TITLE: Is the set of real numbers the largest possible totally ordered set?
QUESTION [27 upvotes]: Because I find any totally ordered set can be "lined up" in a straight line, I'm guessing that the set of all the real numbers is the biggest totally ordered set possible. In the sense that any other totally ordered set is isomorphic (order-preserving) to a subset of real numbers. Is that right?

REPLY [6 votes]: The real line is the largest order separable linearly ordered set. That is, if $(L,\preceq)$ is a linearly ordered set such that a countable set $C\subseteq L$ exists satisfying that for all $x,y\in L$ with $x\prec y$ there is $c\in C$ with $x\preceq c\preceq y$, then there exists a function $f:L\to\mathbb{R}$ such that $f(x)<f(y)$ iff $x<y$ for all $x$ and $y$ in $L$. The existence of such a set $C$ is also necessary. A proof of these facts is given here. 
Here is yet another counterexample without order separability: Let $(W,\preceq)$ be an uncountable well-ordered set. Let $W'$ be $W$ without the maximum of $W$ if such a maximum exists. For each $w\in W'$, let $S(w)$ be its immediate sucessor given by $S(w)=\min\{w'\in W:w'\succ w\}$. It is easily seen that $x\prec y$ implies $x\prec S(x)\preceq y\prec S(y)$. So if a strictly increasing function $f:W\to\mathbb{R}$ would exist, the uncountable family of intervals $\{(f(w),f(S(w))):w\in W'\}$ would be disjoint, which cannot be for the same reason pointed out by Haskel Curry in his answer.