Recall, the Archimedean property states that if and is arbitrary, then there exists an integer such that .
Further, recall that the least upper bound axiom states that every nonempty set of real numbers which is bounded above has a supremum.
Now, prove that satisfies the Archimedean property, but not the least-upper-bound axiom.
First, we prove that satisfies the Archimedean property.
Proof. This is immediate since if are arbitrary elements in with , then they are also in since . Thus, by the Archimedean property in we know there is an such that . Hence, the Archimedean property is satisfied in
Next, we prove that does not satisfy the least-upper-bound axiom.
Then is non-empty since . Further, is bounded above by 4 since . (Of course, there are better upper bounds available, but we just need any upper bound.)
Now, we must show that has no supremum in to show that the least-upper-bound property fails in (since this will mean is a nonempty set which is bounded above, but fails to have a least upper bound in ).
Suppose otherwise, say with . We know (I.3.12, Exercise #11). Thus, by the trichotomy law we must have either or .
Case 1: If , then there exists such that (since the rationals are dense in the reals, see I.3.12, Exercise #6). But then, (since ) and implies with , contradicting that is an upper bound for . Hence, we cannot have .
Case 2: If , then there exists such that . But then, , so if we have ; hence, is an upper bound for which is less than . This contradicts that is the least upper bound of . Hence, we also cannot have .
Thus, there can be no such (since by the trichotomy exactly one of must hold, but we have shown these all lead to contradictions).
Hence, does not have the least-upper-bound property
For showing that the set of rationals does not possess the least-upper-bound axiom it would be quicker assume it does, and then apply the result of Exercise 6 to the interval of rational numbers between 0 and sqrt 2. Exercise 6 guarantees that we can find another number closer to sqrt 2 than the supposed supremum.
Aren’t you assuming that the square root of 2 exists in this proof. That is proved only in the next section.
Here’s a proof that does not assume that the square root of 2 exists (I hope the LaTeX works okay):
Let with . Choose so that s – 1/nS(s – 1/n)^2 > 2s – 1/n < ssS$.
Okay, that didn’t work. I tried to use an align* environment and it didn’t appear.
Let’s try this again.
Let with . Choose so that
So is also an upper bound of (because , but , contradicting the claim that is the least upper bound of .
Okay, I give up with the formatting. Just choose n so that 1/n is less than (s^2 – 2)/(2s). Show that (s – 1/n)^2 is greater than 2. Thus s – 1/n is an upper bound of S, but s – 1/n is less than sup S, a contradiction.
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)
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?
No, we can’t do that. The least-upper-bound axiom is not the same as being unbounded above. For instance, is unbounded above but possesses the least upper bound axiom.
The critical thing is that if is the set you want to show satisfies the least upper bound axiom, you need to show that for every nonempty subset which is bounded above has a least-upper-bound in . So, the second proof is showing that fails this since we have a non-empty set that is bounded above, but it’s least upper bound is not actually in (since its least upper bound is which is not a rational number). Does that make sense?