Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 128 (Forcing and the Continuum Hypothesis)

The definable power set of a set $X$ isThus definability is evaluated internally in the structure $(X,\in )$ and parameters from $X$ are allowed.

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The constructible hierarchy is defined by transfinite recursion:when $\lambda$ is a limit ordinal. Its union over all ordinals is the constructible universe $L$.

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A condensation sentence for the constructible hierarchy may be obtained by taking a single conjunction $\sigma$ that expresses a sufficiently strong finite fragment of set theory, the assertion $V=L$, and that the ordinals have no largest member. The finite fragment is chosen strong enough to define the satisfaction relation needed for the $L$-construction and to prove its absoluteness for transitive sets.

If a transitive set $X$ satisfies $\sigma$, let $\lambda =X\cap \mathrm{Ord}$. The absence of a largest ordinal makes $\lambda$ a limit ordinal. Internal $V=L$ says every $x\in X$ belongs to some internally constructed $L_{\beta }$, while transitivity and absoluteness identify that level with the actual $L_{\beta }$. Conversely the finite closure axioms ensure that every $L_{\beta }$, $\beta <\lambda$, belongs to $X$. Hence $X=L_{\lambda }$.

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Choose a successor ordinal $\beta =\delta +1>\alpha$, and then choose a limit ordinal $\gamma >\beta$. The level $L_{\gamma }$ satisfies the condensation sentence for the constructible hierarchy. The level $L_{\beta }$ cannot satisfy it: otherwise condensation would give $L_{\beta }=L_{\lambda }$ for a limit $\lambda$, butwould imply the impossible equality $\beta =\lambda$. Thus the condensation sentence belongs to $T_{\gamma }$ but not to $T_{\beta }$, and $T_{\beta }\ne T_{\gamma }$.

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The language of set theory has only countably many sentences, so there are at most $2^{{\aleph }_0}$ possible complete sets of sentences $T_{\xi }$. Choose $(2^{{\aleph }_0}{)}^{+}+1$ ordinals above $\alpha$. By cardinal pigeonhole, two of their theories agree. Hence some $\gamma >\beta >\alpha$ satisfy $T_{\beta }=T_{\gamma }$.

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With the convention that $p\le q$ means that $p$ is stronger, a set $G\subseteq \mathbb{P}$ is a generic filter over $M$ when it is a filter and meets every dense set $D\subseteq \mathbb{P}$ with $D\in M$. Explicitly, $G$ is upward closed toward weaker conditions, every two members have a common stronger member in $G$, and $G\cap D\ne \varnothing$ for every such $D$.

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For $g\in {\mathbb{N}}^{\mathbb{N}}\cap M$ and $n\in \mathbb{N}$, the setis dense in ${\mathbb{Q}}_0$: extend any finite sequence at one fresh coordinate with the corresponding value of $g$. The generic union $x_0$ meets every $D_{g,n}$, so it agrees infinitely often with every ground-model $g$. Thus $x_0$ is infinitely equal over $M$ and is not eventually different over $M$.

For $g\in {\mathbb{N}}^{\mathbb{N}}\cap M$, conditions of ${\mathbb{Q}}_1$ whose side set contains $g$ form a dense set. Once such a condition enters $G_1$, every later coordinate added to its stem must avoid $g$. Hence the generic union $x_1$ is eventually different from every ground-model $g$, and consequently is not infinitely equal over $M$.

The four answers are therefore

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

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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

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