Supremum and infimum of the following set

Question

I am a beginner at Real Analysis and attempting to prove some of the results in Problems in Mathematical Analysis volume - I by Kaczor and Nowak. I would like someone to verify, if my proof to the below question is rigorous and correct. If it isn't, I would like some hints that would lead me to the correct proof.

1.1.2. Let $A \subset \mathbb{R}$ be a non-empty subset. Define $-A=\{x:-x \in A\}$. Show that \begin{align*} \sup (-A)&=-\inf A \\ \inf (-A)&=-\sup A \end{align*}

Proof.

(1) Let $l \in \mathbb{R}$ be the infimum of the set $A$. From the definition of the greatest lower bound for a set:

(a) $l$ is a lower bound for $A$.

So, $l \le x$ for all $x \in A$. Thus, $-l \ge -x$ for all $x$. So, $-l$ is an upper bound for $-A$.

(b) If $b$ is any lower bound for $A$, $b \le l$. Therefore, $-b \ge -x$ for all elements in $-A$ and $-b \ge -l$.

Hence, we can conclude that $-l$ is the supremum of $-A$.

(2) The proof for the infimum of $-A$ follows similarly.

Answer 1

Rather than commenting on your work, I will comment on your ability to verify your work.

In other to prove that a number is the supremum of a set, there are two things you need to do. If you've shown those two things, then you are done, you don't need any help checking your work.

(This is not to discourage you from asking for help on this website, but you seem to understand how to prove what's being asked of you, so I'm trying to help you figure out that at least in this instance, you can easily verify your work yourself.)

Edit:

Actually, you know what, I think I mislead you above. The issue I see with your proof is that you proved $-l$ is an upper bound for $-A$, but you didn't quite prove that it's the least upper bound. So consider some number $r$ such that $r \geq -x$ for all $x \in A$. (Aside: I think you typed $S$ in a couple places where you meant to type $A$.) We see that $r \geq -x$ implies that $-r \leq x$, so $-r$ is a lower bound for $A$, and therefore $-r \leq l$ since we know $l$ is the infimum of $A$. Thus, $r \geq -l$, so $-l$ is the least upper bound of $-A$ as desired.

Basically, you took any other lower bound $b$ of $A$, showed that $-b$ is an upper bound for $-A$, and showed that $-l$ is no greater than $-b$. This is really close to what you should do, but I don't think it's quite right. To be proper you need to take any other upper bound $r$ for $-A$ as I did above, and then show that $r$ is at least as big as $-l$.

Answer 2

I'm going to be very fussy.

In a) you showed that for every $x \in A$ then $-x \in -A$ and $l\le x$ so $-l \ge -x$. But you haven't prove that $-l \ge y$ for all $y \in -A$. You've only shown it is true for the $y \in A$ so that there is an $x \in A$ so that $y = -x$.

Yes, you can argue that there is no other $y \in A$ except thoose of the form $-x \in A$ but.... well, you have to argue that.

Better to do: For any $y \in -A$ then there is an $x\in A$ so that $y = -x$. $x \le l$ so $-l \ge y$ so $-l$ is an upper bound of $A$.

b) is a little worse. You have argued that for ever lower bound $b$ then $b \le l$ so $-b \ge -l$. But there's nothing there that tells me $-b$ is an upper bound nor that if $-b$ is an upper bound that $-b$ represents an *arbitrary upper bound.

I think this, like my issue of a), is a problem of starting with what you know in $A$ and applying it to $-A$ is a fallacy as one once you select the item $x,b$ in $A$ then deriving $-x, -b$ from that $x,b$ is no longer arbitrary. You need to pick an arbitrary upper bound in $b$ of $-A$. Then show $-b$ is a lower bound of $A$ (because if $b$ is an upper bound of $-A$ then for any $x \in A$ then $-x \in A$ and $b \ge -x$ so $-b \le x$ so $-b$ is a lower bound of $A$.) Then as $-b$ is a lower bound of $A$ then $-b \le l$ so $b\ge -l$. But $-l$ is an upper bound, but $-l$ is equal or less than all other upper bounds so $-l = \sup -A$.

I'm going to be further picky by asking you: Have you proven that if $m$ is a upper bound of $X$ that is equal or lesser than all upper bounds then $m = \sup X$? It's not the definition of $\sup$ so if you havent proven it is an equivalent condition you must before you can state it.

But It is easy to prove it is an equivalent condition so maybe you have proven it.

.... Any way, that is me being extremely fussy. I'd say you were over 98% of the way there and I'm being really picky.