Lectures 2024-03-08

iii/forcing/03_forcing.tex

@@ -89,14 +89,14 @@ \subsection{Forcing posets}
     and if \( \mathbb P \) has a maximal element, so does \( \faktor{\mathbb P}{\sim} \).
 \end{proposition}
 \begin{example}
-    For sets \( I, J \), we let \( \operatorname{Fn}(I, J) \) denote the set of all finite partial functions from \( I \) to \( J \).
-    \[ \operatorname{Fn}(I, J) = \qty{p \mid \abs{p} < \omega \wedge p \text{ is a function} \wedge \dom p \subseteq I \wedge \ran p \subseteq J} \]
-    We let \( \leq \) be the reverse inclusion on \( \operatorname{Fn}(I, J) \), so \( q \leq p \) if and only if \( q \supseteq p \).
+    For sets \( I, J \), we let \( \Fn(I, J) \) denote the set of all finite partial functions from \( I \) to \( J \).
+    \[ \Fn(I, J) = \qty{p \mid \abs{p} < \omega \wedge p \text{ is a function} \wedge \dom p \subseteq I \wedge \ran p \subseteq J} \]
+    We let \( \leq \) be the reverse inclusion on \( \Fn(I, J) \), so \( q \leq p \) if and only if \( q \supseteq p \).
     The maximal element \( \Bbbone \) is the empty set.
-    Then \( (\operatorname{Fn}(I, J), \leq, \varnothing) \) is a forcing poset, and moreover, the preorder is separative.
+    Then \( (\Fn(I, J), \leq, \varnothing) \) is a forcing poset, and moreover, the preorder is separative.
 \end{example}
 \begin{remark}
-    When \( \alpha \) is an ordinal, the forcing poset \( \operatorname{Fn}(\alpha \times \omega, 2) \) is often written \( \operatorname{Add}(\omega, \alpha) \), denoting the idea that we are adding \( \alpha \)-many subsets of \( \omega \).
+    When \( \alpha \) is an ordinal, the forcing poset \( \Fn(\alpha \times \omega, 2) \) is often written \( \operatorname{Add}(\omega, \alpha) \), denoting the idea that we are adding \( \alpha \)-many subsets of \( \omega \).
 \end{remark}
 
 \subsection{Chains and \texorpdfstring{\( \Delta \)}{Δ}-systems}
@@ -111,9 +111,9 @@ \subsection{Chains and \texorpdfstring{\( \Delta \)}{Δ}-systems}
 \end{definition}
 \begin{example}
     \begin{enumerate}
-        \item Consider the tree \( \operatorname{Fn}(\omega, 2) \).
+        \item Consider the tree \( \Fn(\omega, 2) \).
         A chain is a branch through the tree, and an antichain is a collection of points on different branches.
-        \item The set of functions \( \qty{\qty{\langle 0, 0 \rangle, \langle 1, n \rangle} \mid n \in \omega} \) forms an antichain of length \( \omega \) in \( \operatorname{Fn}(I, \omega) \) if \( \qty{0,1} \subseteq I \).
+        \item The set of functions \( \qty{\qty{\langle 0, 0 \rangle, \langle 1, n \rangle} \mid n \in \omega} \) forms an antichain of length \( \omega \) in \( \Fn(I, \omega) \) if \( \qty{0,1} \subseteq I \).
     \end{enumerate}
 \end{example}
 \begin{definition}
@@ -169,18 +169,18 @@ \subsection{Chains and \texorpdfstring{\( \Delta \)}{Δ}-systems}
 \end{proposition}
 \begin{lemma}
     (\( \mathsf{ZFC} \))
-    \( \operatorname{Fn}(I, J) \) has the countable chain condition if and only if \( I \) is empty or \( J \) is countable.
+    \( \Fn(I, J) \) has the countable chain condition if and only if \( I \) is empty or \( J \) is countable.
 \end{lemma}
 \begin{proof}
-    First, we observe that if \( I \) or \( J \) are empty, then \( \operatorname{Fn}(I, J) \) is empty and so trivially has the countable chain condition.
+    First, we observe that if \( I \) or \( J \) are empty, then \( \Fn(I, J) \) is empty and so trivially has the countable chain condition.
     Now let us assume that both \( I \) and \( J \) are nonempty.
 
     Suppose that \( J \) is uncountable.
     Then for any \( i \in I \), the set
     \[ \qty{\qty{\langle i, j \rangle} \mid j \in J} \]
     is an uncountable antichain.
 
-    Now suppose \( J \) is countable, and let \( \qty{p_\alpha \mid \alpha \in \omega_1} \) be a collection of distinct elements of \( \operatorname{Fn}(I, J) \).
+    Now suppose \( J \) is countable, and let \( \qty{p_\alpha \mid \alpha \in \omega_1} \) be a collection of distinct elements of \( \Fn(I, J) \).
     Let \( \mathcal A = \qty{\dom p_\alpha \mid \alpha \in \omega_1} \), which is a collection of \( \omega_1 \)-many finite sets.
     By the \( \Delta \)-system lemma, we can find an uncountable subset \( \mathcal B \subseteq \mathcal A \) with a root \( R \subseteq I \).
     By definition, \( R \subseteq \dom(p_\alpha) \) for all \( \dom p_\alpha \in \mathcal B \), the root \( R \) must be finite.
@@ -203,8 +203,8 @@ \subsection{Dense sets and genericity}
     Let \( I \) be infinite and \( J \) nonempty.
     Then for all \( i \in I \) and \( j \in J \), the following are dense.
     \begin{enumerate}
-        \item \( D_i = \qty{q \in \operatorname{Fn}(I, J) \mid i \in \dom q} \);
-        \item \( R_j = \qty{q \in \operatorname{Fn}(I, J) \mid j \in \ran q} \).
+        \item \( D_i = \qty{q \in \Fn(I, J) \mid i \in \dom q} \);
+        \item \( R_j = \qty{q \in \Fn(I, J) \mid j \in \ran q} \).
     \end{enumerate}
 \end{example}
 \begin{definition}
@@ -248,7 +248,7 @@ \subsection{Dense sets and genericity}
     But all conditions in \( F \) are compatible, so one of \( r \) and \( q \) is not in \( F \).
 \end{proof}
 \begin{proposition}
-    For sets \( I, J \) such that \( \abs{I} \geq \omega \) and \( \abs{J} \geq 2 \), the forcing poset \( \operatorname{Fn}(I, J) \) is a separative partial order without a minimal element.
+    For sets \( I, J \) such that \( \abs{I} \geq \omega \) and \( \abs{J} \geq 2 \), the forcing poset \( \Fn(I, J) \) is a separative partial order without a minimal element.
 \end{proposition}
 \begin{proposition}
     (\( \mathsf{ZFC} \))