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An increasing chain of countable sets has union of cardinality at most ${\aleph }_1$. If the union is uncountable, select ${\aleph }_1$ of its points and one containing member for each. Every countable member misses one selected point, so linear comparison puts it below that point's chosen member. The chosen subfamily is cofinal, and its union has size at most ${\aleph }_1$ by infinite cardinal arithmetic.

If $\kappa$ is a regular cardinal and $\mathbb{P}$ has the $\kappa$-chain condition, then forcing with $\mathbb{P}$ preserves every cardinal number and cofinality at least $\kappa$. Indeed, a proposed surjection $f:\mu \to \lambda$ with $\mu <\lambda$ has its range contained in ${\bigcup }_{\xi <\mu }{B}_{\xi }$ from the possible-values lemma for chain-condition forcing. The regularity of $\kappa$ and infinite cardinal arithmetic make this union have cardinality below the ground-model cardinal $\lambda$, a contradiction.

For an infinite cardinal number $\kappa$, its cofinality isviewing $\kappa$ as an initial ordinal. Equivalently it is the smallest cardinal number which is the domain size of a cofinal map into $\kappa$.

Put $\lambda =2^{\kappa }$ and $\mu =\mathrm{cf}(\lambda )$. Suppose $\mu \le \kappa$, and choose cofinal ordinals ${\alpha }_i<\lambda$ for $i<\mu$. Put $a_i=|{\alpha }_i|<\lambda$. Their union is $\lambda$, so infinite cardinal arithmetic givesApply the König theorem for cardinal numbers to $a_i<a_i^{+}$. Each successor cardinal $a_i^{+}$ is at most $\lambda$, whenceHere the last equalities use the currying law for cardinal exponentiation and square of an infinite cardinal. This contradiction proves the cofinality of a power-set cardinal bound

Put $\mu =\mathrm{cf}(\kappa )$ and choose cofinal ordinals ${\alpha }_i<\kappa$, $i<\mu$. With $a_i=|{\alpha }_i|$, the same union comparison and infinite cardinal arithmetic give ${\sum }_{i<\mu }{a}_i=\kappa$. Since $a_i<\kappa$, their successor cardinals satisfy $a_i^{+}\le \kappa$. The König theorem for cardinal numbers therefore givesThis König bound for the Gimel function applies to regular and singular cardinals alike; choosing cofinal ordinals also covers successor cardinal numbers.

Let $\mathcal{O}$ be the set of strict total orders on $\kappa$. Each is a subset of $\kappa \times \kappa$, so the square of an infinite cardinal givesFor the lower bound, use the set $\kappa \times \{0,1\}$ and order the two-element blocks by their first coordinates, using the well-order of $\kappa$. For every $A\subseteq \kappa$, put $(\alpha ,0)$ before $(\alpha ,1)$ exactly when $\alpha \in A$, reversing them otherwise. This defines a strict total order for each $A$, and different subsets give different orders. Transport these orders to $\kappa$ along a fixed bijection $\kappa \times \{0,1\}\cong \kappa$. Thus $2^{\kappa }\le |\mathcal{O}|$, and the Cantor-Schröder-Bernstein theorem givesIf using the customary axiom of choice development of infinite cardinal arithmetic, its use here is precisely to supply $\kappa \times \kappa \cong \kappa$ and $\kappa \times 2\cong \kappa$. The ordering construction makes no further choices. It proves the cardinality of the family of strict total orders. For $\kappa$ itself an initial ordinal, these particular product identities can also be proved in ZF by well-ordering pairs according to their maximum coordinate, so full axiom of choice is stronger than this instance intrinsically requires.

Let $f$ be given and choose $Y\subseteq \mathbb{R}$ of cardinality ${\aleph }_1$. Each $f(y)$ is countable, so the axiom of choice and infinite cardinal arithmetic giveSince the continuum is larger, choose $x$ outside this union. Then $x\notin f(y)$ for every $y\in Y$. The countable set $f(x)$ cannot contain $Y$, so choose $y\in Y\setminus f(x)$. ThusThe two points are distinct because $x\notin Y$. This proves the countable-valued free-pair criterion and the required direction of the Freiling axiom of symmetry without any measurability assumption on $f$.

For an infinite cardinal number $\kappa$, the gimel function isThe singular cardinals hypothesis asserts that for every infinite singular cardinal $\kappa$,Here ${\kappa }^{+}$ is the successor cardinal. Equivalently, for every infinite singular cardinal,To see why the second formulation adds nothing in the other case, put $\theta =\mathrm{cf}(\kappa )$. If $\kappa \le 2^{\theta }$, then $2^{\theta }\le {\kappa }^{\theta }\le (2^{\theta }{)}^{\theta }=2^{\theta }$ by infinite cardinal arithmetic. Equality $\kappa =2^{\theta }$ is impossible here: the König theorem for cardinal numbers gives $\mathrm{cf}(2^{\theta })>\theta$, whereas $\mathrm{cf}(\kappa )=\theta$. Thus in this case $2^{\theta }\ge {\kappa }^{+}$, as required by the maximum formula. For a strong limit cardinal that is singular, the hypothesis also yields $2^{\kappa }={\kappa }^{+}$: restrictions of a subset of $\kappa$ to a cofinal sequence of smaller cardinals give $2^{\kappa }\le {\kappa }^{\theta }$, while the reverse inequality is immediate. The quantified implication above is the precise general statement.

