What makes R different than Q so that R possesses least upper bound property. Q is also an ordered field.

Definition : An ordered field $(F, +, ·, ≤)$ is said to have the least upper bound property provided that any non-empty subset $A$ of $F$ that is bounded above has a least upper bound.

Lets take $(0,1)$ isn't 1 an upper bound? in $Q$

  • 6
    $\begingroup$ Sure. The problem is a set like $\{ x : x^2 <2 \} $. $\endgroup$
    – Ian
    Sep 29 '16 at 18:55
  • 3
    $\begingroup$ There exist sets for which no rational least upper bound exists (see the comment by @Ian). Ultimately, it comes down to "why would you expect rational least upper bounds to always exist?" and there isn't really an answer to that. From a formal perspective, a better question might be "why does $\mathbb{R}$ possess the least upper bound property?", to which the answer is perhaps slightly boring: it does because we decided it should (based on experience with other mathematical objects), so we made the formal definition of $\mathbb{R}$ with that property specifically in mind. $\endgroup$
    – Will R
    Sep 29 '16 at 19:29
  • $\begingroup$ Any ordered field with the LUB property is isomorphic to $\mathbb R.$ $\endgroup$ Sep 30 '16 at 4:06

Even though ${\mathbb Q}$ is "dense in itself" it becomes evident in mathematical praxis that there are important numbers missing that we need, and are convinced "to be there", and which are not available in ${\mathbb Q}$. It's not just special values like $\sqrt{2}$, or $e$ and $\pi$, but in fact any infinite decimal which is not ultimately periodic, and there are many of these. The completion process invented by Dedekind fills these fissures with a sort of "glue" so that in the end a perfect system results, which is not only "order complete", but allows of all the operations which we are so fond of from primary school arithmetic of fractions.

  • $\begingroup$ When looking at LUB property, In $Q$ I believe we can always find a rational number which is a least upper bound for instance. But we can not find a real number. So I conclude least rational upper bound exists but least upper bound does not exist $\endgroup$
    – user346936
    Oct 8 '16 at 12:53

There are two separate questions here:

Why doesn't $\mathbb{Q}$ have the least upper bound property?


What about $\mathbb{R}$ makes it have the least upper bound property?

First question:

Note that for $\mathbb{Q}$ to have the LUB property, it's not enough for some sets to have least upper bounds (as you note, $(0,1)\cap\mathbb{Q}$ has a least upper bound) - every set needs to have a least upper bound. A good counterexample is given in the comments: $A=\{q\in\mathbb{Q}: q^2<2\}$ has no least upper bound, but is bounded above (e.g. $7$ is greater than every element of $A$). You can construct a number of other examples along similar lines.

This shows that $\mathbb{Q}$ doesn't have the LUB property; as to why it doesn't, one good reason is that it is just too small (even though it's infinite)! It's a good exercise to show that no countable ordered field can have the LUB property.

Second question:

It turns out that this question is slightly backwards. In fact, $\mathbb{R}$ is constructed specially so that it has the LUB! Basically, the idea is to take $\mathbb{Q}$ and "fill in" all the gaps: for every set $A\subseteq\mathbb{Q}$ which is bounded above, we want to add an upper bound of $A$; this gives us a bigger set of numbers, which (with a bit of thought) we can turn into an ordered field, and (with a bit more thought) show satisfies the LUB property.

This larger field - the completion of $\mathbb{Q}$ - is the real numbers! There are several ways to construct the real numbers - Cauchy sequences, Dedekind cuts, and others - but they all come down to starting with $\mathbb{Q}$ and "filling in" the gaps in some reasonable way. So the reason $\mathbb{R}$ has the LUB property, is just: we built it for precisely that purpose!


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