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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.

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    Show that the least upper bound of $X$ hold the condition of the completeness axiom about $c$. – 2017-01-24
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    Okay, so I know that every element of Y is an upper bound of X since all x≤y. The c would have to be the infirma of Y and suprema of X, right? Or no? – 2017-01-24
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    Okay. I'm having trouble coming up with the contradiction. I'm assuming that the contradiction arises from the fact that the set must have a unique least upper bound? I was going to assume that c wasn't the suprema of X, so there was an element x1 such that x1 is greater than or equal to all elements of X. I'm just confused about where the contradiction is coming from. – 2017-01-24
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    OKay, so like this? Suppose there does not exist a c such that x≤c for all elements x. However, by assumption, X is bounded above so there exists an element such that x≤this element. I guess the direct proof of that would just state that since X is nonempty and is bounded above by Y, there exists a unique least upper bound by the least upper bound principle. We can let c=least upper bound, so c=sup(x). Is that right? Then, would you do the same process for Y except use the greatest lower bound principle? But then I would have to use a different variable, say b, right? b=inf(Y)? – 2017-01-24
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    Forget the proof by contradiction, just prove, directly, that $x\le \sup(X)\le y$ for all $x\in X$ and all $y\in Y$. – 2017-01-24
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    Okay. By definition, x ≤ sup(X). Since the sup(X) is an element of x, then the sup(x) ≤ y for all y. Correct? – 2017-01-24
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    Alright. Thanks for explaining. I'm sort of lost now, but it's telling me to not have extended discussions in comments lol – 2017-01-24

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I will show about the strategy of this proof. We want to show that

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

  2. ...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

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 $y\in Y$.$\Box$

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    This makes perfect sense. It's what I was thinking, I just couldn't write it all out in my head. Thank you so much. – 2017-01-24
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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.

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    I'm understanding what you're saying. I'm just confused about how to formally write it out. By the least upper bound principle, there exists a unique least upper bound of X since X is nonempty. We know that all elements of Y are upper bounds of X since all x≤y. I just don't know where to go from there. Are you saying that c=inf(Y)=sup(X)? If so, is that always true? – 2017-01-24
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    i edited the post to explain the end - hope it helps – 2017-01-24
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    While what you said is true, i don't think you want to use the term inf or supp for this exercise, this should be done directly from definitions. – 2017-01-24
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    Okay, so this is what I'm thinking so far. Since X is non empty and is bounded above by Y, there exists a unique least upper bound for X. We can say that that boundary is the element c, so that x≤c for all x in X. Also, since Y is non empty and is bounded below by X, there exists a unique greatest lower bound for Y. We can say that boundary is the element b, so that b≤y for all y in Y. Now, how do I prove that b=c? Am I doing this right? – 2017-01-24