Supremum and Infimum of Bounded Sets Multiplication

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SUMMARY

The discussion focuses on proving the properties of the infimum and supremum of the product of two bounded sets of non-negative real numbers, specifically sets X and Y. The conclusions drawn are: a) inf{X*Y} = infX * infY, and b) sup{X*Y} = supX * supY. The proof for part a) establishes that the infimum of the product set is bounded below by the product of the infima of the individual sets, and similarly for the supremum in part b).

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Hey all,
I started to learn this subject, and i understtod how to find the supremum and infimum of a given set or function.
but I have problem with one question which I can not solve, and I don't know how to start.
This is the quesion:

Given to bounded sets X and Y, which their element are REAL and non negative numbers,
so that X*Y = {x*y: x in X, y in Y}
prove that:

a)inf{X*Y} = infX *infY
b)sup{X*Y}= supX * supY

Can anyone help please?
 
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Welcome, esuahcdss12! (Wave)

Let's start with question a). Let $a = \inf X$ and $b = \inf Y$. Given $x\in X$ and $y\in Y$, $x \ge a$ and $y\ge b$, so that $xy\ge ab$ (which follows from the fact that $x$ and $y$ are nonnegative). Deduce that $\inf(XY) \ge ab$. To obtain $\inf(XY) \le ab$, note that if $\epsilon > 0$, $a + \epsilon$ is not a lower bound for $X$ and $b + \epsilon$ is not a lower bound for $Y$. So there are $x\in X$ and $y\in Y$ such that $a + \epsilon > x$ and $b + \epsilon > y$. Hence $(a + \epsilon)(b + \epsilon) > xy \ge \inf(XY)$. Since $\epsilon$ was arbitrary chosen, deduce that $ab \ge \inf(XY)$.

The argument for question b) is similar.
 

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