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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

1 Answer 1

up vote 4 down vote accepted

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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