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 $\mathcal P\omega$?

  • 1
    $\begingroup$ Whoever gave you this exercise, inform them that the notation should be $\aleph(X)$ for the Hartogs number of $X$. $\endgroup$
    – Asaf Karagila
    Apr 23, 2016 at 18:18
  • $\begingroup$ "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". $\endgroup$ Apr 24, 2016 at 5:40
  • $\begingroup$ 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 $\omega_1$. $\endgroup$
    – akkarin
    Apr 24, 2016 at 15:14

1 Answer 1


If an ordinal can be mapped injectively into a set, then the range of such injection has an induced order isomorphic to your ordinal.

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


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