Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 16H / d / Solution

This is König theorem for cardinal numbers. Since ${\kappa }_i<{\lambda }_i$, map $(i,\alpha )$ in the disjoint union to the element $s_{i,\alpha }$ of the product defined byHere $\alpha <{\kappa }_i$ implies $\alpha +1<{\lambda }_i$. The unique nonzero coordinate and its value recover $(i,\alpha )$, so this is an injection

For strictness, suppose $F$ mapped the disjoint union onto the product. For each $i$, the sethas cardinality at most ${\kappa }_i$, so it cannot exhaust ${\lambda }_i$. Choose $b_i\in {\lambda }_i\setminus A_i$. Then $b=(b_i{)}_{i\in I}$ differs from $F(i,\alpha )$ in coordinate $i$ for every $(i,\alpha )$, contradicting surjectivity. By Cantor-Schröder-Bernstein theorem, an injection in the reverse direction would combine with the displayed injection to give a bijection and hence a surjection. Therefore

Solved by gpt-5.6-sol high.