1$, a closed set of otp $\alpha$.
2. Show without using AC, that there exist a surjective function from $\mathbb{R}$ on $\aleph_1$.

The second one seems pretty obvious, so I'm not sure what I need to prove.

Thanks.

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Do you mean surjective, in question 2? –  Andres Caicedo Feb 8 '11 at 20:49
Oh, yes, fixed the question. –  Pavel Feb 8 '11 at 21:05

For question 2, one way is as follows: We can identify the set ${\mathbb R}\setminus{\mathbb Q}$ of irrationals with the set ${}^\omega\omega$ of functions from $\omega$ to $\omega$, for example, by looking at continued fraction representations. (There are other ways, just playing with the decimal expansions, for example.)

Now, each natural can be seen as coding two, say $a$ codes $(b,c)$ iff $a+1=2^b(2c+1)$. (Again, there are other ways.)

The point is that this gives us a way to associate to each irrational $r$ a binary relation on $\omega$, namely we have associated $r$ with the set $\{(b_0,c_0),(b_1,c_1),\dots\}$. Here, first $r$ was associated with a sequence $(a_0,a_1,\dots)$, and then each $a_i$ was identified with a pair $(b_i,c_i)$.

Ok. If the relation associated to $r$ is a well-ordering of $\omega$ of type $\omega+\alpha$, then map $r$ to $\alpha$. Else, map $r$ to 0. This is a surjection of ${\mathbb R}$ onto $\aleph_1$. We did not use choice. That it is surjective comes from noting that for any $\alpha<\omega_1$ there is a well-ordering of $\omega$ in type $\omega+\alpha$, and the codings described above are reversible, so any such well-ordering is the image of some $r$ under the coding we described.

The set of $r$ that code well-orderings is usually called WO and plays an important role in descriptive set theory; the construction I described is "canonical" in some sense. On the other hand, without choice we cannot prove the dual result that there is an injection from $\omega_1$ into ${\mathbb R}$.

For question 1, one argues by induction that a stationary set contains closed copies of $\alpha+1$ for any $\alpha<\omega_1$. The following is from a set of notes I wrote for a course I taught a while ago:

Let $S\subseteq\omega_1$ be a given stationary set, and argue by induction on $\alpha<\omega_1$ that $S$ contains closed copies $t$ of $\alpha+1$ with $\min(t)$ arbitrarily large. (Of course, if the result holds, this must be the case: Given any $\gamma<\omega+1$, notice that $S\setminus(\gamma+1)$ is stationary, so it must contain a closed copy $t$ of $\alpha+1$, and $\min(t)<\gamma$.)

This strengthened version holds trivially for $\alpha$ finite or successor, by induction. So it suffices to show it for $\alpha$ limit, assuming it holds for all smaller ordinals. Define a club $C\subseteq\omega_1$ with increasing enumeration $\{\gamma_\beta:\beta<\omega_1\}$ as follows:

Let $(\alpha_n:n<\omega)$ be strictly increasing and cofinal in $\alpha$. Since $S$ contains closed copies $A_n$ of $\alpha_n+1$ for all $n$, with their minima arbitrarily large, by choosing such copies $A_n$ with $\min(A_{n+1})>\max(A_n)$ and taking their union, we see that $S$ must contain copies of $\alpha$, closed in their supremum, with arbitrarily large minimum element. (I am not claiming that $A=\bigcup_n A_n$ built this way has order type $\alpha$. For example, if $\alpha=\omega+\omega$ and $\alpha_n=\omega+n$, then $A$ would have order type $\omega^2$; but for sure $A\subset S$ is closed in its supremum and has order type at least $\alpha$. So a suitable initial segment of $A$ is as wanted.)

Let $\gamma_0$ be the supremum of such a copy of $\alpha$. At limit ordinals $\beta$, let $\gamma_\beta=\sup_{\delta<\beta}\gamma_\delta$. Once $\gamma_\beta$ is defined, find such a copy of $\alpha$ inside $S$ with minimum larger than $\gamma_\beta$, and let $\gamma_{\beta+1}$ be its supremum.

The set $C$ so constructed is club, so it meets $S$. If they meet in $\gamma_0$ or in a $\gamma_{\beta+1}$, this immediately gives us a closed copy of $\alpha+1$ inside $S$. If they meet in a $\gamma_\beta$ with $\beta$ limit, let $(\beta_n:n<\omega)$ be strictly increasing and cofinal in $\beta$, and consider an appropriate initial segment of $A=(\bigcup_n A_n)\cup\{\gamma_\beta\}$, where $A_n$ is a closed copy of $\alpha_n+1$ in $S\cap[\gamma_{\beta_n},\gamma_{\beta_{n+1}})$.

Let me add that if one is familiar with the method of forcing, there is a nice argument for question 1. (This is perhaps overkill, but a nice one in any case.)

Namely, given any stationary subset $S$ of $\omega_1$, there is a forcing poset that adds a club subset of $S$ while not adding any new countable sequences of countable ordinals. This means that the proper initial segments of the club are subsets of $S$ that are in $V$, and of course are closed in their supremum, so this immediately gives the result.

The poset is the natural one: Conditions are closed initial segments of $S$, and the order is end-extension. It takes a bit of work to see that this poset indeed does the work; this is usually presented in the context of proper forcing, since this is an example of an $S$-proper poset. This is an argument that goes back to the mid-seventies and is due to Baumgartner-Harrington-Kleinberg. It is specific to $\omega_1$, there are no natural generalizations for stationary subsets of larger cardinals.

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