Related to my answer on this question : set of values of finite measure, exhaustion method
I would like to know if for an arbitrary total order $K$, there exists a sequence $\{k_n\}$ such that for all elements $k \in K$, there exists $k_n$ such that $k \le k_m$. In some way, it would allow you to "go to infinity".
The reason why I needed this is because I was trying to use Zorn's Lemma in order to answer a question on MSE, and I came up with this set $$ \bigcup_{k \in K} E_k $$ where $E_{k_1} \subseteq E_{k_2}$ whenever $k_1 \le k_2$ and $K$ is a total order. The $E_k$'s were elements of a $\sigma$-field, so if I have the sequence $k_n$, I can re-write this union as a countable union and show that the above set is in the $\sigma$-field. (That is what I needed.)
Any suggestions?