Math 104 HW Feb 24 #6 Redone

Since the midterm is already over, I guess you probably don't want to be bored so I will keep it brief:
Suppose U is an open set in R, for x in U, define the orbit of x or orb(x) as the maximal open interval containing x that is a subset of U. Here "maximal" means it is a superset of all open intervals satisfying the two conditions (contain x and is a subset of U).
Before we go on, we need to show such a "max" actually exists: let A={s in R|(s,x) is a subset of U} and B={t in R|(x,t) is a subset of U}, and let a=inf(A) and b=sup(B), then (a,b) is the "max" we want.
OK I will supply some details:
First, it is an open interval containing x.
Second, it is a subset of U. Suppose not, then some y between a and b and different from x (since x is assumed to be in U) is not in U. Suppose y>x, then since y is not an upper bound of B, there exist some t>y such that (x,t) is a subset of U, but this implies y is in U, contradiction. y<x similar. So (a,b) must be a subset of U.
Third, it is max: suppose (c,d) contains x and is a subset of U, then if c<a, then c doens't belong to A, so (c,x) contains something that is not in U, contradiction. So c>=a. Similarly, d<=b.
Fourth, the max is unique by set properties. So we are justified to say "the maximal".
Now clearly U is the union of the orb(x)'s for all x in U. We need to show the orbits are disjoint:
First, show orb(x)=orb(y) if y is in orb(x): orb(x) contains y implies orb(x) is a subset of orb(y) since y is the "max" containing y, but this implies x is in orb(y), so by the same reason as above, orb(y) is a subset of orb(x). So orb(x)=orb(y).
Now if orb(x) and orb(y) have z in common, then orb(x)=orb(z)=orb(y). In other words, if two orbits are not disjoint, then they are the same. Done!