Cofinality of a set

Question

Let $(I, \leq)$ be a directed set. Suppose that $I$ has uncountable cofinality and that one can decompose $I$ as a countable union $I = \bigcup_{n = 0}^\infty I_n$, is it true that at least one of the $I_n$ is cofinal in $I$?

Answer 1

No: let $I := \omega_1 \times \omega \space$ with the product order, i.e. $(\alpha, n) \le (\beta, m)$, if $\alpha \le \beta$ and $ n\le m$. It is easy to see that $(I, \le)$ is directed and has cofinality $\omega_1$. It is $I = \bigcup_{n \in \omega} (\omega_1 \times \{n\})$, but no $\omega_1 \times \{n\}$ is cofinal in $I$.

On the positive side, we have that at least one of the $I_n$ is unbounded:
Assume not and choose for each $n \in \omega$ an upper bound $x_n$ of $I_n$. Then $\{x_n: n \in \omega\}$ is a countable set, which is cofinal in $I$.