I have the following definition for regular cardinals:

An infinite cardinal $$\kappa$$ is called singular if there exists an increasing transfinite sequence $$\{\alpha_\nu :\nu< \vartheta\}$$ of ordinals $$\alpha_\nu < \kappa$$ whose length $$\vartheta$$ is a limit ordinal less than $$\kappa$$ and $$\kappa=\sup\{\alpha_\nu :\nu< \vartheta\}$$. An infinite cardinal that is not singular is called regular.

Now, I'm trying to prove that the condition of increasing sequence does not matter in the following sense:

Let $$\kappa$$ be a regular cardinal, $$\vartheta$$ a limit ordinal less than $$\kappa$$ and $$\{\alpha_\nu :\nu< \vartheta\}$$ a sequence of ordinals such that $$\alpha_\nu < \kappa$$ for every $$\nu < \vartheta$$, then $$\sup\{\alpha_\nu :\nu< \vartheta\} < \kappa$$.

I don't know if the result is true but it seems intuitive to me that I can reorder the transfinite sequence in a way that it has the same terms but is now increasing, so its supremum would be the same and by the regularity of $$\kappa$$ it would follow the result I want to prove.

The problem is that I don't know how to formally do that reorder of the sequence. Can anyone help me with this problem?

• Hint: Extract by induction from $\alpha_\nu$ an increasing subsequence with the same sup as $\alpha_\nu$ Feb 22, 2021 at 16:41
• @AlessandroCodenotti Could you please elaborate how to do it? I'm a little confuse on how to properly write it down when working with ordinals. And I'd like to have a fully worked and detailed example to understand the concepts better. Feb 22, 2021 at 18:17
• $sup$ of a collection of ordinals is independent of the order its elements are enumerated in! Feb 23, 2021 at 23:09

About the method brought up in the comments, you asked for an example: let $$\kappa=\aleph_1$$, $$\vartheta=\omega$$, and $$(\alpha_\nu)_{\nu<\omega}$$ be a sequence of countable ordinals such as $$(1, \omega, \omega+4, \omega+2, \varepsilon_0, \omega^{495}, \varepsilon_3, \ldots)$$. Let's extract from this sequence each "new highest ordinal" (each ordinal that's greater than all previous ones in the sequence.) The sequence of these is $$(1, \omega, \omega+4, \varepsilon_0, \varepsilon_3, \ldots)$$, and since each of these is greater than all ordinals before it, this sequence is increasing. Since it's a subsequence of $$(\alpha_\nu)_{\nu<\omega}$$ it must have order type $$\le\omega$$.
In general, if we have $$(\alpha_\nu)_{\nu<\vartheta}$$, extract a subsequence $$(\beta_\nu)_{\nu<\theta\le\vartheta}$$ from it by transfinite induction on $$\nu$$: assuming $$\beta_\mu$$ is defined for all $$\mu<\nu$$, set $$\beta_\nu=\alpha_{\textrm{min}\{\gamma<\vartheta:\forall(\mu<\nu)(\beta_\nu<\alpha_\gamma)\}}$$, stopping this procedure when there are no more such $$\gamma<\vartheta$$ to choose. This subsequence must be increasing by definition, and it has an order type $$\le\vartheta$$.