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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 $A=B$.

It suffices to obtain a model of ZFC without weakly inaccessible cardinals. Starting with any model of ZFC, pass to its constructible universe, which satisfies ZFC and the Generalized continuum hypothesis. If it has no inaccessible cardinal, use that model. Otherwise pass to its rank segment at its least inaccessible cardinal $\delta$. This segment satisfies ZFC, retains the Generalized continuum hypothesis, and has no inaccessible cardinals. Under the Generalized continuum hypothesis, every weakly inaccessible cardinal is strongly inaccessible: if $\lambda <\kappa$ and $\kappa$ is a limit cardinal, then $2^{\lambda }={\lambda }^{+}<\kappa$. Thus in either case the resulting model $N$ has no weakly inaccessible cardinals.

Work inside $N$. If $\alpha$ is a nonzero limit ordinal, let $\lambda ={\sup }_{\beta <\alpha }{\sigma }_{\beta }$. It is an uncountable limit cardinal. If it were regular, it would be a weakly inaccessible cardinal, which is impossible in $N$. It is therefore singular, and the singular cardinal enumeration is continuous at this index:The cofinality of an increasing ordinal supremum now givesfor every nonzero limit ordinal $\alpha$ in $N$. Hence $N$ satisfies the negation of the proposed existential assertion. By the soundness theorem for first-order logic, consistency of ZFC prevents ZFC from proving that assertion. The model construction is a relative-consistency argument; it does not assume that consistency alone supplies a countable transitive model. Here, as usual, a limit ordinal excludes zero.

Weakly inaccessible cardinals are unbounded below the given cardinal. In fact, they form a stationary set there. Let $\kappa$ be the weakly Mahlo cardinal, so it is a regular uncountable limit cardinal and the set $R=\{\alpha <\kappa :\alpha =\mathrm{cf}(\alpha )\}$ is stationary.

The set $C$ of uncountable limit cardinals below $\kappa$ is a club set. For unboundedness, above any starting point choose a strictly increasing countable sequence of cardinal numbers below $\kappa$; its supremum remains below $\kappa$ by regularity and is an uncountable limit cardinal. For closure, a limit of such limit cardinals is again a limit cardinal.

Every $\alpha \in R\cap C$ is an uncountable regular cardinal and a limit cardinal, hence a weakly inaccessible cardinal. The intersection of a stationary set with a club set is stationary, because its intersection with any further club is nonempty. ThereforeThis proves the stronger form of the requested conclusion.

A weakly inaccessible cardinal is an uncountable regular cardinal that is a limit cardinal. It need not be a strong limit cardinal. Under the Generalized continuum hypothesis, it is a strongly inaccessible cardinal, since every smaller infinite $\lambda$ satisfies $2^{\lambda }={\lambda }^{+}<\kappa$.

A weakly inaccessible cardinal $\kappa$ is weakly Mahlo if $\{\alpha <\kappa :\alpha =\mathrm{cf}(\alpha )\}$ is a stationary set. Intersecting this set with the club set of uncountable limit cardinals shows that the weakly inaccessible cardinals below $\kappa$ are stationary, hence unbounded.