Proof that the Order topology on $\mathbb{R}$ has the same basis as the Euclidean topology I want to prove that the Order topology on $\mathbb{R}$ has the same basis as as the Euclidean topology on $\mathbb{R}$.
Assume that the only thing we know about the order topology is that it has the open rays as its subbase. My problem is that at some point I have to make some kind of claim that the base must consist only of sets of the form:


*

*$(a,b)$

*$(c,\infty)$

*$(-\infty,d )$


No matter how I do this, it becomes messy. It is easy to prove, but it becomes messy. How can I prove this tidily, without having to write out a whole load of combinations of interesections of open rays?
 A: Given a subbasis for a topological space, you can construct a basis by taking all finite intersections of the subbasis elements.  It's not hard to show that the intersection of two open rays is an either empty, an open ray, or an open interval. Moreover, the interesection of an open interval with an open ray is either empty or another open interval.  Since we're only considering intersections of finitely many open rays, it follows by induction that the elements of our basis are exactly the elements you claim.
Edit:  We want to show that the intersection of two open rays is either empty, an open ray, or an open interval.  It's really not hard to do directly.  First suppose the rays point in the same direction.  WLOG they are $(a,\infty)$ and $(b,\infty).$  Then it's clear that one ray contains the other, so the intersection is just $(\max(a,b),\infty).$  If the rays point in opposite directions, then they can be written as $(a,\infty)$ and $(-\infty,b).$ If $b\leq a,$ they obviously don't intersect.  Otherwise, their intersection is $(a,b).$ This is maybe a little hard to digest in words, but everything will become clear if you draw a picture for each case.
