If $C=\{x^2,x\in S\}$, show that $\sup(C)=\max\{\sup(S)^2,\inf(S)^2\}$

real-analysissupremum-and-infimum

Let $S\subset \mathbb{R}$ non empty and bounded. Prove that:

  1. $S$ can be chossen to satisfy $\sup (S)\neq \max{S}$.
  2. If $C=\{x^2,x\in S\}$, show that $\sup(C)=\max\{\sup(S)^2,\inf(S)^2\}$.

For (1) I do not understand it very well. If $S$ is finite, then it is not true. In the infinite case, i do not know if i have to give an example.

For (2) It got $\sup (C)\leq \max\{\sup(S)^2,\inf(S)^2\}$. If $\sup (C)< \max\{\sup(S)^2,\inf(S)^2\}$, I get the cases:

a) If $\sup(S)^2>\inf (S)^2$.Then if $\alpha=\sup(C)\geq 0$, then
$$\alpha<\sup(S)^2\Longrightarrow \sqrt\alpha<|\sup(S)|$$
Which gives 2 more cases:

a.1) $\sup(S)\geq 0$, then
$$ \sqrt\alpha<\sup(S)\Longrightarrow \exists a\in S,\sqrt\alpha<a\Longrightarrow \alpha<a^2,a \in S (Contradiction).$$

a.2) $\sup(S)< 0$, then
$$ \sqrt\alpha<-\sup(S)=\inf(-S)\Longrightarrow \forall a\in S,\sqrt\alpha<-a\Longrightarrow \alpha<a^2,a \in S (Contradiction).$$

In the same way if $\inf(S)^2>\sup(S)^2$

Is it ok?

Thank you

Best Answer

It is straightforward to show the following:

  1. If $A,B$ are disjoint and non empty then $\sup A \cup B = \max ( \sup A, \sup B)$.
  2. If $f$ is non decreasing and continuous then $\sup f(A) = f(\sup A)$.
  3. If $f$ is non decreasing and continuous then $\sup f(A) = f(\inf A)$.

Let $S_+ = \{ x \in S | s \ge 0 \}$ and $S_- = \{ x \in S | s < 0 \}$.

We have $S = S_+ \cup S_-$, so $C = S_+^2 \cup S_-^2$.

If $S_+$ is non empty, let $f(x) = \begin{cases} 0 , & x <0 \\ x^2,& \text{otherwise} \end{cases} $, then $\sup f(S_+) = \sup S_+^2 = f(\sup S_+) = (\sup S_+)^2 = (\sup S)^2$.

If $S_-$ is non empty, let $g(x)=f(-x)$, then $\sup g(S_-) = \sup S_-^2 = g(\inf S_-) = (\inf S_-)^2 = (\inf S)^2$.

If $S_-$ is empty then $0 \le \inf S \le \sup S$ and $\sup S^2 = (\sup S)^2$ (from 2.).

If $S_+$ is empty then $\inf S \le \sup S \le 0$ and $\sup S^2 = (\inf S)^2$ (from 3.).

If both are non empty then $\sup S^2 = \max( \sup S_+^2, \sup S_-^2 ) = \max( (\sup S)^2, (\inf S^2 ))$ (from 1.) and so we have the desired result.

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