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cbarker1

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**Prove this Proposition 1.2.6 v in Introduction to Real Analysis by Jiri Lebl**

Dear Everybody,

I need some help with seeing if there are any logical leaps or errors in this proof.

The theorem states:

$A\subset\Bbb{R}$ and $A\ne\emptyset$

If $x<0$ and A is bounded below, then $\sup(xA)$=$x(\inf(A))$

Proof: Suppose $x<0$ and A is bounded below. Let d be a lower bound of A. Then, for all $a\in A$, $a\ge d$. So $ax\le dx$ for some $x< 0$. So dx is an upper bound of xA. In particular, if ${d}_{0}=\inf A$, then $\sup(Ax)\le{d}_{0}x=x \inf A$.

Let f be an upper of Ax. Then, for all $a\in A$, $ax<f$ for some $x<0$. Since x is negative, the inequality will switch to greater than to yield: $a>\frac{f}{x}$. So $\frac{f}{x}$ is a lower bounded of A. In particular, if ${f}_{0}=\sup (Ax)$, then $\inf A\le \frac{{f}_{0}}{x}$

$\inf A\le \frac{\sup(Ax)}{x}$

$x \inf A \ge \sup(Ax)$

Therefore, $\sup(Ax)=x(\inf A)$ QED

Thanks,

Cbarker1

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