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

[cbarker1]

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

 

[jvicens]

Dear Cbarker1,

Thank you for your post. After reviewing your proof, I do not see any logical leaps or errors. Your proof is well-structured and follows the correct steps to prove Proposition 1.2.6. Here are some comments and suggestions to make your proof clearer and more concise:

1. In the first line, you can specify that A is a non-empty subset of $\Bbb{R}$.

2. In the second line, you can clarify that d is a lower bound of A.

3. In the fourth line, you can specify that dx is an upper bound of xA, as this is not explicitly stated.

4. In the seventh line, you can explain why $\frac{f}{x}$ is a lower bound of A by using the fact that x is negative.

5. In the eighth line, you can clarify that $f_0$ is an upper bound of Ax.

6. In the ninth line, you can mention that $\frac{f_0}{x}$ is a lower bound of A and use it to show that $\inf A \leq \frac{f_0}{x}$.

7. In the final line, you can mention that $\frac{\sup(Ax)}{x} \leq \inf A$ and use it to show that $x\inf A \geq \sup(Ax)$.

Overall, your proof is well-written and logically sound. I hope my suggestions are helpful in making your proof clearer. Keep up the good work!