# Prove that the given set is bounded above in $\mathbb{Q}$ but does not have a Supremum in $\mathbb{Q}$

Let $S=\{r\in \mathbb{Q}\mid r\leq\sqrt{3}\}$.

Now prove that $S$ is bounded above in $\mathbb{Q}$ but it does not have a supremum in $\mathbb{Q}$.

The following is the proof I came up with, but I do not feel confident with it. Please let me know what is incorrect or could be written better.

Proof.
Let $m=\sqrt{3}$.

Then their exists an $m$ that belongs to the set of real numbers $\mathbb{R}$ such that $r\leq m$ for all $r$ that belong to the set $S$. Hence, $S$ is bounded above in $\mathbb{Q}$.

Now, suppose that there exists a smaller upper bound in $S$, $t$. Then $t<\sqrt{3}$, and as $t$ is an upper bound of $s$, $t\geq s$ for all $s\in S$.

Therefore, $s\leq t<\sqrt{3}$, and $s\leq \sqrt{3}\leq t$, which implies $\sqrt{3}\leq t\leq \sqrt{3}$, which is impossible.

Therefore, $S$ does not have a supremum in $\mathbb{Q}$.

Edit (added by Arturo Magidin, taken from comment made by OP)

2nd Attempt:
Proof Let $m=\sqrt{3}$. Then $m\geq r$ for all $r$ that belong to $S$. Hence, $S$ is bounded above by $m$ in $\mathbb{R}$. Therefore, $S$ is bounded above in $\mathbb{Q}$. Now, suppose $m=\sup(S)$ in $\mathbb{R}$. Let $t$ be an upper bound in $\mathbb{Q}$ and $t\leq m$. Hence, there exists $t$ that belongs to $\mathbb{Q}$ such that $m>t\geq r$ for every $r$ that belongs to $S$. Then there would exist an $r$ that belongs to $S$ such that $r>t$. Which contradicts $t\geq r$. Therefore, $S$ doesn't have a supremum in $\mathbb{Q}$.

3rd Attempt:
Proof
Since for every $r \in S$ $r<3$ and $3 \in \mathbb{Q}$, we know $S$ is bounded above in $Q$.
To prove it doesn't have a supremum in $\mathbb{Q}$, I will use contradiction.
Suppose m were the supremum of $S$ in $\mathbb{Q}$, then m does not equal $\sqrt{3}$, and $m\in \mathbb{Q}$.

If $m\lt \sqrt{3}$, by Archmedian's Property, their exists $t$ that belongs to $\mathbb{Q}$ such that $m\lt t\lt \sqrt{3}$.
This is a contradiction, because $m$ isn't an upper bound.
If $m\gt\sqrt{3}$, by Archmedian's Property, their exists $u$ that belongs to $\mathbb{Q}$ such that $\sqrt{3}\lt u\lt m$
This is also a contradiction.
Hence, $S$ does not have a supremum in $\mathbb{Q}$

