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I wonder if this statement holds. I think it doesn't (if the cardinal of T is very large), but have no idea how to find a counter example.

Let $T$ be a totally ordered set, does there exists a sequence $(t_n)_{n \in \omega}$ such that $ \forall t \in T, \exists n, t \leq t_n$

Thank you !

Edit : I meant $(t \leq t_n)$, not $t < t_n$

TuTor
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1 Answers1

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Let $T$ be the first uncountable ordinal. Then for whatever $(t_n)_{n\in\omega}$ you pick, the set $$ \{\,t\in T\mid\exists n\in\omega\colon t<t_n\,\}=\bigcup_{n\in\Bbb \omega}\{\,t\in T\mid t<t_n\,\}$$ is a countable union of countable (because proper) subsets, hence is countable, hence is not all of $T$.

Hagen von Eitzen
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  • This answer assumes that the first uncountable ordinal has no maximal element. Is it obvious ? – TuTor Feb 04 '20 at 17:48
  • @TuTor: Yes. It is obvious. Such maximal element would be a maximal countable ordinal, and there's no such thing. – Asaf Karagila Feb 04 '20 at 18:34