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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?

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  • $\begingroup$ Hint: Extract by induction from $\alpha_\nu$ an increasing subsequence with the same sup as $\alpha_\nu$ $\endgroup$ Feb 22, 2021 at 16:41
  • $\begingroup$ @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. $\endgroup$
    – Eparoh
    Feb 22, 2021 at 18:17
  • $\begingroup$ $sup$ of a collection of ordinals is independent of the order its elements are enumerated in! $\endgroup$
    – BrianO
    Feb 23, 2021 at 23:09

1 Answer 1

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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 $\le\vartheta$.

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