Prove Sup S = c: Step-by-Step Guide

[HaLAA]

Homework Statement

S≡{x|x∈ℝ,x≥0,x² < c} Prove Sup S = c

Homework Equations

The Attempt at a Solution

Since x in the set real numbers, there are two cases for x: x < 1 or 0 <= x <=1

if 0 <= x <=1, then x < c + 1 since c is positive.
if x < 1, then x^2 < c < x*c + x = x(c+1)
thus x < c+1, by the completeness axiom, S has a least upper bound, let's called it b.

I am stuck at this step, can anyone give me a hit or a suggestion to keep going. Thanks

 

[Mark44]

Homework Statement

S≡{x|x∈ℝ,x≥0,x² < c} Prove Sup S = c

Homework Equations

The Attempt at a Solution

Since x in the set real numbers, there are two cases for x: x < 1 or 0 <= x <=1

It would seem to me that the two cases are $0 \le x \le 1$ and $x > 1$. Is there some reason you're not considering $x > 1$?

if 0 <= x <=1, then x < c + 1 since c is positive.
if x < 1, then x^2 < c < x*c + x = x(c+1)
thus x < c+1, by the completeness axiom, S has a least upper bound, let's called it b.

I am stuck at this step, can anyone give me a hit or a suggestion to keep going. Thanks

 

[nuuskur]

Set S consists of all real numbers $x$ such that $x\geq 0$ and $x^2 < C$.
We know $C>0$, for if $C<0$, there exists no real number that satisfies $x^2 < C$ and if $C=0$, then $x^2 < 0$, which is also impossible.
Provided that $C>0$ then we can say set S is non-empty and then apply the axiom:
If a non-empty subset has an upper bound, then it has a least upper bound.

If we establish $b$ as the candidate for the least upper bound, then $b$ satisfies certain critera:
1) $x\leq b$ for every $x\in S$ (in other words, $b$ has to be an upper bound)
2)For every $\varepsilon >0$, there exists an $x_0$ such that $b-\varepsilon < x_0$ (meaning $b$ is the least of upper bounds)
EDIT:
The second criterion could also be written as:
2a)For every $y\in\mathbb{R}$ such that $y < b$, there exists an $x_0\in S$ such that $y < x_0$
2 and 2a mean the same thing.

We can agree that $b$ is an upper bound, now fix an arbitrary $\varepsilon >0$ and show that $b$ is indeed the least of upper bounds and in the finale it follows immediately after that $\sqrt{C} = b$.

 

[davidmoore63@y]

Surely sup S = sqrt c not c ? Unless I'm missing something ?

 

[nuuskur]

Surely sup S = sqrt c not c ? Unless I'm missing something ?

I meant $\sqrt{C}$ thanks for the correction :)

 

[davidmoore63@y]

ok then so the proof of the OP would proceed as follows:

suppose Sup S = sqrt c + epsilon where epsilon > 0. Then (Sup S - epsilon /2)^2 > c (contradicting the definition of Sup S)
suppose Sup S = sqrt c - epsilon. Then (sup S + epsilon/2)^2 <c (again contradicting the definition of Sup S)
so Sup S = sqrt c

 

[nuuskur]

ok then so the proof of the OP would proceed as follows:

suppose Sup S = sqrt c + epsilon where epsilon > 0. Then (Sup S - epsilon /2)^2 > c (contradicting the definition of Sup S)
suppose Sup S = sqrt c - epsilon. Then (sup S + epsilon/2)^2 <c (again contradicting the definition of Sup S)
so Sup S = sqrt c

Well, hurts my eyes without TeX, for one, but more importantly, do not post complete solutions. I merely referred to the definition in an attempt to hint at how to proceed.

 

[davidmoore63@y]

ok noted

 

[HaLAA]

It would seem to me that the two cases are $0 \le x \le 1$ and $x > 1$. Is there some reason you're not considering $x > 1$?

That is a typo