Yes. By Hartogs' theorem, there is an ordinal that has no injection into $R$. The minimal such ordinal is the smallest well-ordered cardinal not injecting into $R$. It is naturally well-ordered by the usual order on ordinals. None of this needs AC.
One can think very concretely about the order as follows: Consider all subsets of $R$ that are well-orderable. By the Axiom of Replacement, each well-order is isomorphic to a unique ordinal. Let $\kappa$ be the set of all ordinals that inject into $R$ in this way. One can show that $\kappa$ itself does not inject into $R$, and this is the Hartogs number for the reals.
More generally, of course, there is no end to the ordinals, and they are all canonically well-ordered, without any need for AC.
But in terms of the remarks in your "motivation" paragraph, there are linear orders that do not map order-preservingly into $R$ that are not larger than $R$ in cardinality. For example, the ordinal $\omega_1$ cannot map order-preservingly into $R$, since if it did so, then there would be an uncountable family of disjoint intervals (the spaces between the successive ordinals below $\omega_1$), but every such family is countable by considering that the rationals numbers are dense. Another way to see this is to observe that the real line has countable cofinality for every cut, but $\omega_1$ has uncountable cofinality.
Lastly, there is a subtle issue about your request that the order by "larger than $R$". The examples I give above via Hartog's theorem are not technically "larger than $R$", although they are not less than $R$ in size. The difficulty is that without AC, the cardinals are not linearly ordered, and so these two concepts are not the same. But you can turn the Hartogs argument into a strict example of what you requested by using the lexical order on $R\times\kappa$, where $\kappa$ is the Hartog number of $R$. This order is strictly larger than $R$ in size, and it is canonically linearly ordered by the lexical order.
Something very close to François' conditions achieves the
desired if-and-only-if version of the theorem for partial
orders, providing an induction-like characterization of the complete partial orders, just as Pete's theorem characterizes the complete total orders.
Suppose that $(P,\lt)$ is a partial order. We say that it
is complete if every subset $A$ has a least upper bound
$sup(A)$ and a greatest lower bound $inf(A)$. This implies
that $(P,\lt)$ has a least and greatest element, and in
this case, it is easy to see that completeness is equivalent to the assertion that
every set has a greatest lower bound (the least upper bound of a set with upper bounds is the greatest lower bound of its upper bounds). The complete partial
orders are exactly the complete
lattices.
For the purpose of this question, let us define that
$S\subset P$ is inductive if the following occur:
- (1) $S$ is downward closed: $y\lt x\in S\to y\in S$;
- (2) $S$ has no largest element, except possibly the largest element of $P$;
- (3) If $d=sup(A)$ for some $A\subset S$, then $d\in S$.
In (2), by a largest element, I mean an element that is
larger than all other elements (in distinction with
maximal element, a weaker concept). Conditions (2) and
(3) are both slightly weaker than François' conditions. The
difference in (2) is that we no longer assume that a
condition $x\in S$ can be extended in any particular
direction, and the difference in (3) is that we are not
assuming here that $P$ is complete, but only that $S$
contains the suprema of its subsets, when these suprema exist.
Lastly, define that a partial order $(P,\lt)$ has
partial-order induction if the only inductive set is all
of $P$.
Theorem. For any partial order $(P,\lt)$, the
following are equivalent:
- $(P,\lt)$ is complete.
- $(P,\lt)$ has least and greatest elements and satisfies partial-order induction.
Proof. For the forward implication, suppose $(P,\lt)$ is
complete. It follows that $(P,\lt)$ has least and greatest
elements, since these are the sup and inf of the emptyset.
Suppose that $S\subset P$ is inductive. By (3) we know
$sup(S)\in S$, which would make it the largest element of
$S$, contrary to (2), unless $sup(S)$ is largest in $P$, in which case
$S=P$ by (1). So $(P,\lt)$ has partial-order induction.
Conversely, suppose that $(P,\lt)$ has least and greatest
elements and satisfies partial-order induction. We want to
prove completeness. Suppose $B\subset P$ is nonempty and
has no greatest lower bound. Let $S$ be the set of lower
bounds of $B$. This is closed downwards, and so (1) holds.
Since $B$ has no greatest lower bound, it follows that $S$
has no largest element, and so (2) holds. Finally, if
$A\subset S$ and $sup(A)$ exists in $P$, then it remains a
lower bound of $B$, since every element of $B$ is an upper
bound of $A$ and $sup(A)$ is the least upper bound of $A$.
Thus, (3) holds and so $S$ is inductive, contrary to the
fact that it contains no elements of $B$. Thus, $B$ must
have a greatest lower bound after all, and so $(P,\lt)$ is
complete. QED
Note that when $(P,\lt)$ is a total order, then these
concepts reduce to the conditions Pete mentions in the
question (adding a least and greatest element if
necessary). Thus, this theorem seems to be the desired
generalization.
(I have deleted my other answer to the question, as it was
based on a misunderstanding of the question.)
Best Answer
Let's think about the countable case like this: think of the binary tree $2^{\lt\omega}$, which has size $\omega$, but has $2^\omega$ many branches. Each branch describes a cut in the natural lexical order on the nodes, and so we have a countable linear order with $2^\omega$ many cuts.
So consider a cardinal $\kappa$ and the tree $2^{\lt\kappa}$, which admits a similar lexical order. If this tree has size $\kappa$, then we get a linear order of size $\kappa$ with $2^\kappa$ many cuts. And such a family of cuts turns into a chain just as you point out in the question.
Thus, if $B$ has size $2^{\lt\kappa}$, then we can find $A$ of size $2^\kappa$. In particular, whenever $2^{\lt\kappa}=\kappa$, then $P(\kappa)$ has a chain of size $2^\kappa$, as you desire.
In particular, under the GCH, the phenomenon will occur for every infinite cardinal, since GCH implies $2^{\lt\kappa}=\kappa$ for all infinite cardinals $\kappa$.
Lastly, let me point out that this argument method can be turned into a characterization. Namely, the sets $B$ for which there is an $A$ as you desire are exactly the sets $B$ of size $\kappa$ for which there is a linear order on $\kappa$ with $2^\kappa$ many cuts. The one direction we've established, and for the converse, when there is such a chain $A$ in $P(B)$ of that size, then we may place a pre-order on $B$ according to the order in which points are added to sets in the chain (and extend this to a linear order). Each element of $A$ gives a cut in this order, and so we have a linear order on $\kappa$ with $2^\kappa$ many cuts.
So the phenomenon occurs for exactly those sets $B$ of size $\kappa$ for which there is a linear order on $\kappa$ with $2^\kappa$ many cuts.