Prove that the Completeness Axiom follows from the Least Upper Bound Principle. We've proved that the Least Upper Bound principle can be proven from the completeness axiom, however now we have to prove that the completeness axiom can be proved from the least upper bound principle (showing that the statements are equivalent). For definitions, the Least Upper Bound principle states, "Every nonempty set of real numbers that is bounded above has a unique least upper bound." The completeness axiom is as follows, "If X and Y are nonempty subsets of R such that x≤y for all x∈X and y∈Y, then there exists c ∈ R such that x≤c ≤y for all x∈X and y∈Y."
I'm confused as to how to proceed. I'm assuming that there are two sets X and Y that are nonempty and such that x≤y for all x and y. I'm just not sure where to go from there.
 A: Since $Y$ is non-empty, there is $y\in Y$ which is an upper-bound of $X$ - so the upper bound principle applies. This gives you a unique minimal upperbound of $X$. This is your $c$, now you need to prove it satisifies $x\leq c$ and $c\leq y$ for all $x,y$ in their respective set.
edit: $c$ has exactly two properties: First, it is an upper bound of $X$, so $x\leq c$ for all $x$. It also is the smallest number with that property - but all $y$ satisfy $c\leq y$ right? So you get $c\leq y$ for all $y$ as well.
A: I will show about the strategy of this proof. We want to show that 


*

*The existence of a least upper bound on bounded sets by above imply that...

*...if we have have two sets $X$ and $Y$ such that $x\le y$ for all $x\in X$ and all $y\in Y$ then exists a $c$ such that $x\le c\le y$ for all $x\in X$ and all $y\in Y$
Then we want show that for $S:=\sup(X)$ holds 
$$x\le S\le y,\quad\forall x\in X,\forall y\in Y\tag{1}$$ provided that $x\le y$ for all $x\in X$ and all $y\in Y$.
Because $S$ is the supremum of $X$ then the LHS of (1) is clear, that is, $x\le S$ for all $x\in X$ by the definition of supremum. Then to conclude the proof we must show that $S\le y$ for all $y\in Y$.
Then we show that if $S\le y,\forall y\in Y$ is not true then it must be the case that exists some $y_0\in Y$ such that $y_0<S$. But then this would imply that $S$ is not the supremum of $X$, that is, $x\le y_0<S$ for all $x\in X$, a contradiction.
Hence if $S$ is the supremum of $X$ it must be the case that $S$ is a lower bound of $Y$, that is, that $x\le S\le y$ for all $x\in X$ and for all $y\in Y$.$\Box$
