1

Let \(\mathbf{M}\) be a ctm of ZFC and \(f:\omega\to2\). Suppose \(f\) is a Cohen real over \(\mathbf{M}\), \(g:\omega\to2\) and \(g\in\mathbf{M}\). Prove that \(f+g\) is a Cohen real over \(\mathbf{M}\).

Let \(f\) be a Cohen real over \(\mathbf{M}\) and fix \(g:\omega\to2\) in \(\mathbf{M}\). For any \(p\in\operatorname{Fn}(\omega,2)\) we denote \[p' = g\upharpoonright(\operatorname{dom}(p)) + p\] which is the pointwise xor of \(p\) with \(g\) at all points in which \(p\) is defined. Then naturally for any set \(D\subseteq \operatorname{Fn}(\omega,2)\) we denote \[D' = \left\{p': p\in D\right\}\] and since xor’ing twice does nothing we see that \(p'' = p\) and as a result \(D'' = D\).

Observation. Whenever \(D\subseteq {\mathbb{P}}\) is dense, \(D'\) is dense too. Let \(p\in\operatorname{Fn}(\omega,2)\), by density of \(D\) let \(q\in D\) such that \(q\leq p'\), then \(q' \leq p'' = p\) and \(q' \in D'\).

To see that \(f+g\) is a Cohen real, let \(D\in \mathbf{M}\) be a dense subset of \({\mathbb{P}}\), we want to show that \(D\) meets \(G_{f+g}\). It is shown that \(D'\) is dense, and by replacement \(D'\in\mathbf{M}\) too, so as \(G_f\) is generic over \(\mathbf{M}\), \(G_f\) meets \(D'\). Let \(p\in G_f \cap D'\), then \(p' \in D'' = D\), and it follows from the definition of \(p'\) and fact that \(p\in G_f\) that \(p' \in G_{f+g}\).

2

Let \(\mathbf{M}\) be a ctm of ZFC and suppose \(f:\omega\to2\) is a Cohen real over \(\mathbf{M}\). Suppose \(Y\in\mathbf{M}\) is an infinite subset of \(\omega\). Show that both \(Y\cap X_f\) and \(Y\setminus X_f\) are infinite.

Suppose \(Y\cap X_f\) is finite, we show that \(f\) is not a Cohen real over \(\mathbf{M}\). The case where \(Y\setminus X_f\) is finite is similar.

Suppose \(Y\cap X_f\) is finite then \(Y\) (and hence \(\mathbf{M}\)) knows about infinitely many \(n\) where \(f(n) = 0\). Let \(\left\lvert Y \cap X_f\right\rvert = l\), use this rather large formula to define the following dense set \[ D = \left\{p: \left(\exists n_0,n_1,\dots,n_l<\omega\right)\left(\overline{n}\in\operatorname{dom}(p) \land \bigwedge_{i\ne j} n_i\ne n_j \land \bigwedge_{i=0}^l p(n_i) = 1 \iff n_i\in Y\right)\right\}. \] The density of \(D\) follows from \(Y\) being infinite, and \(D\in\mathbf{M}\) too. Now \(G_f\) cannot meet \(D\) as \(f\) is \(1\) in \(Y\) at at most \(l\) points.

3

Assume \(\textsf{MA}_{\omega_1}\). Let \(({\mathbb{P}},\leq_{\mathbb{P}}, 1_{\mathbb{P}})\) be a c.c.c. poset and let \(\kappa\) be an infinite cardinal. Define \({\mathbb{Q}}\) to be the set of all \(f: \kappa\to{\mathbb{P}}\) such that \(\left\{\alpha<\kappa: f(\alpha) \ne 1_{\mathbb{P}}\right\}\) is finite. For \(f,g\in{\mathbb{Q}}\), define \(f\leq_{\mathbb{Q}}g\) iff \(\left(\forall \alpha<\kappa\right)\left(f(\alpha)\leq_{\mathbb{P}}g(\alpha)\right)\). Show that \({\mathbb{Q}}\) satisfies c.c.c..

