1. ## Completeness Axiom

Let A be a nonempty set of the reals which is bounded below, and let -A be the set of all -x such that x is in A.

Prove inf(A) = -sup(-A)

I know there must exist an alpha such that alpha = inf (A) by the completeness axiom, implying x is greater than alpha for all x in A.

But I don't know what to do next. Thanks!

2. Originally Posted by Zero266
Let A be a nonempty set of the reals which is bounded below, and let -A be the set of all -x such that x is in A.

Prove inf(A) = -sup(-A)

I know there must exist an alpha such that alpha = inf (A) by the completeness axiom, implying x is greater than alpha for all x in A.

But I don't know what to do next. Thanks!
Hello friend

What exactly are you having difficulties with? This is a pretty straightforward result. How about I get you started and you finish?

Let $E\subset \mathbb{R}$ be nonempty set bounded below, by the completeness axiom we know that $\inf E=\xi$ exists. So we can see that for any $x\in E$, $\xi and if $x>\xi$ the $x$ is not a lower bound of $E$...now multiply $E$ by negative one, noting that multiplying a set by negative one is equivalent to multiplying each element by negative one. What does this do to $(1)$? What can we conclude from this?

I hope that helps...any problems just ask

P.S. Frome one analysis novice to another I have found that problems like this can look like dragons but almost always end up being puppies...exhaust every idea...leave no stone unturned...some other trite expression...just get your hands dirty.

3. I am extremely new to all of this. I'm still a little confused. Here is what i have. there exists an alpha such that alpha = inf(A) and alpha is less than or equal to x for all x in A (completeness axiom)

Then I said, this implies -alpha is greater than or equal to -x (ordered field ?)

Then i said, for all -x in -A, where x is also in A, -alpha = sup(-A)

But i know that's wrong because -alpha is really just an upper bound of -A. I can't seem to get the sup in this whole mess. It's driving me crazy.

4. Originally Posted by Zero266
I am extremely new to all of this. I'm still a little confused. Here is what i have. there exists an alpha such that alpha = inf(A) and alpha is less than or equal to x for all x in A (completeness axiom)

Then I said, this implies -alpha is greater than or equal to -x (ordered field ?)

Then i said, for all -x in -A, where x is also in A, -alpha = sup(-A)

But i know that's wrong because -alpha is really just an upper bound of -A. I can't seem to get the sup in this whole mess. It's driving me crazy.
Note that for some reason I had $\xi instead of $\xi\leqslant x$ (sorry for the accident).

If we multiply by negative one we get some number $-\xi$, by the properties of ordered fields we then have that $y<-\xi~~\forall y\in -E$ and with just one more observation we have by def. that $-\xi$ is $\sup -E$

Can you handle it from there?

5. Yes I can handle it from there. That's what i had, but my problem is why can u assume (-alpha) is the sup(-A), clearly it is, but not by definition. -alpha would be just an upper bound of (-A). sup(-A) exists by definition, but something is bothering me when you equate that to -alpha.

If you let beta = sup(-A), then -alpha = beta and the proof can be completed easily, but what definition are u using to skip the step where beta is shown to equal -alpha

6. Originally Posted by Zero266
Yes I can handle it from there. That's what i had, but my problem is why can u assume (-alpha) is the sup(-A), clearly it is, but not by definition. -alpha would be just an upper bound of (-A). sup(-A) exists by definition, but something is bothering me when you equate that to -alpha.

If you let beta = sup(-A), then -alpha = beta and the proof can be completed easily, but what definition are u using to skip the step where beta is shown to equal -alpha
Im not using any definition. What I did was merely a self-imposed convention, if you don't like it like that feel free to change it.