Intuition on Axiom of Completeness (Lower Bounds) In my textbook (Abbott) the Axiom of Completeness is defined as follows: 

Every nonempty set of real numbers that is bounded above has a least upper bound. 

The book then asserts that by this axiom, there is no need to discuss the case when a set is bounded below. The argument used is that if $A$ is bounded below, and we consider the  set of lower bounds to $A$, the infimum of $A$ is simply the supremum of the set of lower bounds. I'm a bit confused about this, because intuitively, it seems like this argument "approaches" the supremum/infimum from a different side on the number line (increasing towards it vs. decreasing towards it).  
 A: This does indeed take proof, but it is true - the idea roughly being that whether you go towards a number from above or from below, you're still going towards the same thing.
So let's prove (most of) it!
The statement we want to show is:

Suppose $A$ is a nonempty set of real numbers which is bounded below. Then $A$ has a greatest lower bound.

The book tells us to consider the set $$L=\{\mbox{lower bounds of $A$}\}.$$ We know that $L$ has an upper bound - namely, for any $a\in A$ (remember, $A$ is nonempty!) we have $a\ge l$ for all $l\in L$ (why?). So by completeness, we know that $L$ has a least upper bound, call it $g$.
We now claim that $g$ is in fact the greatest lower bound of $A$. To justify this, we need to show two things:


*

*$g$ is in fact a lower bound of $A$.

*If $h$ is a lower bound of $A$ then $h\le g$.
I'll do the first bullet, and leave the second for you as an exercise. Suppose $a\in A$; we want to show $g\le a$. Well, if $g>a$ then every lower bound of $A$ has to be $\le a$; hence, $a$ is an upper bound on the set of lower bounds of $A$. But this can't happen - $g$ is the least upper bound on the set of lower bounds of $A$, so ...
A: Suppose a nonempty set $S$ has a lower bound. Then the set $T$ of all of its lower bounds is nonempty and has an upper bound, since every member of $S$ is an upper bound of $T.$ Therefore $S$ has a least upper bound. The problem now is to show that the least upper bound of $S$ is the greatest lower bound of $T.$ Certainly there can be no lower bound of $T$ that is greater than the least upper bound of $S,$ because, being a lower bound, it would be a member of $S.$ So we then need to show only that it is a lower bound of $T.$ But if some member of $T$ were less than the least upper bound of $S,$ then there would be a smaller upper bound of $S,$ namely anything between those two.
