# Determining the relevant set/field for least upper bounds

My real analysis textbook makes the following statement: a set of real numbers $S_1 = \{ s \in \mathbb{R} | s^2 < 2 \}$ has a least upper bound, as we can show that $sup S_1 = \sqrt{2}.$ On the other hand, it contends that $S_2 = \{ s \in \mathbb{Q} | s^2 < 2 \}$ does not have a sup. Now something I don't understand is, why does sup have to belong to the same field (rational numbers)? Is it part of its definition? Because my book doesn't specify this requirement in the definition of supremum... $\sqrt{2}$ still works but it just isn't a rational number, why does the upper bound have to be rational? A minimum or a maximum (if they exist) of the set would clearly have to be, since they actually would belong to the set, but such treatment of supremum leaves me puzzled.

This provokes a further question of what if I am just given a certain set of numbers, and all of them happen to be in $\mathbb{Q},$ And asked to find a sup if it exists. Do I look only for a rational sup, wouldn't a real sup work too, since we have a containment of rationals in reals, and the set could just be treated as a set of real numbers, which just 'happened' to be rational as well?

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You are right to be puzzled. When one says that a set has (or doesn't have) a supremum, that is always relative to some specific underlying set. Thus the complete assertion is that the set of all rational $s$ such that $s^2<2$ does not have a $\sup$ in the rationals. – André Nicolas Dec 2 '11 at 8:12

The fact that $\mathbb{R}$ and $\mathbb{Q}$ are fields doesn't come into play here; what matters is the set we are looking for a supremum in.

Given an ordered set $T$, and a subset $S$, the supremum of $S$ in $T$ is the least element of $T$ that is greater than or equal to every element of $S$. That is, $x=\sup(S)$ must satisfy

1. $s\leq x$ for all $s\in S$
2. $x\leq y$ for all $y\in T$ such that ($s\leq y$ for all $s\in S$)

Now, depending on what $T$ is, an element satisfying these properties may or may not exist in $T$. If we consider $S_2$ as a subset of $\mathbb{R}$, there is a supremum for $S_2$: it is $\sqrt{2}$. Either you have not mentioned it, or the book leaves it implicit, but the claim is that $S_2$ has no supremum when we are thinking of it as living in $\mathbb{Q}$, i.e. $S=S_2$ and $T=\mathbb{Q}$.

Suppose a rational number $r$ has the property that $s\leq r$ for all $s\in S_2$. If we had $r^2<2$, then $t=2-r^2>0$, and for any $t>0$ there is some $n\in\mathbb{N}$ such that both $\frac{t}{2}>\frac{1}{n^2}>0$ and $\frac{t}{2}>\frac{2}{n}>0$. Then we have $$(r+\tfrac{1}{n})^2=r^2+\tfrac{2}{n}+\tfrac{1}{n^2}<r^2+t=2,$$ so $r$ cannot be a supremum for $S_2$ in $\mathbb{Q}$ because $r+\frac{1}{n}\in S_2$ but $r+\frac{1}{n}\not\leq r$.

Thus, any rational number $r$ has the property that $s\leq r$ for all $s\in S_2$ must also have the property that $r^2\geq 2$. We know that equality is impossible, as $\sqrt{2}$ is irrational, so we must have that $2<r^2$. But by a similar argument, there is some $n\in\mathbb{N}$ such that $2<(r-\tfrac{1}{n})^2$, so that $r-\frac{1}{n}$ also has the property that $s\leq r-\frac{1}{n}$ for all $s\in S_2$; but $r-\frac{1}{n}<r$, so $r$ cannot be a supremum for $S_2$.

Thus, we have shown that there is actually no supremum for $S_2$, when it is considered as a subset of $\mathbb{Q}$; for any element of $\mathbb{Q}$ that is greater than everything in $S_2$, we can find a strictly smaller element of $\mathbb{Q}$ with that same property.

In fact, this argument actually shows that there is no supremum for $S_2$ in any dense subset of $\mathbb{R}$ that is missing $\sqrt{2}$.

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A small quibble with your last sentence: $S_2$ does have a supremum in $(-\infty,\sqrt2)\cup\{42\}\subseteq \mathbb R$, namely $42$. There's no supremum in any dense subset of $\mathbb R$ that misses $\sqrt 2$. – Henning Makholm Dec 2 '11 at 9:01
Ah indeed, thank you for the correction! – Zev Chonoles Dec 2 '11 at 9:04