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Let $(B, \prec)$ be a well ordered set with the ordinal $\omega_1$. Show that every countable subset of B is bounded in $(B, \prec)$. Let A be such a subset. A is a subset of a well ordered set and such have a least element and it's will be the lower bound. I would like a hint on how to approach the upper bound element.

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Here is a sketch. Most of the unproven facts I use below should be easy exercises involving the definition of ordinals and cardinals.

As $B$ has order type $\omega_1$, you may as well assume that $B$ is $\omega_1$.

Let $A \subseteq \omega_1$ be countable. $\alpha \in A \subseteq \omega_1$ and the fact that $\omega_1$ is ordinal (hence transitive), one has that $\alpha \subseteq \omega_1$. $|\alpha| < \aleph_1$ since $\alpha < \omega_1$. Easily check that union of a set of ordinals is an ordinals. So $\bigcup_{\alpha \in A} \alpha$ is an ordinal. Countable union of countable sets are countable. So $|\bigcup_{\alpha \in A} \alpha| < \aleph_1$. So $\beta := \bigcup_{\alpha \in A} \alpha$ is a countable ordinal. Hence $\bigcup_{\alpha \in A} \alpha < \beta + 1 < \omega_1$. Using transitivity of ordinals again, each $\alpha \in A$, $\alpha \leq \bigcup_{\alpha \in A} \alpha < \beta + 1$. $A$ is bounded above.


Just to mention, your title is misleading. Countable sets can be unbounded in uncountable sets. For example, $\{\aleph_n : n \in \omega\}$ is unbounded in the uncountable set $\aleph_\omega$.

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Hint: For each $a \in A$ consider the set $B_a = \{ b \in B : b \prec a \}$. What can we say about these sets?

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There are two facts you need to use:

  1. If $B$ has order type $\omega_1$, then $B$ is uncountable, but each element of $B$ has only a countable set of predecessors.

  2. A countable union of countable sets is countable.

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  • $\begingroup$ Why does every element of B has only a countable set of predecessors? $\endgroup$ – Gyt Feb 1 '14 at 8:56
  • $\begingroup$ Because that's how $\omega_1$ is defined: it's the first uncountable ordinal. As an ordered set, $\omega_1$ can be characterized (up to isomorphism) as an uncountable well-ordered set in which each element has a countable set of predecessors. If $B$ is order-isomorphic to $\omega_1$ then it has the same properties. $\endgroup$ – bof Feb 1 '14 at 17:26

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