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Let $S$ be a set of non-negative real numbers that is bounded above and let $T=\{x^2:x\in S\}$. Prove that if $u = \sup S$, then $u^2=\sup T$.

I think I need to show some of my work. First of all, it is obvious that $u^2$ is an upper bound of $T$ because $u\ge x$ for all $x\in S$, then $u^2\ge x^2$. Then I want to show for any upper bound $v$ of $T$, $v\ge u^2$. But I am stuck at here because what I thought it should be like this since $v$ is an upper bound of $T$, then $v\ge x^2$ for all $x\in S$, then $\frac vx \ge x$, the $\frac vx$ is an upper bound of $S$, so $\frac vx\ge u$, then $v \ge ux$, but then I cannot prove $v\ge u^2$.

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    $\begingroup$ Non-negativity here is key. Why? And for the proof, you need to check the definition of sup $\endgroup$ – Lost Dec 19 '13 at 4:23
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$0 < x < u$ for all $x \in S$, which implies $0 < x^2 < u^2$ for all $x \in S$, so $u^2$ is an upper bound on $T$. Now we must prove it is the least upper bound.

Suppose that it is not the least upper bound, so for some number $\alpha < u^2$ we have $0 < x^2 < \alpha$ for all $x \in S$. Then, we have that $0 < x < \sqrt{\alpha}$ for all $x \in S$. But this contradicts the fact that $u$ is the least upper bound of $S$, because $\sqrt{\alpha} < u$. Therefore, $u^2$ must be the least upper bound on $T$.

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$sup S=a$

Take any element from $y=x^2$ from T

Now as $x<a$ for $x\in S$ it follows $x^2<a^2$ for $x\in S$ (beacuase $x$ and $a$ are both positive.)

Therefore, $a^2$ is an upper bound from T

To show it is the least upper bound.

Proof by contradiction.

If not, let $a_1 $ is an upper bound of T such that $a_1<a^2$. Note: $a_1>0$ as all the terms of T are positive .But then it will follow that $\sqrt a_1 < a$ (beacuase $a^2$ and $a_1$ are both positive. ). Also $a_1$ is an upper bound will say $x^2<a_1$ for $x\in$S .i.e. same as saying $x<\sqrt a_1$ for $x\in S$. Therefore, we have an upper bound $a_1$of S which is less than $a$, that contradicts that $a=sup S$

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    $\begingroup$ I don't understand, Why is this downvoted? $\endgroup$ – Babai Dec 19 '13 at 4:58
  • $\begingroup$ It looks right to me. $\endgroup$ – Hawk Dec 19 '13 at 5:53

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