= Solution
The <cardinal bound for an increasing chain of countable sets> is $\boxed{|X|\leq\aleph_1}$. If $X$ is countable, there is nothing to prove. Otherwise choose $Y\subseteq X$ with $|Y|=\aleph_1$, and for every $y\in Y$ choose an index $i(y)$ such that $y\in X_{i(y)}$.
For any $i\in I$, its countable $X_i$ cannot contain all of $Y$, so choose $y\in Y\setminus X_i$. Since $I$ is a <linear order>, comparison of $i$ and $i(y)$ forces $i<i(y)$: the other direction would imply $y\in X_i$. Therefore $X_i\subseteq X_{i(y)}$. The selected family is cofinal among the original sets, and
$$
X=\bigcup_{y\in Y}X_{i(y)}.
$$
By <infinite cardinal arithmetic>, this union of $\aleph_1$ countable sets has <cardinality> at most $\aleph_1$. No <well-order> or <cofinality> assumption on $I$ was used; a <linear increasing union of countable sets> has the same bound even for an arbitrary linear index order.
Back to article page
