Countable subset bounded in uncountable set 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. 
 A: Hint: For each $a \in A$ consider the set $B_a = \{ b \in B : b \prec a \}$.  What can we say about these sets?
A: There are two facts you need to use:


*

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

*A countable union of countable sets is countable.
A: 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 $\aleph_\omega$. 
