Proof. Let
be the set of all totally ordered subsets of
.
is not empty, since the
empty set is an element of
.
Partial order 
by
inclusion. Let
be a
chain (of elements) in
. Being each totally ordered, the
union of all these elements of
is again a totally ordered subset of
, and hence an element of
, as is easily verified. This shows that
, ordered by inclusion, is inductive. The result now follows from Zorn's lemma.
