I was recently going through a proof of the result that:
"Prove the existence of $\sqrt b$ for any real number $b\geq 0.$"
The proof I encountered was:
Define the set $T = \{t\in\Bbb R : t^2 < b\}$ and let $\alpha =\sup T$ which we know exists because $T$ is non-empty (it contains $0$) and bounded above. We’ll show $\alpha^2= b$ by ruling out the possibilities $\alpha^2\lt b$ and $\alpha^2\gt b.$
First assume $\alpha^2\lt b$ and observe, $(\alpha+\frac 1n)^2= \alpha^2+\frac 1{n^2}+2\frac{\alpha}{n}\lt \alpha^2+\frac 1{n}+2\frac{\alpha}{n}.$ From Archimedian Property, we choose $n_0$ large enough so that $\frac{1}{n_0}\lt\frac{b-\alpha^2}{2\alpha+1}.$ This implies $\frac{2{\alpha} + 1}{n_0} < b-\alpha^2$, and consequently that $(\alpha+\frac 1n_0)^2\lt b.$
Thus, $\alpha+\frac1{n_0}\gt \alpha,$ contradicting the fact that $\alpha$ is an upper bound for $T.$
We conclude that $\alpha ^2< b$ cannot happen.
Now, let’s assume $\alpha^2\gt b.$ This time, we have, $(\alpha-\frac 1n)^2= \alpha^2+\frac 1{n^2}-2\frac{\alpha}{n}\gt \alpha^2-2\frac{\alpha}{n}.$
This time pick $n_0$ large enough so that $\frac{1}{n_0}\lt\frac{\alpha^2-b}{2\alpha+1}.$
With this choice of $n_0,$ we have $(\alpha-\frac 1{n_0})^2\gt b.$
This means $\alpha-\frac 1{n_0}$ is an upper bound for $T.$ But then $(\alpha-\frac 1{n_0})\lt \alpha = \sup T,$ which leads to a contradiction because all upper bounds of $T$ should be greater than
or equal to the supremum $\alpha.$ Thus, $\alpha^2 > b$ is not a possibility and we are left with $\alpha^2 =b$ as desired.
However, I dont understand how they are saying, "This means $\alpha-\frac 1{n_0}$ is an upper bound for $T$". I feel this needs more clarity. This is because, any real number $x\in \Bbb R$ such that $x^2\gt b,$ implies $x\notin T,$ but this does not nnecessarily, mean that $x$ is a upper bound of $T.$ For example, if $b=2,$ then $-\sqrt 3\notin T,$ and $(-\sqrt 3)^2=3\gt 2,$ But this does not mean that $-\sqrt 3$ is an upper bound of $T,$ for $1\in T$ and $-\sqrt 3\lt 1.$ So, I feel that we can't really, assert that $(\alpha-\frac 1{n_0})$ is an upper bound of $T.$
Any possible explanation regarding this issue would be really very helpful.
Best Answer
Your definition of $T$ is not correct. It should be $T=\{t∈R:t^2\le b\}$ since they did not exclude $b=0$ and they are saying that $T$ contains $0$.
As a consequence, $\alpha\ge0.$
"Now, let’s assume $\alpha^2\gt b.$" Then, $\frac1{n_0}\le\frac{\alpha^2-b}{2\alpha+1}$ by choice of $n_0,$ and $\frac{\alpha^2-b}{2\alpha+1}\le\alpha$ because $\alpha\ge0$ and $b\ge0.$ Therefore, $$\alpha-\frac1{n_0}\ge0.$$
"$(\alpha-\frac 1{n_0})^2\gt b.$ This means $\alpha-\frac 1{n_0}$ is an upper bound for $T.$" Indeed, $$\forall t\in T\quad\alpha-\frac 1{n_0}\ge t$$ because $\alpha-\frac1{n_0}\ge0$ and: