Complete exercises

Signed-off-by: zeramorphic <[email protected]>

iii/mtncl/01_substructures.tex

@@ -25,10 +25,22 @@ \subsection{Homomorphisms and substructures}
         \item Let \( \mathcal L \) be the language of groups.
         Then \( (\mathbb N, +, 0) \) is a substructure of \( (\mathbb Z, +, 0) \), but it is not a subgroup.
         \item If \( \mathcal M \) is an \( \mathcal L \)-structure, \( X \) is the domain of a substructure of \( \mathcal M \) if and only if it is closed under the interpretations of all function symbols.
-        In particular, it should also contain all of the constants in the language. % (exercise)
+        The forward implication is clear.
+        If \( f \) is a function symbol of arity \( n \) and \( X \) is closed under \( f^{\mathcal M} \), \( \eval{f^{\mathcal M}}_{X^n} \) is a function \( X^n \to X \) interpreting \( f \) on the domain \( X \), as required.
+        In particular, any substructure should also contain all of the constants in the language.
         \item The substructure \emph{generated by} a subset \( X \subseteq \mathcal M \) is given by the smallest set that contains \( X \) and is closed under the interpretations of all function symbols in \( \mathcal M \).
-        This is denoted \( \langle X \rangle_{\mathcal M} \), and one can check that for infinite \( \mathcal L \) (but not necessarily infinite signature), % (exercise)
+        This is denoted \( \langle X \rangle_{\mathcal M} \), and one can check that for infinite \( \mathcal L \) (but not necessarily infinite signature),
         \[ \abs{\langle X \rangle_{\mathcal M}} \leq \abs{X} + \abs{\mathcal L} \]
+        We prove this by iteratively closing up \( X \) by applying interpretations of function symbols to elements of \( X \), and then taking the union of the resulting sets.
+        At each stage, for each function symbol \( f \) of arity \( n \), we add at most \( \abs{X}^n \leq \abs{X} \cdot \aleph_0 \) new elements.
+        So in a single stage, we add at most \( \abs{X} \cdot \aleph_0 \cdot \abs{\mathcal L} = \abs{X} \cdot \abs{\mathcal L} \) new elements to \( X \).
+        Repeating this \( \omega \) times, the final set has size at most
+        \begin{align*}
+            \abs{X} + \abs{X} \cdot \abs{\mathcal L} + \abs{X} \cdot \abs{\mathcal L}^2 + \cdots &= \abs{X} (1 + \abs{\mathcal L} + \abs{\mathcal L}^2 + \cdots) \\
+            &\leq \abs{X} (\abs{\mathcal L} + \abs{\mathcal L} + \abs{\mathcal L} + \cdots) \\
+            &= \abs{X} \cdot \abs{\mathcal L} \cdot \aleph_0 \\
+            &= \abs{X} \cdot \abs{\mathcal L}
+        \end{align*}
         We say that \( \mathcal M \) is \emph{finitely generated} if there exists a finite subset \( X \subseteq \mathcal M \) such that \( \mathcal M = \langle X \rangle_{\mathcal M} \).
         \item Consider
         \[ (\mathbb R, \cdot, -1) \vDash \neg \exists x.\, (x^2 = -1) \]