# No uncountable subset can be well-ordered $\iff$ Hartogs ordinal is $\omega_1$

I'm looking at an exercise which starts as follows:

Suppose that no uncountable subset of $\mathcal P\omega$ can be well-ordered (equivalently, the Hartogs ordinal $\gamma(\mathcal P\omega)$ is $\omega_1$).

In order to proceed with the exercise I should probably understand this equivalence.

I guess that "no uncountable … can be well-ordered" means "there is one that can't be well-ordered", otherwise any finite set would be a counterexample (as they don't have such uncountable subsets yet their Hartogs ordinal is surely not $\omega_1$). Having this said, I get that the existence of an uncountable subset that can't be well-ordered implies that $\mathcal P\omega$ doesn't inject into $\omega_1$. However, why does it imply that $\omega_1$ doesn't inject into $\mathcal P\omega$?

• Whoever gave you this exercise, inform them that the notation should be $\aleph(X)$ for the Hartogs number of $X$. Apr 23, 2016 at 18:18
• "No uncountable subset of P(w) can be well-ordered" does NOT mean "There exists one..." Finite sets are not uncountable and are not counter-examples. Countable means "finite or countably infinite". Apr 24, 2016 at 5:40
• Are you sure? Then for a finite set "no uncountable subset can be well-ordered" would be true while the Hartog ordinal of a finite set is not $\omega_1$. Apr 24, 2016 at 15:14

In this case we are talking about the least uncountable ordinal, or $\omega_1$.