What is the Supremum of a Set in ℝ?

[mtayab1994]

Homework Statement

Let T be a set such that:
$$T=\{t\in\mathbb{R}/t^{2}<2\}$$

Homework Equations

a) Justify the existence of a real number a such that a=Sup(T)

b) Prove that the proposition $$a^{2}<2$$ is false.

c) Suppose that $$a^{2}>2$$. Prove that we can find a contradiction with a=Sup(T).

d) Conclude that finally $$a^{2}=2$$

The Attempt at a Solution

a) Ok we have $$t^{2}<2$$ which implies t<√2. There for $$t\in]-\infty,\sqrt{2}[\subset\mathbb{R}$$ , and since T is a non empty set bounded from the top. Hence by the completeness axiom there exists an a in ℝ such that a=Sup(T).

b) I have a feeling that I will have to use a proof by contradiction, but I don't know how to start.

c) I started this one with letting a and b be two upper bounds of T such that a≠b and let ε>0:

Let ε'=a-b>0 . Therefore there exists an x in T such that: a-ε'<x. Therefore, after substituting ε' with its value we get that there exists an x in T such that b<x. Hence we reach a contradiction with b being an upper bound of T therefore b<a. And finally a=Sup(T)

d) Well from b) we have that a^2<2 is a false statement and from c) we found that a^2>2 contradicts with a=Sup(T), which leaves us with a^2=2 and therefore a=Sup(T).

Thank you for your help in advanced.

 

[jbunniii]

a) Ok we have $$t^{2}<2$$ which implies t<√2. There for $$t\in]-\infty,\sqrt{2}[\subset\mathbb{R}$$ , and since T is a non empty set bounded from the top. Hence by the completeness axiom there exists an a in ℝ such that a=Sup(T).

OK, although you could have worded it better. Your claim is that $T$ is a nonempty set bounded from above. So give an example of an element in $T$, and give an explicit upper bound. You provided $\sqrt{2}$ as an upper bound, and while it's "obvious" that this is correct, a line or two to prove it would be a good idea.

b) I have a feeling that I will have to use a proof by contradiction, but I don't know how to start.

If $a^2 < 2$, can you find an element of $T$ which is greater than $a$?

c) I started this one with letting a and b be two upper bounds of T such that a≠b and let ε>0:

Let ε'=a-b>0 .

I don't see how $a-b > 0$ follows from your assumption.

d) Well from b) we have that a^2<2 is a false statement and from c) we found that a^2>2 contradicts with a=Sup(T), which leaves us with a^2=2 and therefore a=Sup(T).

The conclusion is that $a^2 = 2$. You already defined $a = \sup(T)$ in part (a), so there's no "therefore" about it.

 

[pasmith]

Homework Statement

Let T be a set such that:
$$T=\{t\in\mathbb{R}/t^{2}<2\}$$

Homework Equations

a) Justify the existence of a real number a such that a=Sup(T)

b) Prove that the proposition $$a^{2}<2$$ is false.

c) Suppose that $$a^{2}>2$$. Prove that we can find a contradiction with a=Sup(T).

d) Conclude that finally $$a^{2}=2$$

The Attempt at a Solution

a) Ok we have $$t^{2}<2$$ which implies t<√2.

This is putting the cart before the horse: the object of the question is to prove that there actually exists a positive real number $a$ such that $a^2 = 2$.

 

[mtayab1994]

OK, although you could have worded it better. Your claim is that $T$ is a nonempty set bounded from above. So give an example of an element in $T$, and give an explicit upper bound. You provided $\sqrt{2}$ as an upper bound, and while it's "obvious" that this is correct, a line or two to prove it would be a good idea.


If $a^2 < 2$, can you find an element of $T$ which is greater than $a$?


I don't see how $a-b > 0$ follows from your assumption.


The conclusion is that $a^2 = 2$. You already defined $a = \sup(T)$ in part (a), so there's no "therefore" about it.

B) No there is no element in T that's greater than a since a is the sup(T). Is that all?

 

[jbunniii]

B) No there is no element in T that's greater than a since a is the sup(T). Is that all?

Let me rephrase the question. Suppose $a^2 < 2$. Can you find a number $t \in T$ such that $t > a$, and therefore $a$ cannot be the supremum of $T$?

 

[mtayab1994]

Let me rephrase the question. Suppose $a^2 < 2$. Can you find a number $t \in T$ such that $t > a$, and therefore $a$ cannot be the supremum of $T$?

Supposing that a^2<0 and that b is an upper bound of T. We want to prove that a<b.

So by proof by contradiction we suppose that b<a and let ε=a-b>0. So by definition there exists an x in T such that a-ε<x. By substituting ε with it's value we get that b<x , and that contradicts with our assumption of a<b (b is an upper bound). Finally a^2<2 is false and a=sup(T). Is this ok now?

 

[jbunniii]

Supposing that a^2<0 and that b is an upper bound of T. We want to prove that a<b.

I think you mean $a^2 < 2$.

So by proof by contradiction we suppose that b<a and let ε=a-b>0. So by definition there exists an x in T such that a-ε<x. By substituting ε with it's value we get that b<x , and that contradicts with our assumption of a<b (b is an upper bound). Finally a^2<2 is false and a=sup(T). Is this ok now?

I don't see where you used the assumption that $a^2 < 2$.

I don't think you want to introduce $b$ as an upper bound of $T$. Try this instead: Suppose that $a$ is any real number such that $a^2 < 2$. Then I claim that there is some $t \in T$ (not necessarily an upper bound of $T$) such that $a < t$. Therefore $a$ is not even an upper bound of $T$, so it cannot be the supremum of $T$. All you need to do is prove the claim.

 

[mtayab1994]

I think you mean $a^2 < 2$.


I don't see where you used the assumption that $a^2 < 2$.

I don't think you want to introduce $b$ as an upper bound of $T$. Try this instead: Suppose that $a$ is any real number such that $a^2 < 2$. Then I claim that there is some $t \in T$ (not necessarily an upper bound of $T$) such that $a < t$. Therefore $a$ is not even an upper bound of $T$, so it cannot be the supremum of $T$. All you need to do is prove the claim.

Ok so I let there be two elements of T (a and t) such that a≠t, and we want to prove that a<t<2.

So with a proof by contradiction we'll say that t<a<2 and let ε=t-a>0. When we plug in ε to the previous inequality we get that ε+a<a⇔ ε<0 which contradicts with it being greater than 0. I'm pretty sure this isn't it though.

 

[jbunniii]

Ok so I let there be two elements of T (a and t) such that a≠t, and we want to prove that a<t<2.

No, first of all, you want $a < t < \sqrt{2}$, not $a < t < 2$. And you can't "let there be" two such elements. Start with an element $a \in T$ (i.e. $a^2 < 2$). You want to prove that there EXISTS an element $t\in T$ such that $a < t$.

 

[jbunniii]

Hint: if $a^2 < 2$, then there exists a rational number of the form $p^2/q^2$ such that $a^2 < p^2/q^2 < 2$. (Why?)

 

[mtayab1994]

Hint: if $a^2 < 2$, then there exists a rational number of the form $p^2/q^2$ such that $a^2 < p^2/q^2 < 2$. (Why?)

Well like you said and if we have p and q two co-prime integers then when we square root them we get: a<p/q<√2 and multiply by q and we get aq<p<q*√2. I was thinking that's going to say that p isn't an integer but you can't say from what I've reached.