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I need a little help regarding one of the basic properties of $\omega _1$.

Namely, I am supposed to show that for any $\alpha \in \omega_1$ there exists $\beta \in \omega_1\cap A$ such that $\alpha < \beta$.

Here $A$ stands for a set $A:=$ { $\alpha$ $ |$ $ \alpha \in Ord$ $\wedge$ $cof(\alpha)=\omega$ }.

What I am trying to show is that for any $\alpha \in \omega_1$ $\exists \beta \in \omega_1 \cap Lim$, where $Lim$ is the class of all limit ordinals such that $\alpha < \beta$.

This will be enough since every countable limit ordinal has cofinality $\omega$ which is not so hard to show.

My problem is, how to show that this defined $\beta$ indeed exists, i.e. $\beta$ being $Lim$.

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    $\begingroup$ How about letting $\beta = \sup \{ \alpha + 1, \alpha + 2, \cdots \}$? $\endgroup$ – Clive Newstead Mar 18 '17 at 12:22
  • $\begingroup$ Okay, it is a limit ordinal, but what is the guarantee for being strictly < than $\omega_1$? $\endgroup$ – edward_scissorhands Mar 18 '17 at 12:26
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    $\begingroup$ It's a countable union of countable sets, so is countable. $\endgroup$ – Clive Newstead Mar 18 '17 at 12:29
  • $\begingroup$ Thanks! It makes perfect sense now. $\endgroup$ – edward_scissorhands Mar 18 '17 at 12:30
  • $\begingroup$ I'll convert my comments into an answer. $\endgroup$ – Clive Newstead Mar 18 '17 at 12:30
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Define $\beta = \sup \{ \alpha + n \mid n < \omega \}$. Then $\beta > \alpha$ and $\beta$ has cofinality $\omega$; moreover, it is less than $\omega_1$ since it is a countable union of countable sets, so is countable.

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