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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 $2$?

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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!

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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.

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