[proofplan] We prove each closure property by bounding the relevant [transitive closure](/theorems/1493) by a finite union of sets already known to have cardinality below $\theta$. Since $\theta$ is infinite, finite unions of sets of cardinality below $\theta$ again have cardinality below $\theta$; we prove this elementary cardinal estimate explicitly. For function evaluation, we use the Kuratowski coding $(a,b) := \{\{a\},\{a,b\}\}$ and observe that the value $b = f(a)$ lies inside the transitive closure of an ordered pair belonging to $f$, hence inside the hereditary content of $f$. [/proofplan] [step:Record the finite union cardinal estimate] We first record the elementary estimate used throughout. In this proof, $\operatorname{tc}(s)$ denotes the transitive closure of a set $s$ excluding $s$ itself, that is, the set of all objects reachable from $s$ by one or more membership steps. [claim:Finite unions below an infinite cardinal remain below it] Let $\theta$ be an infinite cardinal. If $A_1,\dots,A_n$ are sets with $|A_i| < \theta$ for each $i \in \{1,\dots,n\}$, then \begin{align*} \left|\bigcup_{i=1}^n A_i\right| < \theta. \end{align*} [/claim] [proof] For each $i \in \{1,\dots,n\}$, let $\kappa_i := |A_i|$. Since $n$ is finite, the set $\{\kappa_1,\dots,\kappa_n\}$ has a maximum cardinal; denote it by $\kappa$. Then $\kappa < \theta$ and $|A_i| \leq \kappa$ for every $i$. The union admits a surjection \begin{align*} \{1,\dots,n\} \times \kappa &\longrightarrow \bigcup_{i=1}^n A_i \end{align*} after choosing, for each $i$, a surjection from $\kappa$ onto $A_i$ when $A_i$ is nonempty, and ignoring empty fibres. Hence \begin{align*} \left|\bigcup_{i=1}^n A_i\right| \leq n \cdot \kappa. \end{align*} If $\kappa$ is finite, then $n \cdot \kappa$ is finite, hence $n \cdot \kappa < \theta$ because $\theta$ is infinite. If $\kappa$ is infinite, then $n \cdot \kappa = \kappa < \theta$. Therefore \begin{align*} \left|\bigcup_{i=1}^n A_i\right| < \theta. \end{align*} [/proof] [/step] [step:Bound the transitive closure of the unordered pair] Let $x,y \in H_\theta$. By definition of $H_\theta$, \begin{align*} |\operatorname{tc}(x)| < \theta \quad\text{and}\quad |\operatorname{tc}(y)| < \theta. \end{align*} Define \begin{align*} A := \operatorname{tc}(x) \cup \operatorname{tc}(y) \cup \{x,y\}. \end{align*} The set $A$ is transitive enough to contain every element obtained by iterating membership downward from $\{x,y\}$: the only elements of $\{x,y\}$ are $x$ and $y$, and the elements below $x$ and $y$ lie in $\operatorname{tc}(x)$ and $\operatorname{tc}(y)$ respectively. Hence \begin{align*} \operatorname{tc}(\{x,y\}) \subseteq A. \end{align*} By the finite union estimate, \begin{align*} |A| < \theta. \end{align*} Therefore \begin{align*} |\operatorname{tc}(\{x,y\})| \leq |A| < \theta, \end{align*} so $\{x,y\} \in H_\theta$. [guided] We want to show that $\{x,y\}$ is hereditarily of size below $\theta$. The only new objects introduced by forming the unordered pair are the two top-level elements $x$ and $y$ themselves; everything below them was already present in their transitive closures. Since $x,y \in H_\theta$, the definition gives \begin{align*} |\operatorname{tc}(x)| < \theta \quad\text{and}\quad |\operatorname{tc}(y)| < \theta. \end{align*} We define the controlling set \begin{align*} A := \operatorname{tc}(x) \cup \operatorname{tc}(y) \cup \{x,y\}. \end{align*} Every membership chain starting from $\{x,y\}$ first reaches either $x$ or $y$. Once it reaches $x$, all further members are contained in $\operatorname{tc}(x)$; once it reaches $y$, all further members are contained in $\operatorname{tc}(y)$. Thus \begin{align*} \operatorname{tc}(\{x,y\}) \subseteq A. \end{align*} The set $A$ is a finite union of two sets of cardinality below $\theta$ and one finite set. Since $\theta$ is infinite, the finite union estimate gives \begin{align*} |A| < \theta. \end{align*} Therefore \begin{align*} |\operatorname{tc}(\{x,y\})| \leq |A| < \theta. \end{align*} This is exactly the