For all x in the set of positive rational numbers, L(x) is a lower cut if L(x) ;= {a/b is in the set of positive rational numbers such that a/b < x}.
- L is a non-empty subset of Q+
- L is bounded above
- L has no maximum element
- If a/b is in the set L and 0 < c/d < a/b then c/d is in the set L
A set with these four properties is a lower cut of Q+. We know that L is not an empty set because from another proof that I wrote out before, if a/b < c/d are rational numbers then the set of rational numbers x/y such that a/b < x/y < c/d is not finite. Therefore there exists at least one element and in fact there is an infinite set of elements in L because of this proof. L is bounded above by x, whatever positive rational number that is decided upon. L has no maximum element because there is an infinite set of elements in the set of L that is between 0 and x. As for the last requirement for a set to be a lower cut of the positive rational numbers, it seems quite obvious that if a/b is in the set L and 0 < c/d < a/b then c/d is in the set L.
The function x |→ L(x):Q+ → R+ is one-one. For every element in the set of L, there is only one positive real number that matches the element in L.