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

[STEMucator]

Homework Statement

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

Homework Equations

I'll split the question into 2 parts.

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?

 

[Dick]

Homework Statement

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

Homework Equations

I'll split the question into 2 parts.

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?

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.

 

[LCKurtz]

Homework Statement

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

Homework Equations

I'll split the question into 2 parts.

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.

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.)

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.

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.

 

[STEMucator]

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.

(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 $\sqrt{2} \notin \mathbb{Q}$.

 

[Dick]

(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 $\sqrt{2} \notin \mathbb{Q}$.

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.

 

[STEMucator]

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.

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.

 

[micromass]

I'm thinking I have to show m is an integer.

Exactly.

 

[STEMucator]

Exactly.

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, let's say $sup(A) = m$.

Now let's 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 what's going on around it, but do I have to be particular about it?

 

[Dick]

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, let's say $sup(A) = m$.

Now let's 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 what's going on around it, but do I have to be particular about it?

Suppose m is not an integer. What then?

 

[micromass]

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

 

[STEMucator]

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

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, let's 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).

 

[Dick]

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, let's 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).

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?

 

[STEMucator]

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?

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.