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 ${\omega }_1<{\aleph }_{{\omega }_1}$.

The Freiling theorem givesFirst assume the Continuum hypothesis and fix a well-order $(r_{\alpha }{)}_{\alpha <{\omega }_1}$ of the real numbers. Define $f(r_{\alpha })=\{r_{\beta }:\beta \le \alpha \}$. Each value is countable. For any pair, one index is at most the other, so at least one of $x\in f(y)$ or $y\in f(x)$ holds. The same holds for $x=y$, since $x\in f(x)$. This violates the Freiling axiom of symmetry.

Conversely, assume $|\mathbb{R}|>{\aleph }_1$ and let $f$ assign a countable set to each real. Choose $X\subseteq \mathbb{R}$ of cardinality ${\aleph }_1$. By infinite cardinal arithmetic,Choose $y$ outside this union. Since $f(y)$ is countable and $X$ is uncountable, choose $x\in X\setminus f(y)$. Then $y\notin f(x)$ and $x\notin f(y)$, with $x\ne y$. This proves the Freiling axiom of symmetry and completes both implications.

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.

Assume $M$ satisfies the countable chain condition for forcing. We prove cardinal preservation by chain-condition forcing using the possible-values lemma for chain-condition forcing, and give the argument explicitly for ccc.

Let $\kappa$ be an uncountable cardinal of $M$. If it ceased to be a cardinal, some ordinal $\mu <\kappa$ would admit a surjection onto $\kappa$ in an extension. By the forcing theorem, there would be a condition $p$ and a forcing name $\dot{f}$ such thatInside $M$, for each $\xi <\mu$ choose a maximal antichain in a forcing order below $p$ deciding the ordinal value of $\dot{f}(\xi )$. Conditions deciding that value are dense below $p$. The chain condition makes the chosen antichain countable in $M$, so the corresponding set $B_{\xi }\subseteq \kappa$ of possible decided values is countable in $M$.

A generic filter containing $p$ meets the downward closure of every such maximal antichain, so its value $f(\xi )$ belongs to $B_{\xi }$. Therefore its whole range lies in the ground-model setThe last inequality is infinite cardinal arithmetic and uses only that $\kappa$ is uncountable and $\mu <\kappa$; no regularity of $\kappa$ is needed. Some ordinal in $\kappa \setminus B$ exists already in $M$ and cannot occur in the range, contradicting the forced surjectivity.

Finite cardinals cannot collapse, and $\omega$ cannot become finite: a finite domain has finite image, and transitive models agree on the natural numbers. Old non-cardinals cannot become cardinals because their old bijections persist. Thus

Let $\lambda \ge \kappa$ be a cardinal number of $M$. If forcing collapsed $\lambda$, some condition would force a surjective function $\dot{f}:\mu \to \lambda$ for an ordinal $\mu <\lambda$. Replacing $\mu$ by its ground-model cardinality lets us assume that $\mu$ is a cardinal.

For each $\xi <\mu$, choose in $M$ a maximal antichain in a forcing order deciding $\dot{f}(\xi )$. The chain condition for forcing makes its size less than $\kappa$, so the set $B_{\xi }$ of possible values of $\dot{f}(\xi )$ has size below $\kappa$. This is the possible-values lemma for chain-condition forcing. Every interpreted range is contained in $B={\bigcup }_{\xi <\mu }{B}_{\xi }$. If $\mu <\kappa$, the fact that $\kappa$ is a regular cardinal gives $|B|<\kappa \le \lambda$; if $\kappa \le \mu <\lambda$, infinite cardinal arithmetic gives $|B|\le \mu <\lambda$. Either way $B$ cannot contain the range of a surjection onto $\lambda$, a contradiction. By cardinal preservation by chain-condition forcing, $\mathbb{P}$ preserves every cardinal at least $\kappa$.

Apply König theorem for cardinal numbers with $I=\omega$, ${\kappa }_i={\aleph }_i$, and ${\lambda }_i={\aleph }_{i+1}$. Standard infinite cardinal arithmetic giveswhereas every factor in the product is at most ${\aleph }_{\omega }$. Hence

Now suppose $2^{{\aleph }_{\omega }}={\aleph }_{\omega }$. The exponent laws for infinite cardinals would givecontradicting the strict inequality. Therefore