• @enlgmatic: Add it to the body as an edit marked as an edit, not as a comment, and not twice (once here, once as a response to my answer). – Arturo Magidin Mar 30 '11 at 2:29
• If you are trying to prove something in $\mathbb{Q}$ you should not make reference to $\mathbb{R}$-unless you want to prove $\mathbb{R}$ exists and has the properties you need. Otherwise, you can just show that given any upper bound in $\mathbb{Q}$, there is a smaller one, also in $\mathbb{Q}$, so the first was not a supremum. – Ross Millikan Mar 30 '11 at 3:31
• @enlgmatlc: The fact that $S$ is bounded above in $\mathbb{Q}$ is now good (except for the fact that you have a dangling sentence; these should not be two separate sentences, but two clauses of the same sentence, connected by a comma). The second part doesn't do anything (and you're still trying to bring in the reals into the picture), and doesn't say who $y$ is. Right now, the final sentence doesn't go anywhere or prove anything. – Arturo Magidin Mar 30 '11 at 19:55
• @enlgmatlc: Your text disappeared because you are not using mark-up correctly; < and > outside of dollar signs are HTML mark-up, and the renderer tries to interpret them as such. Please look at other people's postings to see how to render math formula correctly; in particular, $\mathbb{Q}$ produces the nice $\mathbb{Q}$. – Arturo Magidin Mar 30 '11 at 20:26
• @enlgmatic: That's better, though it depends a bit on what you understand to be "the Archimedean property" (note spelling). For instance, one way to phrase the Archimedean property is that for all $\delta\gt 0$ and all $M\gt 0$ there is a natural number $n$ such that $n\delta\gt M$; but this by itself would not guarantee that you can also make $n\delta$ *smaller$than some$N\gt M$. But if you've proven that given any$0\lt M\lt N$, you can find a rational$q$with$M\lt q\lt N$, then that's what you want to use. – Arturo Magidin Mar 30 '11 at 20:28 ## 4 Answers Let me give you an example to show you why talk about the real numbers and the supremum in the reals is really not a good idea in general. Consider the set$\mathfrak{Q}=\mathbb{Q}\cap \Bigl((-\infty,0)\cup[1,\infty)\Bigr)$. That is,$\mathfrak{Q}$consists of all rationals that are either negative, or greater than or equal to$1$. Now let$\mathcal{S}=\{q\in\mathbb{Q}\mid q\lt 0\}$, the negative rationals. Notice that$\mathcal{S}$is a subset of$\mathfrak{Q}$, and is bounded above in$\mathfrak{Q}$, since$1\in\mathfrak{Q}$, and for all$s\in \mathcal{S}$,$s\leq 1$. Now, let me take a parallel argument to the one you are attempting to "show" that$S$does not have a supremum in$\mathfrak{Q}$: Take$m=0$; then$m$is the supremum of$\mathcal{S}$in$\mathbb{R}$. Now suppose that$t\in\mathfrak{Q}$is an upper bound on$\mathcal{S}$, and that$t\lt 0$. Then because$t\lt 0$and$0$is the supremum of$\mathcal{S}$in$\mathbb{R}$, there exists$s\in \mathcal{S}$such that$t\lt s\leq 0$. So$t$is not an upper bound for$\mathcal{S}$in$\mathfrak{Q}$, a contradiction. "Therefore",$\mathcal{S}$does not have a supremum in$\mathfrak{Q}$. Of course, the argument is completely false: because$1\in\mathfrak{Q}$is the supremum of$\mathcal{S}$in$\mathfrak{Q}$: note that$s\leq 1$for all$s\in \mathcal{S}$; and if$t\in\mathfrak{Q}$is strictly smaller than$1$, then because it is in$\mathfrak{Q}$it must also be strictly smaller than$0$, so there exists$s\in \mathcal{S}$with$t\lt s\leq 1$. That is: • For all$s\in \mathcal{S}$,$s\leq 1$. • For all$t\in\mathfrak{Q}$, if$t\lt 1$then there exists$s\in \mathcal{S}$such that$t\lt s$. These two properties show that$1$is the supremum of$\mathcal{S}$in$\mathfrak{Q}$. Notice that$1$is not the supremum of$\mathcal{S}$in$\mathbb{R}$, or even in$\mathbb{Q}$. But it is the supremum of$\mathcal{S}$in$\mathfrak{Q}$. So you should really not use the supremum of your$S$in$\mathbb{R}$, except perhaps as a "behind the scenes" guide to what you want. It could be, at least in principle, possible for a subset of$\mathbb{Q}$to have a supremum in$\mathbb{Q}$which is different from its supremum in$\mathbb{R}$(the supremum in$\mathbb{Q}$would have to be larger than the supremum in$\mathbb{R}$, just as it is in my example above). First: Since you are saying what$m$is, you should not then follow it up with "There exists an$m$that belongs to the real numbers." You would just say "Then$m$belongs to the real numbers and..." Second: That said, that part of the argument is incorrect. To show that a subset$S$of$\mathbb{Q}$is bounded above in$\mathbb{Q}$, you need to exhibit an element$m$of$\mathbb{Q}$such that$s\leq m$for all$s\in \mathbb{Q}$. Showing an element not in$\mathbb{Q}$does not show the set is bounded above in$\mathbb{Q}$. So your argument about$m$doesn't work. Third: You cannot conclude that an upper bound of$S$in$\mathbb{Q}$will be strictly smaller than$\sqrt{3}$. And if you assume it, then you are assuming your own contradiction: You know that if$t$is an upper bound to$S$in$\mathbb{Q}$, then by virtue of being in$\mathbb{Q}$,$t$is also in$\mathbb{R}$and an upper bound to$S$. And