Let $\gamma$ be an initial infinite ordinal. Show that $\gamma=\omega_{\alpha}$ for some ordinal $\alpha$

The definition I use for $\omega_{\alpha}$ as follows: Suppose $\omega_{\beta},\aleph_{\beta}$ have been defined for all $\beta<\alpha$. By transfinite induction, we define $$\omega_{\alpha}=\{\gamma \ \text{is ordinal} : |\gamma|\le\aleph_{\beta} \ \text{for some} \ \beta<\alpha\}$$

  • $\begingroup$ I would say that is by definition. What is your definition of "$\omega_\alpha$"? $\endgroup$ – Henning Makholm Jun 8 '16 at 14:54
  • $\begingroup$ @Henning Makholm, I have edited my question :) $\endgroup$ – Rohan Jun 8 '16 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.