I'm having trouble proving this:
let be $A,\ B\subset \Bbb{R}$ intervals and $AB\ =\ \{ab:a \in A,\ b \in B\}$
show that: $\sup(A)\sup(B) \le \sup(AB)$, $\inf(A)\inf(B) \ge \inf(AB)$
Show that product of supreme compared to supreme of the product (intervals)
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real-analysis
1 Answers
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Hint: Show that if $a \in A$ is "sufficiently close" to $\sup(A)$ and $b \in B$ is "sufficiently close" to $\sup(B)$, then $ab$ can be made "arbitrarily close" to $\sup(A) \sup(B)$. Conclude that any upper bound of $AB$ must be at least $\sup(A)\sup(B)$.
A similar idea applies to the inf. Be mindful that $A,B$ may contain negative numbers.