TITLE: Can every set be expressed as the union of a chain of sets of lesser cardinality?
QUESTION [5 upvotes]: If a set $S$ has countably many elements $\{x_n\}$, it can be expressed as a union of a chain of finite sets
$$ \{x_1\} \subset \{x_1,x_2\} \subset $$
But what about a set of arbitrary cardinality $S$? Can we express it as a union : 
$$S=\bigcup_{i\in I}S_i$$
where $I$ is a totally ordered set of some cardinality and each $|S_i|$  has cardinality less than $|S|$?

REPLY [6 votes]: Yes, assuming the axiom of choice. Suppose that $S$ is infinite. Then $|S|=\kappa$ for some infinite cardinal $\kappa$ so there is a bijection $f:\kappa\to S$. For each $\alpha<\kappa$ let $S_\alpha=f[\alpha]=\{f(\xi):\xi<\alpha\}$; since $\kappa=\bigcup_{\alpha<\kappa}\alpha$, clearly $S=\bigcup_{\alpha<\kappa}S_\alpha$, and $S_\alpha\subseteq S_\beta$ whenever $\alpha\le\beta<\kappa$.
Without the axiom of choice it need not be possible. In particular, it cannot be done with a sequence of proper subsets if $S$ is an amorphous set. (As Asaf points out in the comments, it’s possible with a two-element sequence if one is $S\setminus\{p\}$ for some $p\in S$, and the other is $S$ itself.)