# DLO’s II – uniqueness of rationals

Now that we’ve proven that the rationals exist, we would like to know if it would make a difference had we picked another representative of the countable DLO’s without endpoints as our rationals. Would any candidate have been suitable or is our choice of $\left<\mathbb{Q},\prec_\mathbb{Q}\right>$ special? It turns out that there is no such difference, as given by the theorem:

Theorem (Cantor). Let $\left$ and $\left$ be countable dense linear orderings without endpoints. Then $\left\cong\left$.

Proof. Since both $A$ and $B$ are countable, we list them as $A=\{a_i\mid i<\omega\}$ and $B=\{b_i\mid i<\omega\}$. Our plan is to construct a series of partial isomorphisms $f_i:dom(f_i)\cong ran(f_i)$, which satisfies $f_i\subset f_{i+1}$ for all $i<\omega$ as well as $dom(f_i)\subset A$ and $ran(f_i)\subset B$. By doing this, it is straightforward to check that $f:=\bigcup_{i<\omega} f_i$ would grant the desired isomorphism $f:A\cong B$. We define the $f_i's$ recursively; start by defining $f_0:=\emptyset$. Now assume that $f_i$ has been defined. We define $f_{i+1}$ through a method called back-and-forth. Firstly, define $g:\left\to\left$ satisfying $dom(g)=dom(f_i)\cup\{a_i\}$.

Claim. There is a $j<\omega$ such that if $g=f_i\cup\{\left\}$, then $g:\left\cong\left$.

Proof of claim. If $a_i$ is the greatest element of $dom(g)$, then because $\left$ has no right endpoint, we can pick $b_j$ to be bigger than every $b\in ran(f_i)$. Likewise if $a_i$ is the least element of $dom(g)$. If neither, we can find $a_k,a_l$ such that $a_k\prec_A a_i\prec_A a_l$ and every other $a_j\in dom(g)$ is either greater than $a_l$ or less than $a_k$, which is possible due to $dom(g)$ being finite. Then $f_i(a_k)\prec_B f_i(a_l)$ with no $b_j\in ran(f_i)$ satisfying $f_i(a_k)\prec_B b_j\prec_B f_i(a_l)$. Since $\left$ is dense, we can pick a $b_k\in B$ such that $f_i(a_k)\prec_B b_k\prec_B f_i(a_l)$. Now set $b_j:=b_k$.

$\lozenge$

Now define $h:\left\to\left$ satisfying $ran(h)=ran(g)\cup\{b_i\}$. Then with the same argument, mutatis mutandis, as in the previous claim, we find that there is $a_j\in A$ satisfying that if $h=g\cup\{\left\}$ then $h:\left\cong\left$. Now, set $f_{i+1}:=h$. Then by the previously stated argument, $f:=\bigcup_{i<\omega}f_i:\left\cong\left$.

$\blacksquare$

As we’re moving on from the rationals to the reals, one might wonder how to characterize the reals from the rationals exactly. There is a notion of completeness to be defined; that is, the notion that captures, say, $\sqrt{2}$. Or, can it even be defined directly? Interestingly enough, it can be proven that this notion cannot be defined by direct means (that is, through a first-order sentence). To explain why, we state the following corollary without proof:

Corollary. Let $\left$ and $\left$ be two DLO’s without endpoints. Then every sentence $\sigma$ (in the language of $\{<\})$ is true in $\left$ iff it is true in $\left$.

If we now assume that $\sigma$ was a sentence such that $\sigma$ was true in some structure $\left$ iff $\prec_M$ was a complete order, then by the corollary, either every DLO without endpoints is completely ordered or none is. But since we know that $\left<\mathbb{Q},\prec_\mathbb{R}\right>$ is a DLO without endpoints which aren’t complete, as well as $\left<\mathbb{R},\prec_\mathbb{R}\right>$ is a DLO without endpoints which is complete, we arrive at a contradiction. So “completeness” can thus not be defined through first-order logic!. As for the question how we can be sure that the reals even exist, you’ll have to be patient for the next blog post!