Homework 1 Strategy 1

So I'm attempting to do the homework assignment by the first strategy Dr McClure wrote on the board yesterday (show $sup(B) = inf(S)$ where $B = \{x\in \mathbb{R}: x \leq y, \forall y\in S\}$).

The way I'm currently showing $sup(B)$ is a lower bound of $S$ is by saying $x \leq Sup(B) \leq y, \forall x \in B$ and $y \in S$. But... that doesn't work. Right?

Basically, I don't think that proves $sup(B) \leq y, y\in S$. But I just wanted to make sure my lack of faith in myself is correct.

I'm a little confused about your $B$. From my impression of Mark's explanation $B$ would be something like $B = -S = \{ -x : x \in S \}$.

Oooooh yeeeah, you're in the danger zone, gbrock, but do not fear, for the Macho Madness is entering the danger zone too.

You might consider negating your set $S$ as violincounter has; from there you should prove that there is some least upper bound, maybe $s$ for supremum, by proving the following true:

(i) $s$ is an upper bound for $B$;
(ii) if $b$ is any upper bound for $B$, then $s\leq{b}$.

After proving the two statements above, you know you have a supremum of the negated set $B$, and you must remember that $S\subset{\mathbb R}$, meaning that we have a field for which we can assume addition and multiplication of real numbers is commutative, associative, and distributive. This will be useful for when you start negating $sup(B)$ with the inequalities of $x\leq{sup(B),\forall{x}\in{B}}$.

One must only think of the glory of victory!

@gbrock I think what you are showing is exactly what you need to show by the definition of a lower bound. As long as you explain why you know those inequalities are true, that would show $\sup{B}$ is a lower bound.

@violincounter Your way is one way Mark said you could prove the problem, he also mentioned @gbrock's method as another way. Both are valid if done correctly...

@gbrock i as well was wondering how to go about transitioning from finding the $Sup(B)$ and then relating it to $y\in S$. I came up with the following,

"Since for all $x\in B$ we have $x\leq y$ for all $y\in S$, thus the $Sup(B)\leq y$ for all $y\in S$"

I'm not the best with proofs and correct me if it looks wrong, but i feel like this works, and then to finish it i just had to say how the $Sup(B)$ was a lower bound for $S$ and then say how it was the greatest lower bound.

Hey, Brock!
I approached it the same way you did I believe. On my second attempt I figured it was easier to show that the set of lower bounds has an supremum and then work from there. Later I saw in the book that 1.3.3 kind of prods you in this direction also. Glad to see I'm not alone.

@kmaclean I too had that to start out with but then I realized that this statement assumes that $\sup(B) \in B$ which is not necessarily true for all supremum. For example, for $S=\{x\in \mathbb{R} \mid x<4\}$, the $\sup(S)\notin S$. Thus, for your explanation to be correct you must show $\sup(B)\in B$.

@jmincey seems to have hit the nail on the head here.

I'm not the best with proofs either, @kmaclean ... We'll just have to drag ourselves through this course, I guess.

Ohhhhh I get it. And the supremum of these lower bounds is the infimum because it's the largest lower bound.

Deep down inside I knew that's why this worked, but you just made it far more clear to me. Thanks!