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A forcing preserves cofinalities over $M$ if every generic extension hasThe two values are compared as ordinals, using preservation of ordinals by forcing. Equivalently the forcing asserts that no ground ordinal acquires a smaller cofinality. The assertion concerns all ordinal cofinalities, not just that one specified cardinal remains uncountable.

Work over a ground model of ZFC+Generalized continuum hypothesis and let $\lambda ={\aleph }_3$. Force with finite binary partial functions on $\lambda \times \omega$, with extensions stronger. The Delta-system lemma thins any uncountable family of finite domains to an uncountable family with one common root. Only finitely many binary assignments on that root occur, so two conditions agree there and their union is a common extension. Hence the forcing has the countable chain condition for forcing and preserves cardinals and cofinalities.

The generic union yields $\lambda$ distinct reals. Totality at each coordinate is dense, and for two different coordinates it is dense to assign different values at some unused natural-number position. Thus the extension satisfies $2^{{\aleph }_0}\ge \lambda$.

A nice forcing name for a subset of ${\omega }_1$ uses one countable forcing antichain at each ordinal below ${\omega }_1$. Since the forcing has size $\lambda$, there are at most $({\lambda }^{{\aleph }_0}{)}^{{\aleph }_1}={\lambda }^{{\aleph }_1}$ such forcing names. Ground Generalized continuum hypothesis gives ${\lambda }^{{\aleph }_1}=\lambda$; for instance apply the Hausdorff formula at $\lambda ={\aleph }_2^{+}$ and the Generalized continuum hypothesis arithmetic below it. Therefore $2^{{\aleph }_1}\le \lambda$ in the extension. Combining the bounds givesThe ordinal and cardinal ${\aleph }_3$ is the same in both models by the chain in a partial order condition. The forcing theorem formalizes this construction as the requested relative-consistency implication. A countable transitive ground model is a convenient presentation, not an additional consequence silently derived from mere consistency. This is the Cohen forcing two-level continuum plateau.

A cofinal map $f:\delta \to \alpha$ has a range cofinal in $\alpha$:For a limit $\alpha$, this is equivalent to $\sup f[\delta ]=\alpha$. For a successor $\alpha$, its largest element must occur in the range. Monotonicity is an additional property and is not part of this definition. The cofinality of $\alpha$ is the least ordinal that admits such a cofinal map into $\alpha$.

Write $\mathfrak{c}=2^{{\aleph }_0}$. Its cofinality is ${\aleph }_1$, so choose a strictly increasing sequence of infinite cardinal numbers $({\kappa }_{\xi }{)}_{\xi <{\omega }_1}$ below $\mathfrak{c}$ with supremum $\mathfrak{c}$. Such a sequence is obtained by refining a cofinal sequence and choosing larger cardinals recursively; $\mathfrak{c}$ is a singular cardinal, hence a limit cardinal.

By the axiom of choice, fix a bijection $b:\mathfrak{c}\to \mathbb{R}$. SetEvery $A_{\xi }$ is infinite, $|A_{\xi }|={\kappa }_{\xi }<\mathfrak{c}$, and distinct indices give distinct cardinalities. The cofinality of the sequence ensuresThus $|\mathcal{A}|={\aleph }_1$. Pairwise different sizes means different sizes for distinct members; without that qualification the parenthetical condition would contradict the case $A=B$.

The least index is $\boxed{\alpha ={\omega }_1}$, withIndeed, for every nonzero limit ordinal $\beta$, continuity of the aleph number enumeration and the cofinality of an increasing ordinal supremum giveIf $\beta <{\omega }_1$ is a nonzero limit ordinal, this cofinality is $\omega$, so ${\aleph }_{\beta }$ is a singular cardinal. The other infinite cardinal numbers below ${\aleph }_{{\omega }_1}$ are successor-indexed or ${\aleph }_0$, and are regular cardinals. For a successor cardinal ${\mu }^{+}$, a cofinal sequence of length at most $\mu$ would express ${\mu }^{+}$ as a union of at most $\mu$ sets of size at most $\mu$, contradicting infinite cardinal arithmetic.

There are ${\aleph }_1$ nonzero countable limit ordinals, and each countable initial segment contains only countably many. In increasing order they therefore have order type ${\omega }_1$. These are precisely the indices of the uncountable singular cardinals below ${\aleph }_{{\omega }_1}$. Hence every ${\sigma }_{\alpha }$ with $\alpha <{\omega }_1$ has countable cofinality, and the next one is ${\aleph }_{{\omega }_1}$, whose cofinality is ${\omega }_1<{\aleph }_{{\omega }_1}$.

Yes. A cardinal fixed point of the singular cardinal enumeration is obtained by countable iteration. Start with ${\kappa }_0={\aleph }_0$ and putThe singular cardinal enumeration is strictly increasing and satisfies ${\sigma }_{\alpha }\ge \alpha$ for every ordinal $\alpha$; the latter follows by transfinite induction for any strictly increasing ordinal-valued enumeration. If equality occurs at some ${\kappa }_n$, that cardinal number is already a witness. Otherwise the sequence is strictly increasing. Put $\kappa ={\sup }_{n<\omega }{\kappa }_n$. This is an uncountable singular cardinal of cofinality $\omega$.

