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Let $f: \omega_1\to \mathbb{R}$ be one-one. Let $g:[\omega_1]^2\to 2$ be such that for any $\alpha<\beta<\omega_1$, $g(\{\alpha, \beta\})$ is $0$ when $f(\alpha)<f(\beta)$, and $1$ otherwise. $g$ yields the following partial orders on finite subsets of $\omega_1$.

$\mathbb{P_\mu} = \{p\in [\omega_1]^{<\omega}: \forall \alpha, \beta\in p(\alpha\not = \beta \to g(\{\alpha, \beta\}) = \mu)\}$

The exercise is to show that neither $\mathbb{P_0}$ nor $\mathbb{P_1}$ are ccc. The order is $p\leq q$ iff $p\supseteq q$.

I know I'm missing something very simple, but I've spent an unhealthy amount of time on this now, and I need to move on!

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  • $\begingroup$ You provided us with the set $\Bbb P_{\mu}$. But what is the partial order on this set? $\endgroup$ Apr 29, 2016 at 10:30
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    $\begingroup$ Also: How much time you spend on a given project is pretty irrelevant to us . If you need to move on, move on. If you want to solve this problem, this is the right place to look for assistance. $\endgroup$ Apr 29, 2016 at 10:32
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    $\begingroup$ @Stefan: Surely the order must be $\subseteq$. $\endgroup$ Apr 29, 2016 at 10:35
  • $\begingroup$ @Stefan Thanks, I've edited to include the order, which is the one Brian thought. $\endgroup$
    – user104955
    Apr 29, 2016 at 10:41

2 Answers 2

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Here's a forcing argument (not how Kunen wants you to do it since forcing is covered in chapter 4): First, by throwing away all the rational intervals in which $f[\omega_1]$ is countable, we can choose $X \subseteq \omega_1$ such that $\omega_1 \setminus X$ is countable and for every $x \in f[X]$, $|[x, \infty) \cap f[X]| = \omega_1$. Let $Q_0 = \{p \in P_0 : p \subseteq X\}$. Note that for any two conditions $p, q \in Q_0$, $p, q$ are compatible in $Q_0$ iff they are compatible in $P_0$ (as witnessed by their union). Hence it is enough to show that $Q_0$ is not ccc. But this is clear since $Q_0$ adds an increasing $\omega_1$-sequence of reals and hence an injection from $\omega_1$ to rationals so that $\omega_1$ is collapsed. The argument for $P_1$ is identical.

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  • $\begingroup$ Could you outline how $Q_0$ -forcing adds an increasing $\omega_1$-sequence of reals? It's probably obvious but I am very rusty. $\endgroup$ May 2, 2016 at 2:24
  • $\begingroup$ First check that the sets $D_{\alpha} = \{p \in Q_0: (\exists \beta > \alpha)(\beta \in p)\}$ are dense in $Q_0$ for each $\alpha < \omega_1$ - This is why we passed to $Q_0$ from $P_0$. So the union $Y$ of the conditions in any $Q_0$-generic filter over $V$ is uncountable. It follows that the sequence $\langle f(\alpha) : \alpha \in Y \rangle$ is an increasing $\omega_1$-sequence of reals. $\endgroup$
    – hot_queen
    May 2, 2016 at 2:45
  • $\begingroup$ Thank you for that..............+1 $\endgroup$ May 2, 2016 at 3:40
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    $\begingroup$ Nice. @GME: I told you there was a nice forcing answer by showing $\omega_1$ is collapsed. :-) $\endgroup$
    – Asaf Karagila
    May 6, 2016 at 23:30
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This one's tricky - I was convinced there was an error for a minute!

First, let me sketch why I thought it was c.c.c. It's a $\Delta$-system argument. Let's look at $\mathbb{P}_0$. Suppose there were an uncountable antichain $A=\{p_\eta: \eta\in\omega_1\}$. Then by the $\Delta$-system lemma, we can assume WLOG that $A$ is a $\Delta$-system with root $r$, and (by pigeonhole) $p_\eta\setminus r$ has size $n$ for some fixed $n$ (regardless of $\eta$).

Now let's look at the special case where $\vert r\vert=1$ and $n=1$; that is, there is some $\alpha\in\omega_1$ such that each $p_\eta$ has the form $\{\alpha, \beta_\eta\}$, and $\beta_\eta=\beta_\delta\iff\eta=\delta$. Again WLOG by pigeonhole, we may assume $\alpha<\beta_\eta$ and $\beta_\eta<\beta_\delta$ for each $\eta<\delta<\omega_1$.

So what? Well, since each $p_\eta$ is a condition in $\mathbb{P}_0$, we must have $f(\alpha)<f(\beta_\eta)$ for each $\eta$. Since the $p_\eta$s form an antichain, we must also have $f(\beta_\eta)>f(\beta_\delta$ for $\eta<\delta$ (since otherwise $\{\alpha, \beta_\eta, \beta_\delta\}$ would be a condition extending both $p_\eta$ and $p_\delta$).

But this means that $\{f(\beta_\eta): \eta\in\omega_1\}$ is a strictly decreasing sequence of reals of order-type $\omega_1$, which can't happen.

Of course, I've only considered a very special case here; but it looks plausible that this should work in general. So, it's a reasonable guess that this is c.c.c.


But that's wrong - the combinatorics changes completely once we have length-two petals! I'll describe a specific $f$ with an easily-describable uncountable antichain; it's not hard to show that this happens in general.

Here's what I want from my $f$:

  • For each limit $\lambda>0$, $f(\lambda)=-f(\lambda+1)$, both are in $(-1, 1)$, and $f(\lambda)<0$.

  • $f(0)=-7$.

Now let $$p_\eta=\{0, \eta, \eta+1\}$$ for $0<\eta<\omega_1$ a limit ordinal. It's easy to check that $p_\eta\in \mathbb{P}_0$, but the set $\{p_\eta: \eta<\omega_1\mbox{ is a limit }\}$ is an antichain! Suppose $\eta<\delta$ are limit ordinals. Then we have two possibilities:

  • Case 1: $f(\eta)<f(\delta)<f(\delta+1)<f(\eta+1)$. In this case, we have $g(\delta+1, \eta+1)=1$, so $p_\eta\perp p_\delta$.

  • Case 2: $f(\delta)<f(\eta)<f(\eta+1)<f(\delta+1)$. In this case, we have $g(\delta, \eta)=1$, so $p_\eta\perp p_\delta$.

For general $f$, the key lemma you need to prove is:

If $A\subseteq\mathbb{R}$ is uncountable, then there is a family $\{P_\eta: \eta<\omega_1\}$ of disjoint pairs of elements of $A$ such that - if $P_\eta=\{r_\eta<s_\eta\}$ - we have $r_\eta<r_\delta\iff s_\delta<s_\eta$ for all $\eta,\delta<\omega_1$ (basically, we have an uncountable nested family of pairs).

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  • $\begingroup$ Thanks, Noah! It's nice to see what you tried, but also nice to know I'm not a total dunce for not knowing how to do it! $\endgroup$
    – user104955
    May 2, 2016 at 19:59

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