# Can the order relation on $\mathbb{R}$ be recovered from the topology?

It's well-known that the usual topology on $\mathbb{R}$ is induced by the order relation, since the open intervals $\{(a, b) : a < b\}$ are a base for the topology.

I am wondering if you can go the other way. If you took $\mathbb{R}$ with its collection of open sets, but you forgot the order, could you recover it?

Clearly the answer is no, because the reverse order $a <^\prime b \equiv_{\text{def}} b < a$ induces the same topology. But in this case $<^\prime$ and $<$ are order-isomorphic via $x \mapsto -x$.

So the question becomes, can you recover the isomorphism class of the usual order from just the collection of open sets. That is, if you have some uncountable set with the usual topology on the reals, can you create an order relation isomorphic to the usual one?

The order can be reconstructed up to isomorphism, using the fact that each point of $\Bbb R$ is a cut point.
Fix distinct points $p,q\in\Bbb R$ and arbitrarily define $p<q$. For any $x\in\Bbb R\setminus\{p,q\}$ exactly one of the following holds:
• $p$ and $q$ are in different components of $\Bbb R\setminus\{x\}$;
• $x$ and $q$ are in different components of $\Bbb R\setminus\{p\}$;
• $p$ and $x$ are in different components of $\Bbb R\setminus\{q\}$.
In the first case $p<x<q$; in the second, $x<p<q$; and in the third, $p<q<x$. If $y$ is any other point of $\Bbb R\setminus\{p,q\}$, you can use the same technique to determine whether $y<p$, $p<y<q$, or $q<y$. If $x$ and $y$ are not in the same one of the sets $(\leftarrow,p)$, $(p,q)$, and $(q,\to)$, the correct ordering between them follows immediately. If they are both in $(\leftarrow,p)$ or $(p,q)$, apply the same technique to $x,y$, and $p$ to determine their order. Finally, if they are both in $(q,\to)$, apply the technique to $x,y$, and $q$.