Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 128 / 3

The Hausdorff formula for cardinal exponentiation states that for infinite cardinals $\kappa$ and $\lambda$,Equivalently,

Solved by gpt-5.6-sol high.

The Fn forcing $\mathrm{Fn}(X,Y,\kappa )$ consists of partial functions $p:X\rightharpoonup Y$ such thatIt is ordered by reverse inclusion: $p\le q$ exactly when $p\supseteq q$, so a stronger condition supplies more values.

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If $M$ regards $\mathbb{P}$ as satisfying the $\kappa$-chain condition, then forcing with $\mathbb{P}$ preserves every cardinal and cofinality at least $\kappa$. If $M$ regards $|\mathbb{P}|$ as $\mu$, then every antichain has size at most $\mu$, so $\mathbb{P}$ has the ${\mu }^{+}$-chain condition. It follows that every cardinal and cofinality at least ${\mu }^{+}$ is preserved in a $\mathbb{P}$-generic extension.

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If $\mathbb{P}$ is ${\aleph }_1$-closed in $M$, a descending sequence deciding successively all entries of a proposed function $\mathbb{N}\to M$ has a common lower bound. Thus the extension contains no new countable sequences of ground-model elements and in particularIt follows that ${\aleph }_1^M$ remains uncountable and hence is preserved. More generally such closure preserves cardinals at most ${\aleph }_1$, but closure alone need not preserve larger cardinals.

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The forcingis ${\aleph }_1^M$-closed, so by closed forcing it adds no new real numbers. Its generic union can be viewed as a sequenceof old reals. For every $r\in \wp (\mathbb{N}{)}^M$, conditions asserting that some unused row equals $r$ are dense: assigning all countably many values of that row is a legitimate condition. Thus the generic sequence surjects ${\aleph }_1^M$ onto the old set of reals.

In $M$, that set had cardinality ${\aleph }_2^M$. The forcing therefore collapses ${\aleph }_2^M$ to ${\aleph }_1^M$, while adding no reals and preserving ${\aleph }_1^M$. Consequently

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The forcing $\mathbb{P}=\mathrm{Fn}({\aleph }_1^N,2,{\aleph }_1^N)$ is ${\aleph }_1^N$-closed, so it adds no reals:

By contrast, ${\aleph }_1^M$ is countable in $N$. The union of the $\mathbb{Q}$-generic filter is a total binary functionFix in $N$ a bijection $e:\mathbb{N}\to {\aleph }_1^M$. Then $r(n)=h(e(n))$ is a real in $N[H]$. It is not in $N$: for any ground-model real $r_0$ and any $p\in \mathbb{Q}$, some coordinate of ${\aleph }_1^M$ is outside $\mathrm{dom}p$, and extending there forces $r$ to differ from $r_0$. Hence $\mathbb{Q}$ adds a new real, and

Solved by gpt-5.6-sol high.