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A Completeness principle in mathematical analysis is a principle by the help of which we can establish (prove) the completeness of an ordered field( About whom I am going to post one question later). (If I am wrong in defining then It will make me immensely happy if someone corrects it). Cauchy's completeness principle, Dedekind's principle, Cantor's principle, Weierstrass's principle, greatest lower bound principle and least upper bound principle are the completeness principles we come across in analysis. We can take anyone of this as an axiom and derive the remaining. Now let us consider proving the Least upper bound principle taking the greatest lower bound principle as an axiom. My proof goes in the following way.

Let us consider an ordered field $F$. Now taking the glbp as an axiom I can say that any non empty subset, $P$, of this ordered field definitely posses an element $k$ ($k$ belongs to $P$) such that all the elements of $P$ are greater than or equal to $k$. Any element of the field less than $k$ serves as a lower bound for $P$. Now let us consider the set $V = \{ -x\mid x\text{ belongs to }P\}$. Now $-k$ serves the least upper bound for $V$. By GLBP axiom there exists a greatest lower bound for $V$ as $V$ is also a non empty subset of $F$. By similar arguments I can assert $P$ also has a least upper bound (which is the negative of the GLB of $V$}. Hence for every non empty subset of $F$ , there exists a least upper bound.

In this way I obtained LUB property from GLB property. But, I have not yet found any analysis book which carries on this task along this lines. It obviously forces me to question : Does my way of proving lack rigor? If yes, then where?

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You're stating the axiom incorrectly. It is not "every nonempty set" that has a greatest lower bound. (For example $\mathbb Z\subset\mathbb R$ has no glb in $\mathbb R$). The correct axiom says that _if_ $P$ is nonempty and $P$ is known to have some lower bound, then $P$ has a greatest lower bound. (It is possible that you're being confused by the axiom for a "well-ordered" set, which sound a bit more like the what you're describing? Of course if you've never heard of well-orderings that can't be it). –  Henning Makholm Sep 19 '11 at 18:38
    
Also, the lower bound does not have to be a member of $P$ itself. For example, the open interval $(0,1)$ has $0$ as a lower bound (in fact this is the greatest lower bound), but $0$ is not itself a member of the interval. –  Henning Makholm Sep 19 '11 at 18:39
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up vote 5 down vote accepted

First: the "least upper bound" and "greatest lower bound" principles are not at all about ordered fields, they are about partially ordered sets; an ordered field is necessarily a partially ordered set, but there are plenty of partially ordered sets that satisfy the least upper bound and greatest lower bound properties without being fields.

Second: You are trying to prove the least upper bound principle, assuming the greatest lower bound principle holds. Let's be clear about what these principles are:

We say a partially ordered set $(X,\leq)$ satisfies the Greatest Lower Bound Principle if and only if every nonempty subset $P$ of $X$ that has a lower bound has a greatest lower bound; that is, if there exists $x\in X$ such that $x\leq p$ for all $p\in P$, then there exists $g \in X$ such that:

  • $g\leq p$ for all $p\in P$; and
  • if $x\in X$ is such that $x\leq p$ for all $p\in P$, then $x\leq g$.

As for the Least Upper Bound Principle, we have the dual definition:

We say a partially ordered set $(X,\leq)$ satisfies the Least Upper Bound Principle if and only if every nonempty subset $P$ of $X$ that has an upper bound has a least upper bound; that is, if there exists $x\in X$ such that $p\leq x$ for all $p\in P$, then there exists $\ell\in X$ such that:

  • $p\leq \ell$ for all $p\in P$; and
  • if $x\in X$ is such that $p\leq x$ for all $p\in P$, then $\ell\leq x$.

Now, let's look at your argument. You are assuming that your field $F$ satisfies the Greatest Lower Bound Principle, and you take a subset $P$ of $F$. You assume it is nonempty, and that there is an element $k$ that is an upper bound. You incorrectly assert that this element $k$ will be in $P$: you don't know that, and you don't need that.

Then you consider the set $-P=\{-p \mid p\in P\}$, and note that $-k$ is a lower bound for $-P$ (you incorrectly assert it is a greatest lower bound; you don't know that). You then deduce that $-P$ has a greatest lower bound (correct). And you stop there. But you need to show that $P$ has a least upper bound, and you have not done so. (Of course, you probably meant to take the greatest lower bound $\ell$ of $-P$, and then show that $-\ell$ is a least upper bound for $P$; but you have not done so). So your argument is incomplete.

Now, there is nothing wrong with the idea behind your argument (provided you fix the errors and fill in the gaps). The reason you may not want to do this is that the fact that the Least Upper Bound Principle and the Greatest Lower Bound Principles are equivalent is true for any partially ordered set. In a partially ordered set, the argument that relies on "multiplying by $-1$" will not in general work, because that operation will not make sense. So rather than use a proof that only works for a certain kind of partially ordered set, we can use a proof which is not any more complicated, but which works in general. And moreover, a proof that highlights that the least upper bound of a set is in fact a greatest lower bound of a different set (and likewise for greatest lower bounds).

Namely:

Suppose $X$ satisfies the greatest lower bound principle, and we want to prove it satisfies the least upper bound principle. Let $P$ be a nonempty set that has at least one upper bound. We need to show it has a least upper bound. Let $B=\{b\in X\mid b\text{ is an upper bound for }P\}$. By assumption, $B$ is not empty. Moreover, since $P$ is not empty, there is a $p\in P$, and so $p\leq b$ for all $b\in B$. That means that $B$ is a nonempty set that is bounded below. By the Greatest Lower Bound Principle, $B$ has a Greatest Lower Bound; call it $g$. We claim that $g$ is also a least upper bound for $P$.

First, note that if $p\in P$ we have $p\leq b$ for all $b\in B$. By the second property of the greatest lower bound, we conclude that $p\leq g$. Therefore, for every $p\in P$ we have $p\leq g$, so $g$ satisfies the first property necessary to be the least upper bound of $P$. To verify it satisfies the second property, if $x\in X$ is such that $p\leq x$ for all $p\in P$, then $x\in B$ by definition. And since $g$ is the least upper bound of $B$, then the first property of the least upper bound tells us that $g\leq b$ for all $b\in B$, and in particular that $g\leq x$. Thus, if $x$ is an upper bound for $P$, then $g\leq x$. This verifies that $g$ satisfies the second property of the least upper bound for $P$. We conclude that $g$ is indeed the least upper bound for $P$.

So we have shown that if $X$ satisfies the greatest lower bound principle, then any nonempty subset that is bounded above has a least upper bound; that is, $X$ also satisfies the least upper bound principle.

A symmetric argument (or applying the above argument to the partially ordered set $(X,\leq^{\rm op})$) shows that if $X$ satisfies the least upper bound principle, then it satisfies the greatest lower bound principle. $\Box$

Once you fill in all the details into your argument, you'll see that it is no shorter and no less difficult than the above one. But the above one has the virtue of not needing any of the field properties that you use (that $a\leq b$ if and only if $-b\leq -a$, for example), only the properties of order, of upper and lower bounds, and the corresponding principle. So it is a nicer proofs, because it assumes less, concludes the same, and is not any harder or longer than the proof you suggest.

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