# Real analysis problem (from Intro to Real by Bartle - Problem 5 from section 2.4)

So, I'm working on the this problem and I got the first half done. Well, I think I got both halves done, but 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:

1. t ≤ u², ∀ t ∈ T
2. ∀ ε > 0, ∃ t ∈ T so that u² - ε < t

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

1. 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 http://webcache.googleusercontent.com/search?q=cache:wdcnWX840y8J:math.osu.edu/~davis.3929/Math4547Analysis1/lecture7/solution.tex+&cd=10&hl=en&ct=clnk&gl=us&client=firefox-a

Can you help, please?

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## 1 Answer

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.

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Sorry, Pedro, but the definition says that it has to be u² - ε < t, not u² − t < ε. – user129360 Feb 17 '14 at 6:11
@user129360 $u^2-\varepsilon <t$ is the same as $u^2-t <\varepsilon$. – Pedro Tamaroff Feb 17 '14 at 6:14
OK, I started to follow you how you got (u - s)(u + s) < ε. Then, how did you conclude that (u - s) < ε/(4u)? – user129360 Feb 17 '14 at 6:31
@user129360 Given $\varepsilon>0$ I picked s in S so that holds using the definition of sup S. – Pedro Tamaroff Feb 17 '14 at 6:35
@user129360 Indeed, we usually work in scratch paper and then translate our ideas formally! One also does reverse engineering. For example, \begin{align}u^2-s^2&<\varepsilon\\(u-s)(u+s)&<\varepsilon\\u-s&< \frac{\varepsilon}{u+s}\\& \leqslant \frac{\varepsilon}{u+u} \\&=\frac{\varepsilon}{2u}\end{align} Not we "only" get $\leqslant$; and we simply change $2$ to $4$ to get $\leqslant \varepsilon/2<\varepsilon$. – Pedro Tamaroff Feb 18 '14 at 0:01