Let \(A\subseteq{\mathbb{Q}}\) be an uncountable antichain. For each \(f\in A\) we consider its support all \(\alpha<\kappa\) such that \(f(\alpha) \ne 1_{\mathbb{P}}\). \[ \left\{\left\{\alpha<\kappa: f(\alpha)\ne 1_{\mathbb{P}}\right\}: f\in A\right\} \] If the family of all supports is countable then uncountably many points in \(A\) have the same (finite) support, call it \(r\), we restrict \(A\) to these points and get a trivial sunflower with no petals.

In the other case we can restrict \(A\) such that the family of supports forms a \(\Delta\)-system with some root \(r\). The root \(r\) cannot be empty, or else some \(f,g\in A\) would have a common extension (that being their pointwise minimum). Then we can project \(A\) faithfully into the poset \({}^r{\mathbb{P}}\) equipped with the same pointwise ordering. Note that \({}^r{\mathbb{P}}\) is the finite product of \({\mathbb{P}}\) with itself \(\left\lvert r\right\rvert\) many times, and by \(\textsf{MA}_{\omega_1}\) and Theorem 22.9 already has ccc, but the projection of \(A\) is an uncountable antichain in \({}^r{\mathbb{P}}\).

4

Assume \(\textsf{MA}_{\omega_1}\). Let \({\mathbb{P}}\) be a c.c.c. poset and suppose \(A\in [{\mathbb{P}}]^{\omega_1}\). Show that there exists \(\left\langle G_n:n<\omega\right\rangle\) such that each \(G_n\) is a filter on \({\mathbb{P}}\) and \(A\subseteq \bigcup_{n<\omega} G_n\).

Following hint, let \({\mathbb{Q}}\) denote the finite support \(\omega\)-product of \({\mathbb{P}}\). By past question \({\mathbb{Q}}\) is ccc. For each \(p\in A\), let \[D_p = \left\{f: \left(\exists n<\omega\right)\left(f(n) = p\right)\right\}\] as the support of every sequence is finite \(D_p\) is dense in \({\mathbb{Q}}\). Invoke \(\textsf{MA}_{\omega_1}\) to get a filter \(G\) on \({\mathbb{Q}}\) such that \(G\) meets \(D_p\) for every \(p\in A\). By meeting every single \(D_p\), we can cover \(A\) using \(\bigcup_{g\in G} \operatorname{range}(g)\). Now define for each \(n<\omega\) \[ G_n = \left\{g(n): g\in G\right\} \] and \(\bigcup_{n<\omega} G_n = \bigcup_{g\in G} \operatorname{range}(g) \supseteq A\) as desired. As the conditions required to be a filter distributes pointwise, each \(G_n\) is a filter.

5

Let \({\mathbb{P}}\) be a countably closed poset in which every condition has two incompatible extensions. Show that \({\mathbb{P}}\) is not ccc.

Using the same construction as two previous homeworks we can embed the set of all binary strings \(2^*\) in \({\mathbb{P}}\) such that for all \(s\in 2^*\), \(p_{s0} \perp p_{s1}\) and both \(p_{s0}, p_{s1} \leq p_s\). For every \(x\in 2^\omega\) we look at the sequence \(\left\langle p_{x\upharpoonright n}: n<\omega\right\rangle\) and as \({\mathbb{P}}\) is countably closed there is a \(p_x\) below the entire sequence. Now for \(x,y\in 2^\omega\), if \(x\ne y\) then \(p_x\perp p_y\), consider the least \(n<\omega\) where \(x(n) \ne y(n)\), if \(p_x\) and \(p_y\) were to have a common extension, then \(p_{s\,x(n)}\) and \(p_{s\,y(n)}\) where \(s = x\upharpoonright n = y\upharpoonright n\) would have a common extension too, which cannot happen.