For every $\alpha <\kappa$, some $n$ has $\alpha <{\kappa }_n$, and thereforeAlso ${\sup }_n{\sigma }_{{\kappa }_n}=\kappa$, so ${\sup }_{\alpha <\kappa }{\sigma }_{\alpha }=\kappa$. At a limit ordinal index, if the supremum of all preceding enumerated cardinals is itself singular, it is exactly the next member: every smaller singular cardinal already has a preceding index. ThusThis argument uses continuity only at a singular supremum; the enumeration need not be continuous at a weakly inaccessible cardinal.

The cardinal bound for an increasing chain of countable sets is $\boxed{|X|\le {\aleph }_1}$. If $X$ is countable, there is nothing to prove. Otherwise choose $Y\subseteq X$ with $|Y|={\aleph }_1$, and for every $y\in Y$ choose an index $i(y)$ such that $y\in X_{i(y)}$.

For any $i\in I$, its countable $X_i$ cannot contain all of $Y$, so choose $y\in Y\setminus X_i$. Since $I$ is a linear order, comparison of $i$ and $i(y)$ forces $i<i(y)$: the other direction would imply $y\in X_i$. Therefore $X_i\subseteq X_{i(y)}$. The selected family is cofinal among the original sets, andBy infinite cardinal arithmetic, this union of ${\aleph }_1$ countable sets has cardinality at most ${\aleph }_1$. No well-order or cofinality assumption on $I$ was used; a linear increasing union of countable sets has the same bound even for an arbitrary linear index order.

Let $\mathrm{Map}(f,\delta ,\kappa )$ be the bounded graph predicate for a function $\delta \to \kappa$. For an infinite cardinal number, regularity is expressed byThe predicate $\mathrm{InfCard}$ is the cardinal predicate in (i), conjoined with the bounded condition that $\kappa$ is a nonzero limit ordinal. For an infinite cardinal number, this latter condition holds automatically; it excludes the finite cardinals. The expression $f(\xi )<\beta$ abbreviates bounded graph membership with the output lying in $\beta$.

Every quantifier after $\forall f$ is bounded, and the cardinal predicate is already ${\Pi }_1$. Their conjunction can be placed in universal form, so this is a Pi-one formula modulo ZF. Its mathematical content is that no function from a smaller ordinal is cofinal in $\kappa$. By the definition of cofinality, that is equivalent to $\mathrm{cf}(\kappa )=\kappa$. Hence regular cardinalhood is Pi-one definable. This uses the standard convention that a regular cardinal is infinite.

An infinite cardinal number $\kappa$ is singular when $\mathrm{cf}(\kappa )<\kappa$. Thus a short cofinal function reaches arbitrarily high ordinals below $\kappa$. For example, ${\aleph }_{\omega }$ has cofinality $\omega$. The complementary case is a regular cardinal.

Write ${\sigma }_{\alpha }$ for the $\alpha$th uncountable singular cardinal in increasing order. It begins with ${\sigma }_0={\aleph }_{\omega }$ and ${\sigma }_1={\aleph }_{\omega +\omega }$. At a nonzero limit ordinal index it is continuous exactly when the supremum of its earlier values is singular. It can jump when that supremum is a weakly inaccessible cardinal. The first member of uncountable cofinality occurs at index ${\omega }_1$, with value ${\aleph }_{{\omega }_1}$.

For a regular infinite $\theta <\kappa$, with $\kappa$ a regular uncountable cardinal number, $S_{\theta }^{\kappa }=\{\delta <\kappa :\mathrm{cf}(\delta )=\theta \}$ is stationary. Build a strictly increasing continuous $\theta$-sequence in any club set, and take its supremum. Closure places it in that club and the cofinality of an increasing ordinal supremum gives cofinality $\theta$. In particular the two disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ show that the club filter on ${\omega }_2$ is not an ultrafilter.

For an uncountable regular cardinal $\kappa$, choose cofinal sequences $c_{\alpha }:\omega \to \alpha$ for the stationary set of ordinals of cofinality $\omega$. Above any bound, some fixed coordinate exceeds the bound on a stationary subset; Fodor lemma makes that coordinate constant on a stationary subset. Thus the stationary constant fibers, over all coordinates, have unboundedly many values. Regularity makes one coordinate have $\kappa$ such values. Its fibers are disjoint stationary sets; adding all leftover ordinals to one piece partitions $\kappa$.

For infinite regular cardinals $\kappa <\lambda$, a $\lambda$-tree whose levels all have size less than $\kappa$ has a cofinal branch. At levels of cofinality $\kappa$, bound all heights distinguishing distinct predecessor chains. The Fodor lemma makes this bound constant on a stationary set; one predecessor at that height occurs stationarily often. The resulting predecessor chains agree below their common heights and unite into a branch through every level. Distinct limit-level nodes with identical predecessor chains do not affect this argument.