So I've been struggling with starting the proof that's due Friday. I have figured out that it's similar to the proof we did in class on Monday and we want to use a negative to flip sup(S) to inf(S). Would anyone mind maybe sharing ideas or how you've started the proof?
Suppose $m$ is a lower bound for $S$. Define cS the way we did in class (c < 0) (See the mark's suggestions below). Show that $cm$ is an upper bound for $cS$. Now invoke the sacred rite of the Axiom of Completeness.
Now that you have access to the supremum of $cS$ you can reference the statements of the properties of the supremum and massage them until an infimum for $S$ pops out.
Hope that helps.
Edit: words
Okay! That's kind of what I was thinking, but I just wasn't sure it was okay to define the c. That helps. Thanks!
I think the violincounter has given good advice. I would recommend that you work with a specific choice of $c<0$, however. Like, $c=-1$ seems natural. Thus, given that $-S$ has a supremum (call it $\beta$) you can $\alpha=-\beta$ and show that $\alpha$ satisfies the condition to be the infimum of $S$.
This is the method that I also used. Now that the homework is all turned in... I'm curious to hear how it worked for you @agibson ?
I figure we can all discuss what we did and critique each other now. There is a bit of my proof that I did that I was not very sure about, tell me what you all think:
Basically I started with the lower-bounded set $S$ and described it being multiplied by $-1$ to make it $-S$. I explained that $-S$ had a supremum (because we proved every upper-bounded set has a supremum). Then my explanation for why the sup($-S$) was the inf($S$) was basically me describing the rules for the supremum, showing that $-S$ is just reversed $S$, and then applying the same rules for the sup($-S$) for the resulting -sup($-S$) in $S$.
Is this how most of you did it?
So, I started off with a similar proof to the one we did in class where I let $\beta=sup(-S)$ and $\alpha=inf(S)$, and I proved that -S has an upper bound. Thus by the axiom of completeness $sup(-S)=\beta$.Then in the next part of my proof I showed that $\beta=-\alpha$ by using an $x\in-S$ when $-x\leq \beta$ thus proving that $-\beta$ is a lower bound for S. I then used this lower bound, m, of S to actually prove that its negative is the upper bound for the -S (relating back to a previous part in my proof) and ultimately I ended up with $\beta\leq-m$ and $-\beta\geq m$. Therefore since m is a lower bound then $-\beta=\alpha=inf(S)$. Hopefully the way I typed this makes sense. I kind of wanted to explain my process more so than just posting my proof.
Here is my proof:
Assume that $S\subset\mathbb R$ is non-empty and bounded below. We hope to show that $S$ has an infimum.
Let $m$ be a lower bound for $S$ and define
$$-S = \{-x\in \mathbb R: x\in S\}.$$
We claim that $-m$ is an upper bound for $-S$. To show this, let $y\in-S$. Thus, $y=-x$ for some $x\in S$. Since $m$ is a lower bound for $S$, we have $m\leq x$, which implies that $y=-x\leq -m$. As $y\in-S$ was arbitrary, $-m$ is an upper bound for $S$.
Since we now know that $-S$ is non-empty and bounded above, we may let $\beta = \sup(-S)$. We claim that $-\beta$ is the infimum of $S$. To prove this, we must show two things:
$-\beta\leq x$ for all $x\in S$.
This is true because $-x\leq\beta$ for $x\in S$, since $\beta$ is an upper bound for $-S$.
If $-\beta < a$, then $a$ is not a lower bound for $S$.
This is true because $-a<\beta$ implies that $-a$ is not an upper bound for $-S$.