Why doesn't the set $\{x:x^2<5\}$ have a supremum in $\mathbb{Q}$? I know that the rational numbers aren't a complete field, but I'm still not understanding how a set can have upper bounds, but no least upper bound in a field.

In $\mathbb{Z}$ for example, $\{x:x^2<5\}=\{-2,-1,0,1,2\}$. It has the set of upper bounds: $[2,\infty)\cup\mathbb{Z}$. So why isn't the least upper bound $2$?

  • $\begingroup$ Hint: what is the supremum of that set in $\mathbb{R}$? $\endgroup$ – Seub Aug 22 '13 at 0:35
  • $\begingroup$ @Seub It's $\sqrt{5}$, but what does that mean for $\mathbb{Q}$ and $\mathbb{Z}$? $\endgroup$ – Ataraxia Aug 22 '13 at 0:36
  • $\begingroup$ $2$ is not the least upper bound because $2$ is not an upper bound: $2.1$ is also a rational number whose square is less than $5$. $\endgroup$ – Michael Hardy Aug 22 '13 at 0:57
  • 1
    $\begingroup$ @MichaelHardy: But $2$ is an upper bound, whereas $1$ isn't... $2$ is a least upper bound in $\mathbb{Z}$! $\endgroup$ – Clive Newstead Aug 22 '13 at 0:58
  • $\begingroup$ Within $\mathbb Z$, $2$ is the least upper bound of the set $\{x\in\mathbb Z : x^2<5\}$. But if the question is "Why isn't $2$ the least upper bound?", that seems to suggest that a context in which $2$ is not the least upper bound was intended. $\endgroup$ – Michael Hardy Aug 22 '13 at 1:02

Suppose $q$ is a rational number which is an upper bound for your set. Since $\sqrt{5}$ is irrational, it must be the case that $q>\sqrt{5}$. But then there exists a rational number $q'$ such that $$\sqrt{5} < q' < q$$ so $q$ couldn't have been a least upper bound.

Added: if you want to work entirely inside $\mathbb{Q}$ then you can get away without referring to $\sqrt{5}$ at all. It's a fact that if $q > x$ for all $x \in \mathbb{Q}$ with $x^2 < 5$, then there exists $q' < q$ with the same property.

As for $\mathbb{Z}$, the set $\{ x \in \mathbb{Z} : x^2 < 5 \}$ is equal to $$\{ -2, -1, 0, 1, 2 \}$$ It's not equal to $\{0, 1, 4 \}$, which is the set of integer squares less than $5$, whereas our set is the set of integers whose squares are less than $5$. (Think for a while about why these aren't the same thing.) And indeed, this set does have a supremum in $\mathbb{Z}$, namely $2$, its largest element.

Added again:

I'd like to stress that your question really asks about two sets, not just one. The notation $\{ x : x^2 < 5 \}$ is not very clear if it's not specified what values $x$ is meant to range over.

The two sets are, respectively:

  • $\{ x \in \mathbb{Q} : x^2 < 5 \}$, the set of rationals whose squares are less than $5$
  • $\{ x \in \mathbb{Z} : x^2 < 5 \}$, the set of integers whose squares are less than $5$

The first of these sets is infinite, and the second is finite!


Note that $\Bbb Z$ is a Dedekind-complete order. Every bounded set has a least upper bound. The reason is that given $k\in\Bbb Z$, the set $\{n\in\Bbb Z\mid n\leq k\}$ is a well-ordered set. So given any bounded set $A$ of integers, the set of upper bounds has a minimal element.

On the other hand, the order in $\Bbb Q$ is dense. So we don't have this property. And indeed sets like $\{x\in\Bbb Q\mid x^2<5\}$ do not have a least upper bound, even if they have upper bound. In order to see this is true, for this particular case anyway, note that if $y$ was the least upper bound of this set, then $x<y$ implies $x^2<5$ and $y<x$ implies $5\leq x^2$. From this follows that $y^2=5$, but we know that no such $y$ exists in the rational numbers.

Finally, note that the title of your question is wrong. In the field $\Bbb Q[\sqrt5]$, which is not complete either, the set given does in fact have a supremum.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.