Refuted: every limit ordinal has an at most countable cofinal subset — $\omega_1$ has none, assuming countable choice

False claim: every limit ordinal has an at most countable cofinal subset (Cofinal subset of an ordinal, Finite, countably infinite, countable, uncountable).

The claim is plausible because every limit ordinal a reader meets first does have one. $\omega$ is cofinal in itself and at most countable; and every at most countable limit ordinal $\lambda$ is cofinal in itself and at most countable, so the claim holds for all of them, and $\omega + \omega$ and $\omega^{2}$ are among them, both being shown at most countable earlier on this page. Whether $\omega^{\omega}$ and $\varepsilon_0$ are at most countable is a question no item on these pages settles, so neither is offered here as an instance.

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ($\mathrm{AC}_\omega$)). The first uncountable ordinal $\omega_1$ (The first uncountable ordinal $\omega_1 := \aleph(\omega)$) refutes the claim: it is a limit ordinal, and no at most countable subset of it is cofinal in it.

The hypothesis is not removable, and the item states it in the title: without a choice principle the refutation itself fails, since consistently with ZF the ordinal $\omega_1$ is the supremum of an $\omega$-sequence of at most countable ordinals. That is recorded in Choice ledger for this page: $\omega_1$ exists in ZF, and the boundedness theorem does not.

What separates $\omega_1$ from the countable limit ordinals. A limit ordinal is always cofinal in itself, so the claim can only fail when the ordinal is itself uncountable. $\omega_1$ is the least uncountable ordinal ($\omega_1$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF), so it is the first place where the claim can fail at all, and under $\mathrm{AC}_\omega$ it does fail there.

The refutation carries the hypothesis it uses. Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$, so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable is stated under $\mathrm{AC}_\omega$ and spends it at exactly one step, so this counterexample inherits the same cost. A page that quotes this item must carry $\mathrm{AC}_\omega$ forward into its own statement; Choice ledger for this page: $\omega_1$ exists in ZF, and the boundedness theorem does not is the ledger, and it names the model in which the conclusion fails outright.

What is deliberately not said at this point in the reading order. In the later vocabulary this item says $\operatorname{cf}(\omega_1)>\omega$, or that $\omega_1$ is regular. The cofinality and regular/singular vocabulary is introduced later in Cofinality $\operatorname{cf}(\alpha)$, and regular and singular cardinals ↗, so this earlier example stays in the subset language of Cofinal subset of an ordinal. Nothing is lost: the applications, such as the non-normality of the deleted Tychonoff plank, use exactly the subset form.