Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 128 / 1

First form the singleton set $\{b\}=F_1(b,b)$ and then use $F_2$ to take a set union:The unused second argument of $F_2$ may be any term. Since $x\cap c=x\setminus (x\setminus c)$, a term using only the prescribed operation symbols isAfter substituting the displayed term for both occurrences of $U$, this is literally a term in $F_1,F_2,F_3$, and its value is $(a\cup \{b\})\cap c$.

Work in the ambient universe and let $a,u\in L$. SupposeFor each $x\in a$, let $y_x$ be this unique witness. The relativization ${\phi }^L$ is a first-order formula, so the ambient Axiom schema of replacement collects the witnesses $y_x$ into a set. Every witness lies in the constructible hierarchy, hence there is an ordinal $\alpha$ such thatFor example, take the supremum of one constructible rank for each witness and then increase it by one.

The set $L_{\alpha }$ itself belongs to $L_{\alpha +1}\subseteq L$. Taking $b=L_{\alpha }$, every $x\in a$ has a witness $y\in b$ satisfying ${\phi }^L(x,y,u)$. Thereforewhich is the stated instance of Replacement.

Suppose $V=L$. If $x\in L_{{\omega }_1}$, then $x\in L_{\alpha }$ for some countable ordinal $\alpha$. The set $L_{\alpha }$ is transitive and countable, so the transitive closure of $x$ lies in a countable set. Thus $x$ is hereditarily countable, proving

Conversely, let $x\in H_{{\aleph }_1}$ and choose a sufficiently large $L_{\theta }$ containing $x$. By the Downward Lowenheim-Skolem theorem, there is a countable elementary substructure $N\prec L_{\theta }$ that contains every member of $\mathrm{TC}(\{x\})$. The Mostowski collapse theorem gives a transitive collapse of $N$, and the condensation lemma for the constructible universe identifies it with $L_{\beta }$ for a countable ordinal $\beta$. Because $N$ contains the transitive closure of $x$ pointwise, the collapse fixes $x$. Thus $x\in L_{\beta }\subseteq L_{{\omega }_1}$. Hence