When can a set have an upper bound but no least upper bound?

Question

So I'm taking real analysis and have noted that one of the benefits of the Dedekind cut is that 'if one of the sets made has an upper bound it also has a least upper bound'.

I don't understand how a set can have an upper bound and no least upper bound, though.

Is what can lead to this declaring a set in the rationals that is bounded above by an irrational number? I don't see any other way for this to be true (and I don't know why you'd ever make that set, or similarly why it's a 'special' property of the Dedekind cut rather than the general case).

Thank you for your time,

Answer 1

$\sup S$ is shorthand for the least upper bound of $S$ in some set $T\supset S$ with respect to an order $\leq$ on $T$. However, we often omit $T$ and $\leq$ when they are obvious from the context.

Consider the case when $T=\mathbb{Q}$ and $\leq$ is the usual order. Since $\sup\left\{ x\in\mathbb{Q}:x^{2}<2\right\} $ has an upper bound but no least upper bound (in $T=\mathbb{Q}$), this supremum does not exist. Hence, $\mathbb{Q}$ does not satisfy the least upper bound property (a.k.a. Dedekind completeness).

However, if instead we had $T=\mathbb{R}$, the above supremum is $\sqrt{2}$.

Another example is given by @Hagen von Eitzen, where we look for $\sup \emptyset$. Regardless of whether $T=\mathbb{Q}$ or $T=\mathbb{R}$, in both of these spaces, $0$ is an upper bound of $\emptyset$. Furthermore, if $x$ is an upper bound of the empty set, so too is $x-1$. Therefore, there exists no least upper bound. We can write this as $\sup\emptyset = -\infty$ (in fact, if $T=\overline{\mathbb{R}}$, the extended reals with the obvious order, this is a precise statement).

Note that this does not contradict Dedekind completeness of the reals because Dedekind completeness only requires bounded nonempty sets to have supremums.

Answer 2

The set $$\left\{\sum_{k=1}^n\frac1{k!}:n\in\Bbb N\right\}$$ is contained in the rationals, and has upper bound. It is not difficult to show that $3$ is an upper bound.

But in the reals, the series converges to $e$, which is not rational. For a relatively easy proof, see here.

Answer 3

Say we have a continuous function $f :\Bbb{Q}\to \Bbb{Q}$ for which we know $f(a)<0<f(b)$ for fixed rational numbers $a<b$. Now set

$$ M:=\{x\mid a\leq x\leq b \text{ and } f(x)<0\}. $$

Clearly,$M$ is bounded above. Suppose that $M$ has a least upper bound $\alpha$. It is not hard to see that this entails $f(\alpha)=0$ (if you don't believe it, I will add the proof).

Thus, if $f$ has no root in $a\leq x \leq b$, then $M$ cannot have a least upper bound (in $\Bbb{Q}$). This happens e.g. for $f(x)=x^2-2$. In this case, the set $M$ reduces to the case for which you write "I don't know why you would ever make such a set".

We have just seen te answer: Such a set pops up naturally if we want to prove the intermediate value theorem which is a very natural and useful theorem but which fails in general for noncomplete sets like $\Bbb{Q}$.