Pr. 5 of chapter 1 of Walter Rudin.

Question

Pr 5. Let $A$ be a nonempty set of real no. Which is bdd below. Let $-A$ be set of all no. $-x; x \in A$. Prove $\inf(A)=-\sup(-A)$.

My attempt: If $x>0, \inf(A) \ge 0$ & by def $\inf(A) \le x$, for all $x\in A$. Then, $-x < 0$, any $\mathbb R^+$ real no. is an upper bdd of set $-A$. Thus it has a LUB. $\sup(-A) \le 0$ & by def $\sup(-A) \ge -x$, for all $x\in A$. So, $x \ge -\sup(-A)$. Combining above inequality to obtain $-\sup(-A) \le (x\in A) \ge \inf(A)$. Similar argument can be done for $x \le 0$. The equality holds when $x=-\sup(-A)=\inf(A)$.

My question is: What if $\inf(A)$ is not an element of set $A$? Then equality doesn't hold, right? So, $\inf(A) \neq-\sup(-A)$. This is my first and only solution came to my mind while solving this problem.

There are plenty of correct sol. on internet. Like sol. to selected exercise by sam blinstein(unfortunately now I can't find link to pdf). In that solution it is shown that, it is not possible $\gamma(r) > -\sup(-A) r\ge x, x\in A$. By step-by-step logical argument. Do I have to do the same thing? If yes, then my sol. is not efficient. I just want to know, why my solution isn't working despite using definition of supremum and infimum.

Answer 1

Your proof is on the right track, but you don't need to consider separate cases like $x>0$.

Also, notation like $-\sup(-A) \le (x\in A) \ge \inf(A)$ is confusing. I think I know what you are trying to say, but you should not write $a \le b \ge c$.

First, before we go into the details of the proof, lets outline the general steps. Your plan seems to be:

1. Show that $-A$ has a supremum.
2. Show that $-\sup(-A)$ is the infimum of $A$.

That's a good plan. Let's break up step 2 into:

• show that $-\sup(-A)$ is a lower bound for $A$
• show that it is the greatest lower bound of $A$

For step 1, rather than take separate cases, such as $x>0$, just use that we know that $A$ has an infimum (since it is non-empty and bounded below). Then, instead of $x > 0$, we just need $x \ge \inf(A)$. Then we know that $-x \le -\inf(A)$. Hence, $-\inf(A)$ is an upper bound on $-A$. Also, we know that $-A$ is non-empty. Therefore, we know that $-A$ has a supremum. Step 1 complete.

You have step 2a in your proof, but I'll just state it slightly differently:

$\forall x \in A$, we know that $-x \in -A$, so that $-x \le \sup(-A)$, and therefore $x \ge -\sup(-A)$. Hence $-\sup(-A)$ is a lower bound on $A$.

For step 2b, it does not matter if $\inf(A)$ is in $A$ or not. We just need to show that $-\sup(-A)$ is the greatest lower bound on $A$. We show that by showing that if some number is greater, then it is not a lower bound:

Let $y > -\sup(-A)$. Then $-y < \sup(-A)$, which means that $\exists z \in -A$ such that $-y < z$

(since otherwise $-y$ would be an upper bound of $-A$ that was less than $\sup(-A)$)

Then we know that $-z \in A$ and $-z < y$, so that $y$ is not a lower bound on $A$.

Therefore $-\sup(-A)$ is the greatest lower bound on $A$.