Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been stuck on this and although I have gone to professors for advice, I can't grasp it. Here I tried re-doing my proof and I can bet it is certainly wrong, but I am burnt out. I would like to know not only my mistakes in the proof, as I have looked at it for an hour and can't seem to go anywhere; I would also like a throrough explanation of this in layman terms, because that is exactly what I need right now. What I know is that the infimum of a set bounded from above is the same as the supremum of the set bounded from below. I have no idea on how to construct the argument thoroughly.

Prove that if the ordered set $S, <$ satisfies the greatest lower bound property, then it satisfies the least upper bound property.

Let $B$ be a nonempty subset of $S$ that is bounded below. Let $a \in A$ be the set of all lower bounds of $B$. By definition $A$ is nonempty. There also exists an element $b \in B$ for all $a$ such that $a \leq b$, so $A$ is bounded above (by $b$). Since $S$ satisfies the greatest lower bound property, $inf \hspace{2 pt} B = \alpha$ exists for all $a$ such that $a \leq \alpha \leq b$. If $\alpha < a$ otherwise, then $a$ is not a lower bound of $B$. But if $\beta < \alpha$, then $\beta \not\in B$, since $\alpha$ bounds $B$ from below. Thus $a \leq \alpha$, and so $A$ has a least upper bound in $S$. QED.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Your proof is mostly OK, except for the fact that it's not a proof of your original statement.

You are trying to prove that any set with greatest lower bounds automatically has least upper bounds. In your proof, you fix an arbitrary subset $B$ of a set $S$ with greatest lower bounds, and then show that a completely different set $A$ has a least upper bound. Instead, you need to show that $B$ has least upper bounds. Let's see how we can alter your proof so that it proves your original statement.

Let $B$ be a nonempty subset of $S$ that is bounded below.

This is (almost) a great place to start. If you're trying to prove something about all objects in a class, it is always good to start by fixing an arbitrary one. But here, you want to show that any non-empty bounded above subset of $S$ has a least upper bound, so you should be fixing $B$ to be a non-empty bounded above subset of $S$.

Let $a\in A$ be the set of all lower bounds of $B$.

First of all, I think you meant to say 'Let $A$ be the set of all lower bounds of $B$'. But even this is not what you want to do. Fixing $A$ to be the set of lower bounds of $B$ can't help you at all, since the only interesting thing you can say about $A$ is that it has a greatest element ($\inf B$), which is equivalent to $B$ having a greatest lower bound. You certainly can't apply the greatest lower bound property to $A$ - you don't even know that $A$ is bounded and $\inf A$, if it exists, is certainly unrelated to $B$.

At some point, you have to apply the greatest lower bound property to some set $C\subset S$, since you can't conclude that $S$ has the least upper bound property if you don't know that it has the greatest lower bound property (e.g., $\mathbb Q$ does not have the greatest lower bound property), but you have to apply it to a set that you know has a greatest lower bound; i.e., a set that is bounded below. Furthermore, it's going to have to be a set of elements that are $\ge$ the elements of $B$; otherwise, it'll have nothing to do with upper bounds of $B$ at all.

The solution, of course, is to fix $C$ to be the set of upper bounds for $B$. Now we're trying to show that $C$ has a least element.

There also exists an element $b\in B$ for all $a$ such that $a\le b$, so $A$ is bounded above (by $b$).

First of all, we aren't interested in saying that $A$ (or $C$) is bounded above, since being bounded above doesn't tell us anything. Being bounded below, of course, tells us that a set has a greatest lower bound. So we want to show that $C$ is bounded below. Secondly, it is ambiguous whether you mean that there is some $b$ that works for all $A$ or whether there is a separate $b(a)$ for each $a$. Both are true, but only the first implies your conclusion.

I'd prefer to write, 'Fix an arbitrary $b\in B$. Then for each $c\in C$, $c$ is an upper bound for $B$ so, in particular, $b\le c$. So $C$ is bounded below by $b$.'

Since $S$ satisfies the greatest lower bound property, $\inf \hspace{2 pt} B = \alpha$ exists for all $a$ such that $a \leq \alpha \leq b$.

This is where you started to go off the rails. Knowing that $B$ has a greatest lower bound can tell you nothing about whether it has a least upper bound. For example, the set $\{3,3.1,3.14,3.141,3.1415,3.14159,\dots\}$ has a greatest lower bound but no least upper bound in $\mathbb Q$.

I think I've given you enough clues for you to finish the proof on your own. For reference, here is my proof of the statement:

Let $B$ be a non-empty subset of $S$ that is bounded above, and let $C$ be the set of all upper bounds for $B$. Fix $b\in B$; then, for each $c\in C$, $c$ is an upper bound for $B$ so, in particular, $b\le c$. So $C$ is bounded below by $b$, which means that it has a greatest lower bound $\gamma$. Since every $b\in B$ is a lower bound for $C$, $\gamma\ge b$ for each $b\in B$. So $\gamma$ is an upper bound for $B$, and is clearly the least upper bound since it is a lower bound for the set of upper bounds. $\Box$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.