Supremum and infimum of a rational set bounded by irrational numbers

Question

Let $S = \left\{x \in \mathbb{Q} \mid 1 \leqslant {x}^2 \leqslant 29 \right\}$

What can we say about the supremum and infimum of this set? Would it be non-existent?

Would it be correct to say the following?

Suppose $ \sup S < \sqrt{29} $ then $ \exists x \in S $ such that $ x > \sup S$

Therfore $ \sqrt{29} \leqslant \sup S$

But on the other hand $\sup S > \sqrt{29}$ then $\exists x \in \mathbb{Q}$ such that $ x < \sup S$

Therefore $\sup S \leqslant \sqrt{29}$

Therefore $\sup S $ can only be equal to $\sqrt{29}$

However $\sqrt{29} \notin \mathbb{Q}$ therefore $\sqrt{29} \notin S$

Therefore there can not be a supremum to this set.

(I am assuming the supremum of a set must belong to the set. Is this correct?)

Answer 1

The supremum, or least upper bound, of a set need not be elements of the set (and for open sets they never are).

Example. Let $T = (0,1)$. Then $\sup T = 1 \not \in S$. That's it.

So in your set, $\sup S = \sqrt{29} \not \in S$. That's it.

....

Okay, but listen up, this is important. If your universal space is $\mathbb Q$ or some other space. Then $\sqrt{29}$ will not exist in your space. So we would say in the space of the rationals $S$ does not have a least upper bound.

We say $\mathbb Q$ does not have the "least upper bound property" because it is possible for there to exist sets that are bounded above but that do not have a supremum. However if your space is a different set of points and is such a space where every set that is bounded above must have a least upper bound. Then we say that space has the "least upper bound property".

Now the entire point of the real numbers and the entire reason for the existence of irrational number is that $\mathbb Q$ does not have the least upper bound property, but $\mathbb R \supset \mathbb Q$ is an extension of $\mathbb Q$ that DOES have the least upper bound property.

And the irrational numbers are nothing more or less than the least upper bounds of bounded sets of rational numbers that do not have a rational least upper bound.

So your set $S$ does not have a rational least upper bound. So it must have an irrational least upper bound. And it does. It is $\sqrt{29}$.

And every irrational number (indeed every real number) is the least upper bound of some set of rational numbers.

.....

Indeed we can actually say: Let $x \in \mathbb R$ and let $S =\{q\in \mathbb Q| q<x\}$; then $\sup S = x$.

Period. That is always true.

Answer 2

Let's consider the set of rational numbers $$\{ r \in \mathbb{Q} \mid r \ge 1 \text{ and } r^2 \le 29\}$$

The supremum of the set equals $\sqrt{29}$. Perhaps it is more interesting to show that there does not exist a supremum of this set in $\mathbb{Q}$. That is in some way obvious. But we may still play with it and show the following:

1. $S$ does not have a largest element.

2. While $S$ has majorants in $\mathbb{Q}$ ( for instance $6$), it does not have a smallest majorant in $\mathbb{Q}$.

3. Let $r = \frac{m}{n}$ in $S$. We have $\left( \frac{m}{n}\right)^2 \le 29$, so $29 n^2 - m^2 \ge 0$. Some number theory shows that we cannot have equality. So $29 n^2 - m^2 \ge 1$. Rewrite it as $$(\sqrt{29} n - m)(\sqrt{29}n + m) \ge 1 \\ \sqrt{29}n - m \ge \frac{1}{\sqrt{29}n + m} \\ \sqrt{29} - \frac{m}{n} \ge \frac{1}{n^2( \sqrt{29}+ \frac{m}{n})}\ge \frac{1}{n^2 \cdot 2 \sqrt{29}} > \frac{1}{12 n^2}$$

We can rephrase this as follows:

If $\frac{m}{n} \in S$ then $\frac{m}{n} + \frac{1}{12 n^2} \in S$. So clearly $S$ does not have a largest element.

2. From the above, the set rationals so that $t \ge s$ for all $s \in S$ is $$T = \{ t \in \mathbb{Q} \mid t \ge 0 \text{ and } t^2 > 29 \}$$

Let $ \frac{m}{n}\in T$. In a similar way as above we can show that $\frac{m}{n}- \frac{1}{2 m n} \in T$. Conclusion: the set $T$ does not have a smallest element.

We conclude that $S$ does not have a supremum in $\mathbb{Q}$.