condition $\{x,y\} \in H_\theta$. [/guided] [/step] [step:Bound the transitive closure of the union] Again let \begin{align*} A := \operatorname{tc}(x) \cup \operatorname{tc}(y) \cup \{x,y\}. \end{align*} If $z \in x \cup y$, then $z \in x$ or $z \in y$. Hence every element below $z$ lies in $\operatorname{tc}(x)$ or in $\operatorname{tc}(y)$. Therefore \begin{align*} \operatorname{tc}(x \cup y) \subseteq \operatorname{tc}(x) \cup \operatorname{tc}(y) \subseteq A. \end{align*} The finite union estimate applied to $\operatorname{tc}(x)$, $\operatorname{tc}(y)$, and $\{x,y\}$ gives $|A| < \theta$, so \begin{align*} |\operatorname{tc}(x \cup y)| < \theta. \end{align*} Thus $x \cup y \in H_\theta$. [guided] The union $x \cup y$ does not create any new hereditary material. Its elements are precisely the objects that already occur as elements of $x$ or as elements of $y$. Let \begin{align*} A := \operatorname{tc}(x) \cup \operatorname{tc}(y) \cup \{x,y\}. \end{align*} If $z \in x \cup y$, then either $z \in x$ or $z \in y$. In the first case, $z$ and all sets obtained by iterating membership below $z$ are contained in $\operatorname{tc}(x)$. In the second case, they are contained in $\operatorname{tc}(y)$. Hence \begin{align*} \operatorname{tc}(x \cup y) \subseteq \operatorname{tc}(x) \cup \operatorname{tc}(y). \end{align*} The right-hand side is contained in $A$. The set $A$ is the finite union of $\operatorname{tc}(x)$, $\operatorname{tc}(y)$, and $\{x,y\}$. The first two sets have cardinality below $\theta$ because $x,y \in H_\theta$, and the last set is finite, hence has cardinality below the infinite cardinal $\theta$. Applying the finite union estimate gives \begin{align*} |A| < \theta. \end{align*} Therefore \begin{align*} |\operatorname{tc}(x \cup y)| \leq |A| < \theta, \end{align*} which proves $x \cup y \in H_\theta$. [/guided] [/step] [step:Extract the value of a function from the hereditary content of its graph] Let $S$ and $T$ be sets, let $f: S \to T$ be a function with $f \in H_\theta$, and let $a \in S$. Define $b := f(a)$. Since $f: S \to T$ is a function and $a \in S = \operatorname{dom}(f)$, the ordered pair $(a,b)$ belongs to $f$. We use the Kuratowski ordered pair \begin{align*} (a,b) := \{\{a\},\{a,b\}\}. \end{align*} Because $(a,b) \in f$, every object hereditarily below $(a,b)$ is hereditarily below $f$. In particular, \begin{align*} b \in \{a,b\} \in (a,b) \in f, \end{align*} so \begin{align*} \operatorname{tc}(b) \subseteq \operatorname{tc}(f). \end{align*} Since $f \in H_\theta$, we have $|\operatorname{tc}(f)| < \theta$. Therefore \begin{align*} |\operatorname{tc}(b)| \leq |\operatorname{tc}(f)| < \theta. \end{align*} Thus $b \in H_\theta$. Since $b = f(a)$, this proves $f(a) \in H_\theta$. [guided] The graph of a function is a set of ordered pairs. We must show that once the whole graph $f$ is hereditarily below $\theta$, each value of the function is also hereditarily below $\theta$. Let $S$ and $T$ be sets, let $f: S \to T$ be a function with $f \in H_\theta$, and let $a \in S$. Define \begin{align*} b := f(a). \end{align*} Because $f: S \to T$ is a function, $S = \operatorname{dom}(f)$ and there exists a unique value $b \in T$ with $(a,b) \in f$; this value is $f(a)$. We use the Kuratowski coding \begin{align*} (a,b) := \{\{a\},\{a,b\}\}. \end{align*} This coding places $b$ two membership steps below the ordered pair: \begin{align*} b \in \{a,b\} \in (a,b). \end{align*} Since $(a,b) \in f$, it follows that $b$ and everything hereditarily below $b$ occur inside the transitive closure of $f$. Formally, \begin{align*} \operatorname{tc}(b) \subseteq \operatorname{tc}(f). \end{align*} Now we use the hypothesis $f \in H_\theta$. By definition, \begin{align*} |\operatorname{tc}(f)| < \theta. \end{align*} The inclusion above gives \begin{align*} |\operatorname{tc}(b)| \leq |\operatorname{tc}(f)| < \theta. \end{align*} Therefore $b \in H_\theta$. Since $b$ was defined to be $f(a)$, we conclude \begin{align*} f(a) \in H_\theta. \end{align*} [/guided] [/step]