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Let A be a non empty bounded subset of the real numbers, let $B = A \cap [0,\infty)$ and $C = A \cap (-\infty,0]$. Assume both $B$ and $C$ are non empty. Let $D = \{a^2 : a \in A\}$

Find $\sup D$ in terms of $\sup B$ and $\inf C$ and justify.

So my reasoning is that it is $\sup D = \max ((\sup B)^2,(\inf C)^2)$ but I am not sure how to justify this formally!

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  • $\begingroup$ If $0\le a \le a'$ then $a^2 \le a'^2$ so if $a \in B$ then $0 \le a \le \sup B$ then $a^2 \le \sup B^2$. Ditto negs. So max(sup B^2, inf C^2) is an upper bound... you can finish this by taking cases. $\endgroup$ – fleablood Oct 27 '16 at 23:08
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\begin{align} \sup(D) &= \sup\{a^2: a \in A\}\\ &= \sup(\{a^2: a \in B\} \cup \{a^2 : a \in C\})\\ &= \max(\sup\{a^2: a \in B\}, \sup \{a^2: a \in C\})\\ &= \max((\sup B)^2, (\inf C)^2). \end{align}

Each step will need justification.

The first two equalities are by definition.

The last equality is due to elements of $B$ being nonnegative, and elements of $C$ being nonpositive. You may need to be more formal if you are being graded rigorously.


The third equality requires you to prove $\sup(E \cup F) = \max(\sup E ,\sup F)$.

Proving "$\ge$":

clearly $\sup E \le \sup (E\cup F)$ and $\sup F \le \sup (E \cup F)$.

Proving "$\le$":

for any $\epsilon>0$ there is an element $x$ of $E\cup F$ such that $x\ge \sup(E \cup F) - \epsilon$. If $x \in E$, then $x \le \sup E$, so $\sup(E \cup F) - \epsilon\le x \le \sup E \le \max(\sup E, \sup F).$

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Otherwise $x \in F$ and then $\sup(E \cup F) - \epsilon \le x \le \sup F \le \max(\sup E, \sup F).$ Since $\epsilon > 0$ is arbitrary, we have $\sup(E \cup F) \le \max(\sup E, \sup F)$.

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  • $\begingroup$ Thank you that makes total sense! $\endgroup$ – Fkins Oct 28 '16 at 15:41

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