therefore$t$will be greater than or equal to the supremum of$S$in$\mathbb{R}$, since$S$is also bounded above in$\mathbb{R}$. So you know that$t\geq \mathrm{sup}_{\mathbb{R}}(S)$. Now,$\mathrm{sup}_{\mathbb{R}}(S)$happens to be$\sqrt{3}$, but you haven't proven that this is the case, so you cannot invoke that. And you don't explain how you get that$\sqrt{3}\leq t$(presumably from the supremum property, but you haven't shown that$\sqrt{3}$is the supremum in$\mathbb{R}$, so you cannot invoke the supremum property for$\sqrt{3}$). Fourth: So if you know anything about a possible supremum of$S$in$\mathbb{Q}$, it is that it is greater than$\sqrt{3}$(the supremum of$S$in$\mathbb{R}$), since it must be greater than or equal, and cannot be equal. So even if your argument about what happens if$t\lt\sqrt{3}$were correct (well, technically it is because your assumption is an impossibility, which is why you get a contradiction), you would not be done; you would still need to consider the possibility that the supremum$t$of$S$in$\mathbb{Q}$is greater than$\sqrt{3}$, instead of smaller. Added. Comment on 2nd attempt. To show that$S$is bounded above in$\mathbb{Q}$, you should produce an element of$\mathbb{Q}$that bounds$S$above. Going around to$\mathbb{R}$does not do it. Your argument is "gappy", in that it assumes without stating that you can take$m$and produce a rational bigger than$m$in order to conclude that there is a rational that bounds$S$above. Well, why not produce one? I could argue that the set$\{1.5\}$of rationals is bounded above in the rationals by saying "Well,$\pi\in\mathbb{R}$is greater than all elements of$\{1.5\}$, and so$\{1.5\}$is bounded above in$\mathbb{R}$, and therefore$\{1.5\}$is bounded above in$\mathbb{Q}$." But isn't it so much better, clearer, and simpler, to just say "$2\in\mathbb{Q}$, and$2$is an upper bound for${1.5}$"? You merely assert that$m$is greater than or equal to all$r\in\mathbb{Q}$; it should be proven. Even assuming the rest of the argument were correct (it isn't), you cannot simply assume that$m=\sup(S)$in$\mathbb{R}$. If you don't prove it, then your argument is contingent on that extra assumption. To complete the proof, you would have to also consider the case in which$m$is not the supremum of$S$in$\mathbb{R}$. Next: You continue to introduce your own contradiction when you assume that$t$is both an upper bound to$S$in$\mathbb{Q}$and that$t\lt m$. That is impossible to begin with because you are assuming that$m$is the supremum of$S$in$\mathbb{R}$: any upper bound to$S$has to be greater than or equal to the supremum of$S$(any upper bound in$\mathbb{Q}$is also in$\mathbb{R}$). So you would still need to consider the possibility that$t\geq m$, which you never do. So even if your argument about$t$were correct (it's not), it would still be incomplete. Finally: you merely assert "there would exist an$r\in S$such that$r\gt t$. Just saying so doesn't prove it. You have to prove that there is such a thing. Here's the issue here: you know that$S$has a supremum in$\mathbb{R}$; call it$m$. If$t$is any element of$\mathbb{Q}$that is an upper bound to$S$in$\mathbb{Q}$, then it is also an upper bound to$S$in$\mathbb{R}$, so by definition of the supremum, you must have$m\leq t$. If$m\lt t$, then by the properties of the rationals, you should show or justify that there has to be a number$t'$which (i) is an upper bound to$S$; (ii) is in$\mathbb{Q}$; and (iii) is strictly between$m$and$t$,$m\lt t' \lt t$. That means that the only number that has a shot at being the supremum of$S$in$\mathbb{Q}$is$m$itself, the supremum of$S$in$\mathbb{R}$(why? You need to explain this). So then you should prove that the supremum of$S$in$\mathbb{R}$is not a rational number, and that will give you a correct argument. • Excellent insight. – copper.hat Oct 10 '13 at 1:04 There is no need to mention the real numbers. As Pacciu says, we have$S=\{ r\in \mathbb{Q}:\ r\leq 0 \text{ or } r^2\leq 3\}$. So$\frac{2}{1}$is an upper bound for$S$since$\frac{2}{1} \gt 0$and if$ r \gt \frac{2}{1}$then$r^2 \gt \left(\frac{2}{1}\right)^2 = 4 \gt 3$. Now suppose$\frac{p}{q}$with$p$and$q$positive integers is an upper bound for$S$, i.e.$p^2 \ge 3 q^2 $. Then$\frac{3q^2+4pq-p^2}{4q^2}$is a smaller upper bound for$S$and so there is no lowest upper bound or supremum for$S$in$\mathbb{Q}$. Incidentally, this gives the sequence$\frac{2}{1}, \frac{7}{4}, \frac{111}{64}, \frac{28383}{16384}, \cdots$, approaching$\sqrt{3}$from above. • I fixed your LaTeX. There was a superfluous $. By the way: It is better to use \gt and \lt instead of > and < because sometimes the parser gets confused by trying to interpret everything in between < and > as HTML. – t.b. Mar 30 '11 at 12:52
• @Theo Buehler: Thank you - I cannot see the rendered LaTeX so I am just guessing what it looks like. – Henry Mar 30 '11 at 13:05
• How do you prove that every upper bound $r$ for $S$ satisfies $r^2 \ge 3$? Yeah, I know it so obvious, but if I am starting out and just learning about $\mathbb Q$, what is the argument? – CopyPasteIt Jan 6 at 15:16

