# 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).

Homework Helper
Gold Member
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.

Homework Helper

## 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?

Homework Helper
Gold Member
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?

Homework Helper
Gold Member
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.

Homework Helper
Gold Member
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##.