Show the sup axiom holds for $\mathbb{Z}$,but not for $\mathbb{Q}$

1. May 1, 2013

Zondrina

1. The problem statement, all variables and given/known data

9. Show that the least upper bound axiom also holds in Z (i.e., each nonempty subset of Z with
upper bound in Z has a least upper bound in Z), but that it fails to hold in Q.

http://gyazo.com/4c0b79cbb1d15cd5edf0c96ec612a55c

2. Relevant equations

I'll split the question into 2 parts.

3. The attempt at a solution

(a). We want to show the sup axiom holds for $\mathbb{Z}$.

Suppose $A \subseteq \mathbb{Z}$ is nonempty and bounded above.

Let $M = max(A)$ so that $a ≤ M, \forall a \in A$ so that M is an upper bound for A.

Now I have a quick question about this before I continue. I've shown that $M$ is an upper bound, but now I have to show that if $U \in \mathbb{Z}$ is any upper bound then $M ≤ U$.

The question is how to go about this. Should I consider the set $-A$ or is there perhaps a more straightforward alternative.

(b) I'm guessing I should assume the contrary that the sup axiom holds for $\mathbb{Q}$ and then arrive at a contradiction?

2. May 1, 2013

Dick

I'm not quite sure what you mean by max(A), don't you mean sup(A)? You know every bounded set has a sup in the real numbers. If the real number M=sup(A), what do you have to show about M?

For b) just give a counterexample.

Last edited: May 1, 2013
3. May 1, 2013

LCKurtz

A is a potentially infinite set. How do you know it has a maximum element? (And your arguments would be easier to read if you would consistently use lower case letters for numbers and upper case letters for sets.)

Why would you even consider -A? Just because it worked in a completely different problem? There is a very very simple argument that you should be able to see.

Last edited: May 1, 2013
4. May 1, 2013

Zondrina

(a) Yes I meant sup(A), and I want to show that $sup(A) = m$ satisfies two things.

$m$ is an upper bound and if $u$ is any upper bound then $m ≤ u$ ( Gonna use lower case letters ).

So suppose $A \subseteq \mathbb{Z}$ is nonempty and bounded above by $u$.

We want to show sup(A) exists, so let $m = sup(A)$. By the definition of the supremum, we know that m is an upper bound for the set A, that is $a ≤ m, \space \forall a \in A$ so that the first condition is met.

Since $u$ is an upper bound for A, we want $m ≤ u$. I seem to be blanking on the second property for some reason.

(b) Hmm a counter example... take :

$B = \{ b \in \mathbb{Q} \space | \space b^2 ≤ 2 \} \subseteq \mathbb{Q}$

Then $sup(B)$ does not exist because $\sqrt{2} \notin \mathbb{Q}$.

5. May 1, 2013

Dick

That's a fine example for (b) but for (a) you are trying to prove the wrong thing. You ALREADY KNOW A has a sup in R. That means m is a least upper bound. You don't have to prove that. You have to prove something else about m.

6. May 1, 2013

Zondrina

Okay so (b) is fine.

(a) Yes if A was a nonempty subset of R bounded above, then sup(A) = m exists since we already know the sup axiom works for R.

Hmm.. I have to prove something else about m knowing this. I'm thinking I have to show m is an integer.

I don't know why or if this is valid, but something is telling me to consider an $ε > 0$ so that $m - ε < m$ and we know that $m - ε$ is not an upper bound for A.

7. May 1, 2013

micromass

Staff Emeritus
Exactly.

8. May 1, 2013

Zondrina

Well then, let me give this a go.

(a). Suppose $A \subseteq \mathbb{Z}$ is non empty and bounded above.

We know that if we view $A \subseteq ℝ$, it will have a supremum by the axiom, lets say $sup(A) = m$.

Now lets go back to viewing $A \subseteq \mathbb{Z}$. Since A is a set of integers, its supremum also has to be an integer, that is if $m \notin \mathbb{Z}$, then it can't be the supremum of the set.

If $m \in \mathbb{Z}$ then we're done.

How do I force $m$ to be an integer. I understand whats going on around it, but do I have to be particular about it?

9. May 1, 2013

Dick

Suppose m is not an integer. What then?

10. May 1, 2013

micromass

Staff Emeritus
Hint: if $m$ is not an integer, prove that there is a greatest integer that is smaller than $m$.

11. May 1, 2013

Zondrina

Okay.

(a). Suppose $A \subseteq \mathbb{Z}$ is non empty and bounded above.

We know that if we view $A \subseteq ℝ$, it will have a supremum by the axiom, lets say $sup(A)=m$ so that m is an upper bound for A.

I claim that since A is a set of integers, even if we view A as a subset of ℝ, m will be an integer.

Suppose that $m \notin \mathbb{Z}$. We want to show there is a greatest integer $n \in \mathbb{Z} \space | \space n ≤ m$.

Notice that we can find the next greatest integer by taking $n = n+1$ and using this construction, there will always be a next greatest integer. Hence there will always be another upper bound we can find, which means that with enough applications of our construction, we can show that m is not an upper bound for A. This is a contradiction because we said earlier that m IS an upper bound for A.

Hence $m \in \mathbb{Z}$ and therefore the sup axiom holds for $\mathbb{Z}$.

(Hopefully this is somewhat correct, it took me awhile to formulate this for some reason).

12. May 1, 2013

Dick

It must take a while to formulate something like that when the correct line of reasoning is so much simpler. Does that actually sound correct to you?! Draw a picture or something and think about the problem again, ok?

13. May 1, 2013

Zondrina

I know it's not correct, those were some ideas I had floating around. Something seems to not be clicking like it usually does ( Head feels groggy and clouded ).

Perhaps I'll go for a walk in the fresh air to clear my head. Then I'll come back, draw my picture and see what I can do.