Maybe you can argue as follows.

It is obvious that $S=\{ r\in \mathbb{Q}:\ r\leq 0 \text{ or } r^2\leq 3\}$. Hence each $p\in \mathbb{Q}$ s.t. $3< p^2$ is an upper bound for $S$; therefore $S$ is bounded from above in $\mathbb{Q}$.

Assume, by contraddiction, that $S$ has the supremum $m\in \mathbb{Q}$; then clearly $m^2\geq 3$ (for $m$ is the least upper bound of $S$), but indeed $m^2=3$: in fact, using an algorithm for the computation of the square root, one can generate a sequence of rational numbers $p_n\in \mathbb{Q}$ s.t. $p_n^2\geq 3$ and $p_n^2-3\leq \frac{1}{n}$; on the other hand $0\leq m^2-3\leq p_n^2-3$, thus $0\leq m^2-3\leq \frac{1}{n}$ for all $n$; therefore $m^2-3=0$ as claimed. But this is a contraddiction, because it is well known that there doesn't exist any $m\in \mathbb{Q}$ s.t. $m^2=3$. Hence $S$ doesn't have a supremum in $\mathbb{Q}$.

We follow the advice offered in other answers - the logic here only entails knowledge of the ordered field of rational numbers, $$\mathbb Q$$, which we take as our universal set.

For any subset $$A$$ of $$\mathbb Q$$ we have the set of all upper bounds,

$$\tag 1 Au = \{ b \in \mathbb Q \, | \, (\forall a \in A) \; a \le b \}$$

Now let $$A = \{ a \in \mathbb Q \, | \, (\exists d \in \mathbb Q ) \; \left[\,(d^2 \lt 3) \land (a \le d)\,\right] \}$$.

We note that $$1 \in A$$ and $$2 \in Au$$. Also, since the equation $$x^2 = 3$$ has no solutions, the set $$A$$ is the same as the set $$S$$ (spookily) defined by the OP.

Using algebra and the quadratic formula (discriminant discriminant $$\gt 0$$) applied to a quartic polynomial, we can assert the following:

Claim 1: if $$p \in A$$ and $$p \gt 0$$ then

$$\quad q = \frac{6 p}{3 + p^{\;2}} \gt p \text{ and } q \in A$$.

It readily follows from this claim that the set $$Au$$ is equal to $$\{ b \in \mathbb Q \, | \, (b \gt 0) \land (b^2 \gt 3)\}$$.

All that remains is to show that $$Au$$ has no minimum element.

Using algebra and factoring a quartic polynomial (discriminant $$= 0$$), we can assert the following:

Claim 2: if $$p \in Au$$ then

$$\quad q = \frac{1}{2} \, (p + \frac{3}{p}) \lt p \text{ and } q \in Au$$.

But this means $$Au$$ has no minimum element and the supremum of $$A$$ doesn't exist.