Proof of the least upper bound

[furi0n]

Homework Statement

LEt S is supset of real numbers and suppose that there is X₀ is member of S such that x₀>=x for all x which is member of S(i.e. x₀ is the maximum of S). show that x₀=supS

Homework Equations

The Attempt at a Solution

Not: this seems too easy question but i can't understand how ı can prove it please help me it's my important homework.. :(

 

[╔(σ_σ)╝]

Well x_0 is the maximum of x. Suppose a was an upper bound of S but a was not equal to x_0.

The a>x_0 right ?

Thus is there any upper bound smaller than x_0 ?

 

[michonamona]

So we define x_0 as a value such that x_0>=x for all x in S. Now, suppose x_0 was not the sup of S (i.e. x_0 is not equal to supS). Well then this must imply the existence of a value b in S such that x_0 < b <= supS. Notice the contradiction? how did we define x_0 again?

 

[HallsofIvy]

Yes, it is an easy question! Since $x_0\ge x$ for all x in the set, it is an upper bound on the set. Is it possible for any other upper bound to be less than $x_0$? No, because $x_0$ is in the set and saying $M< x_0$ would contradict the definition of "upper bound".

 

[furi0n]

thank you everybody, why i ask for this question is to learn how to prove this because we know already everything there is not anything which we can prove but today İ understood Teacher asked this Question in order to understand whether we learn to definition of least upper bound :D thank you again.