1. ## The Completeness Axiom

We have S defined as: S = {x|x in R, x>=0, x^2< c}

How do I show that c+1 is an upper bound for S, and that S has a least upper bound denoted by b??

I tried doing it this way:

If c<1, we have x<1<1+c, i.e, 1+c is an upper bound for S

If c>1, then x<sqrt(c)<c<c+1 and thus, c+1 is an upper bound for S.

I am still unsure if this is the correct method!! Any suggestions?

2. Originally Posted by harish21
We have S defined as: S = {x|x in R, x>=0, x^2< c}

How do I show that c+1 is an upper bound for S, and that S has a least upper bound denoted by b??

I tried doing it this way:

If c<1, we have x<1<1+c, i.e, 1+c is an upper bound for S

If c>1, then x<sqrt(c)<c<c+1 and thus, c+1 is an upper bound for S.

I am still unsure if this is the correct method!! Any suggestions?
Well...this is so easy...it's hard. I'm not entirely sure I know a good way of saying this. Suppose that there existed some $\ell\in S$ such that $c+1<\ell$. Then, since $c,\ell>0$ we have that $\left(c+1\right)^2<\ell^2 which is clearly a contradiction.

Now, since $S$ has an upper bound $c+1$ we see that $S\ne\varnothing$ (e.g. $0\in S$), $S$ is bounded from above, and that $S\subseteq \mathbb{R}$...therefore $\sup\text{ }S$ exists! ta-da!

3. Originally Posted by harish21
We have S defined as: S = {x|x in R, x>=0, x^2< c}
How do I show that c+1 is an upper bound for S, and that S has a least upper bound denoted by b??
Suppose that $\left( {\exists t \in S} \right)\left[ {c + 1 < t} \right] \Rightarrow \quad \left( {c + 1} \right)^2 < t^2 < c$
What is wrong with that?