- #1

mtayab1994

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## Homework Statement

Let T be a set such that:

[tex]T=\{t\in\mathbb{R}/t^{2}<2\}[/tex]

## Homework Equations

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

b) Prove that the proposition [tex]a^{2}<2[/tex] is false.

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

d) Conclude that finally [tex]a^{2}=2[/tex]

## The Attempt at a Solution

a) Ok we have [tex]t^{2}<2[/tex] which implies t<√2. There for [tex]t\in]-\infty,\sqrt{2}[\subset\mathbb{R}[/tex] , 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.