TITLE: Universal order type
QUESTION [10 upvotes]: Every countable order type, such as the countable ordinals, $\mathbb Z$, etc. can be embedded in $\mathbb Q$, so it is universal for countable order types. Is there a universal space for all linear orders of cardinality continuum? Or more generally, a universal space for all linear orders of any given cardinality?

REPLY [6 votes]: In Joel Hamkins' remark, "If CH fails, however, then we cannot necessarily get a saturated model of size $\aleph_1$," one can actually drop the "necessarily." The existence of a saturated linear order of size $\aleph_1$ is in fact equivalent to CH. If $(L, <)$ is a line, then one can show that $L$ is $\aleph_1$-saturated if and only if

For any countable increasing sequence $a_0 < a_1 < \ldots$ there exists $a \in L$ such that $a > a_n$ for all $n$.  Similarly, any countable decreasing sequence $b_0 > b_1 > \ldots$ is bounded below by some $b$. This says there are no countable cofinal sequences in $L$. 
If $a_0 < a_1 < \ldots < b_1 < b_0$ are sequences in $L$, then there exist points $a, b \in L$ such that $a_i < a < b < b_j$ for all $i, j$. This says that our line has no countable gaps. 
If $a_0 < a_1 < \ldots < x < \ldots < b_1 < b_0$, then there exist points $a, b \in L$ such that $a_i < a < x < b < b_j$ for all $i, j$. This says that no $x$ in $L$ can be approached by a countable sequence from either side. 

From this, one can prove that the order $(2^{\omega_1}, <_{lex})$ embeds into the Dedekind completion of any $\aleph_1$-saturated $L$. If such an $L$ has cardinality $\aleph_1$ (that is, if $L$ is saturated in the full sense), then we have that $(2^{\omega_1}, <_{lex})$ has a dense subset of size $\aleph_1$. But any dense subset of $(2^{\omega_1}, <_{lex})$ must have size at least $2^{<\omega_1}$. The equality $\aleph_1$ = $2^{<\omega_1}$ gives CH.  
The saturated order $L$ plays much the same role in $(2^{\omega_1}, <_{lex})$ that $\mathbb{Q}$ plays in $\mathbb{R}$. The combinatorial characterization of saturation given above allows one to naturally generalize Cantor's proof that any countable dense linear order without endpoints is isomorphic to $\mathbb{Q}$. Indeed, if $(L_1, <)$ and $(L_2, <)$ are two orders both satisfying conditions 1–3, and $|L_1| = |L_2| = \aleph_1$, then a Cantor style back-and-forth argument shows that $L_1 \cong  L_2$. One can also see why any such $L$ will be universal: the conditions 1–3 allow any countable partial embedding of an $\aleph_1$-sized linear order into $L$ to be extended without trouble to a full embedding. 
The generalization of the above argument to higher cardinalities also goes through, showing that GCH not only implies but is equivalent to the statement "for every cardinal $\kappa$, there exists a saturated linear order of size $\kappa^+$."