$AC_\omega$ is weaker than $AC$, though not provable in ZF without $AC$?

Question

According to the wikipedia article on the Axiom of countable choice,

The axiom of countable choice ($AC_ω$) is strictly weaker than the axiom of dependent choice (DC), (Jech 1973) which in turn is weaker than the axiom of choice (AC). Paul Cohen showed that $AC_ω$ is not provable in Zermelo–Fraenkel set theory (ZF) without the axiom of choice (Potter 2004).

But if $AC$ is required to prove $AC_ω$, in what sense is it weaker?

Answer 1

Weaker here means that it does not prove as much. Since $\sf AC$ proves $\sf AC_\omega$, it also proves everything that follows from $\sf AC_\omega$. On the other hand, $\sf AC_\omega$ does not prove $\sf AC$.

So by weaker we mean, here, $$\{\varphi\mid\sf ZF+AC_\omega\vdash\varphi\}\subsetneq\{\varphi\mid ZF+AC\vdash\varphi\}.$$