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Prove that the rationals satisfy the Archimedean property, but not the least-upper-bound axiom.

Recall, the Archimedean property states that if x > 0 and y \in \mathbb{R} is arbitrary, then there exists an integer n > 0 such that nx > y.
Further, recall that the least upper bound axiom states that every nonempty set S of real numbers which is bounded above has a supremum.
Now, prove that \mathbb{Q} satisfies the Archimedean property, but not the least-upper-bound axiom.


First, we prove that \mathbb{Q} satisfies the Archimedean property.
Proof. This is immediate since if x,y are arbitrary elements in \mathbb{Q} with x>0, then they are also in \mathbb{R} since \mathbb{Q} \subseteq \mathbb{R}. Thus, by the Archimedean property in \mathbb{R} we know there is an n \in \mathbb{Z}_{>0} such that nx > y. Hence, the Archimedean property is satisfied in \mathbb{Q}. \qquad \blacksquare

Next, we prove that \mathbb{Q} does not satisfy the least-upper-bound axiom.
Proof. Let

    \[ S = \{ r \in \mathbb{Q} \mid r^2 \leq 2 \}. \]

Then S is non-empty since 1 \in S. Further, S is bounded above by 4 since r^2 \leq 2 \ \implies \ r^2 \leq 16 \ \implies \ r \leq 4. (Of course, there are better upper bounds available, but we just need any upper bound.)
Now, we must show that S has no supremum in \mathbb{Q} to show that the least-upper-bound property fails in \mathbb{Q} (since this will mean S is a nonempty set which is bounded above, but fails to have a least upper bound in \mathbb{Q}).
Suppose otherwise, say b = \sup S with b \in \mathbb{Q}. We know b^2 \neq 2 (I.3.12, Exercise #11). Thus, by the trichotomy law we must have either b^2 < 2 or b^2 > 2.
Case 1: If b^2 < 2, then there exists r \in \mathbb{Q} such that b < r < \sqrt{2} (since the rationals are dense in the reals, see I.3.12, Exercise #6). But then, r^2 < 2 (since r < \sqrt{2} \ \implies \ r \cdot r < \sqrt{2} \cdot \sqrt{2} \ \implies \ r^2 < 2) and r \in \mathbb{Q} implies r \in S with r > b, contradicting that b is an upper bound for S. Hence, we cannot have b^2 < 2.
Case 2: If b^2 > 2, then there exists r \in \mathbb{Q} such that b > r > \sqrt{2}. But then, r^2 > 2, so if s \in S we have s < r; hence, r is an upper bound for S which is less than b. This contradicts that b is the least upper bound of S. Hence, we also cannot have b^2 > 2.
Thus, there can be no such b^2 \in \mathbb{Q} (since by the trichotomy exactly one of b^2> 2, \ b^2 < 2, \ b^2 = 2 must hold, but we have shown these all lead to contradictions).
Hence, \mathbb{Q} does not have the least-upper-bound property. \qquad \blacksquare

3 comments

  1. Fiestita says:

    Hi Rori, i don’t get why you say (Case 2) : b^2 > 2 —> exists r (rational) such that b > r > sqrt(2)
    We know that b= -2 satisfies b^2 > 2 but don’t satisfies that b > r > sqrt(2)

  2. martin says:

    For the second proof, could we say quickly that the set P (positve integers) is in Q so it’s unbounded above that’s why it doesn’t satisfy axiom 10?

    • RoRi says:

      No, we can’t do that. The least-upper-bound axiom is not the same as being unbounded above. For instance, \mathbb{R} is unbounded above but possesses the least upper bound axiom.

      The critical thing is that if X is the set you want to show satisfies the least upper bound axiom, you need to show that for every nonempty subset S \subseteq X which is bounded above has a least-upper-bound in X. So, the second proof is showing that \mathbb{Q} fails this since we have a non-empty set that is bounded above, but it’s least upper bound is not actually in \mathbb{Q} (since its least upper bound is \sqrt{2} which is not a rational number). Does that make sense?

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