Prove that $\sup(S^2)=\sup(S)^2$ where $S$ is a nonempty bounded set of nonnegative real numbers.

Question

It seems like this statement is common sense so I don't know how to prove it using mathematical logic. S={x : x \in S }, and S^2={x^2 : x \in S }. Would it suffice to say that because the numbers in the set are nonnegative, that taking the supremum before or after squaring the set is irrelevant? How do I say this as a proof?

Answer 1

Let $s:=\sup S$. Then for all $x\in S$ we have $x\le s$. Squaring, we obtain $x^2\leq s^2$ and thus $\sup(S^2)\leq s^2=\sup(S)^2$.

Let $\epsilon>0$. Choose $x\in S$ with $s\leq \epsilon + x$. Then $$s^2\le (\epsilon+x)^2=\epsilon^2 +2\epsilon x + x^2\leq \epsilon^2+2\epsilon x +\sup(S^2) \leq \epsilon^2 + 2\epsilon s + \sup(S^2)$$ Letting $\epsilon\to 0$, we obtain $s^2\leq \sup(S^2)$.

Thanks to @Daniel Schepler for the useful comment.

Answer 2

The fact that the numbers are nonnegative is essential: indeed $$ (\sup\{-2,1\})^2=1,\qquad \sup\{(-2)^2,1^2\}=4 $$

Suppose $S$ consists of nonnegative numbers (and is not empty). We can assume that $S$ contains positive numbers, for the case $S=\{0\}$ is trivial. Let $c=\sup S$, so $c>0$.

Let $b$ be an upper bound of $S$. Then, for $a\in S$, we have $a\le b$, so also $a^2\le b^2$. Hence $b^2$ is an upper bound of $S^2$. In particular, $c^2$ is an upper bound of $S^2$.

Is it the least upper bound? Let $\varepsilon>0$: we need to find $a\in S$ such that $a^2>c^2-\varepsilon$, which is the same as $c^2-a^2<\varepsilon$.

How can we do it? We need $(c+a)(c-a)<\varepsilon$. Since, for every $a\in S$ it holds $a\le c$, we can first find $a\in S$ such that $$ a>c-\frac{\varepsilon}{2c} $$ so $$ c^2-a^2=(c+a)(c-a)\le 2c(c-a)<2c\frac{\varepsilon}{2c}=\varepsilon $$ and we're done.