Set of ordinals less than the first uncountable ordinal and countability I am trying to solve the following question from Royden's Real Analysis (3rd edition, Chap. 1, Problem 32).
Let $Y$ be the set of ordinals less than the first uncountable ordinal, i.e., $Y= \{ x\in X: x<\Omega \}$. Show that every countable subset $E$ of $Y$ has an upper-bound in $Y$ and hence a least upper-bound.
I have the following question: If $Y$ is assumed to be the set of ordinals less than the first uncountable ordinal, then shouldn't $Y$ be countable by definition? and so, every subset of a countable set is countable? and then, $\Omega$ is an upper-bound to each $E \subset Y$?
 A: Each member of $Y$ (or, equivalently, each order type shorter than $Y)$ is countable, but $Y$ itself is the shortest possible uncountable well-ordering. 
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Added later in response to discussion in the comments: It's worth noting that this problem requires some use of the axiom of choice. Here's one approach: $S=\bigcup_{e\in E} \{x \mid x\lt e\}$ is a countable union of countable sets, so is countable (this uses AC). Since $\Omega$ is uncountable, there exists $b\in\Omega\setminus S;$ any such $b$ must be an upper bound for $E.$
It's consistent with ZF (without AC) that $\aleph_1$ is cofinal with $\omega,$ in which case the statement of Royden's problem is false.
A: You're making an unfounded generalization from the part to the whole. In the set $\{\{0\}, \{1\}\}$, every element has cardinality $1$; but the set as a whole has cardinality $2$. Likewise, the fact that every member of $Y$ is countable says nothing whatsoever about the cardinality of $Y$ - to take an extreme example, the "set" of all singletons (sets of cardinality exactly $1$) is so large that it isn't even a set (it's a proper class).
It's certainly the case that every subset of a countable set is countable. But your final step doesn't work either - if $E \subseteq Y$ is countable, it's true that $\Omega$ is an upper bound on $E$, but it isn't an upper bound in $Y$. $Y$ is the set of countable ordinals; $\Omega$ is by definition not a countable ordinal, and so is not in $Y$.
