Completeness Axiom: The Completeness Axiom states for every non-empty subset $A$ of $\mathbb{R}$ that $A$ has a least upper bound.
There are many equivalent definitions to this. But I am looking for the exact axiom above that makes sense of the following scenario and not looking for an equivalent definition here.
Scenario: Consider $A=\{x\in \mathbb{Q}| x^2<2\}$. It is well known that this set does NOT have a least upper bound. This is a fact!
That being said....
- $A$ is non-empty
- $A\subseteq \mathbb{R}$
$\therefore$ By the Completeness Axiom, it should have a least upper bound.
What should I be including in the Completeness Axiom differently to avoid this contradictory logic? I feel like there is something missing in this Axiom that is a little more detailed here but isn't written in most textbooks.
$\endgroup$ 32 Answers
$\begingroup$You have missed a hypothesis in the statement of the completeness axiom: $A$ must be nonempty and bounded above. For example, $\mathbb{Z}$ is a nonempty set of reals without a least upper bound, which isn't a problem since $\mathbb{Z}$ doesn't have any upper bounds at all. This doesn't affect the specific example you're asking about, but it is important.
Actually, the set $S=\{x\in\mathbb{Q}: x^2<2\}$ does have a least upper bound ... in the real numbers. It is true that:
$S$ doesn't have a maximum element, and
$S$ doesn't have a least upper bound in the rational numbers,
but neither of those facts contradicts the existence of a least upper bound of $S$ in $\mathbb{R}$: that is, a real number $r$ such that $r$ is $\ge$ every element of $S$ and no $s<r$ is $\ge$ every element of $S$. This $r$ is just $\sqrt{2}$.
$\endgroup$ 4 $\begingroup$Scenario: Consider $A=\{x\in \mathbb{Q}| x^2<2\}$. It is well known that this set does NOT have a least upper bound. This is a fact!
Not exactly. It does not have an upper bound in $\mathbb Q$.
"Having an upper bound" is not a property that exists in isolation. To speak of an upper bound, you need three things:
- A set $X$
- A set $A\subseteq X$
- An ordering $\leq$, defined on $X$.
For your set, if $X=\mathbb Q$, then the set $A=\{x\in \mathbb{Q}| x^2<2\}$ does not have an upper bound.
However, if $X=\mathbb R$, then the same set does have an upper bound, and that upper bound is $\sqrt 2$. It is relatively easy to show both properties an upper bound must have:
- $\sqrt{2}$ is an upper bound of $A$.
- If $y<\sqrt{2}$, then $y$ is not an upper bound of $A$.
