Bartle, Introduction to Real Analysis (2011 4 ed), Section 2.4, Exercise 5, p 45. So, I'm working on the this problem and I got the first half done. I'm not sure about the second part and need help. Here is the problem:



*Let S be a set of nonnegative real numbers that is bounded above and let $T:=\{x² : x ∈ S\}$. Prove that if $u = \sup S$, then $u² = \sup T$.


So, we need to show that u² = sup T by:

*

*$t ≤ u², ∀ t ∈ T$

*$∀ ε > 0, ∃ t ∈ T$ so that $u² - ε < t$.

Proof: Assume that u = sup S is true. Then,

*

*Since u = sup S, it follows that s ≤ u. Squaring both sides, s² ≤ u². And since t = s², it follows that t ≤ u².


So, here is where I got stuck. I don't know how to construct the ε so that in the end I get u² - ε < t. I saw a solution to this problem where they suggested constructing ε/(2u) > 0, but it just didn't make sense to me. You can see the solution here.
 A: Given $\varepsilon >0$, you want to find $t\in T$ such that $$u^2-t<\varepsilon$$
Now, you know $t=s^2$ for some $s\in S$; so you want $$u^2-s^2<\varepsilon$$ for some $s\in S$. This means we want to find $s\in S$ so that $(u-s)(u+s)<\varepsilon$.
Now note $u$ is a fixed positive number (we can assume it is positive, else all elements of $S$ and hence all elements of $T$ are zero, since every element of $S$ is greater or equal to zero). Moreover, we know that for any $s\in S$, $s\leqslant u$, since $u$ is the supremum of $S$. This in turn means that $$\tag 1 u+s\leqslant u+u=2u$$ for any $s\in S$.
Since $u$ is the supremum of $S$, given $\dfrac{\varepsilon}{4u} >0$ there exists $s\in S$ for which $$\tag 2 u-s<\dfrac{\varepsilon}{4 u}$$ Consider now the element $s^2\in T$. Then using $(1)$ and $(2)$ we get $$u^2-s^2=(u-s)(u+s)<\frac{\varepsilon}{4u}(u+s)\leqslant \frac{\varepsilon}{4u}2u=\frac \varepsilon 2<\varepsilon$$
Since you already showed $u^2$ is an upper bound of $T$, the above shows it is the least upper bound of $T$, as we wanted.
A: I don't think you need $\epsilon$? Here's Bartle's partial solution.

If $x∈S$, then $0≤x≤u$, so that $x^2 ≤u^2$ which implies $\sup T ≤u^2$. If t is any
upper bound of T, then $x∈S$ implies $x^2 ≤t$ so that $x≤\sqrt t$. It follows that $u≤
\sqrt t$, so that $u^2 ≤t$. Thus $u^2 ≤\sup T$.

A: $ \forall t \in{T}, \exists s\in{S}, t = s^{2}$
also, since $ U = sup(S), \forall{s}\in{S}, U \ge s$, thus $U^2 \ge s^2$
Then, $U^2 \ge s^2 = t$, therefore $U^2$ is an Upper Bound of $T$
Now suppose that $U^2 \ne sup(T)$, then $\exists V$ such that $V \lt U^2$, and $\forall{t}\in{T}, V \ge t$
We now know that $U^{2} \gt V \ge t \ge 0, \forall{t}\in{T}$
Thus $U \gt \sqrt{V} \ge \sqrt{t} \ge 0$
Thus $U \gt \sqrt{V} \ge s \ge 0$, since $\sqrt{t}$ maps to an $s$
This says that $\sqrt{V}$ is an upperbound of $S$, which is less than $U$
However, since $U = sup(S)$, no such $\sqrt{V}$ exists, and $\therefore \nexists V$ where $V\lt U^2$ and $\forall{t}\in{T}, V \ge t$
Finally, since $U^2$ is an upperbound of $T$ and no lesser upperbounds exist, $U^2 = sup(T)$
