Thread: 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!

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 be nonempty set bounded below, by the completeness axiom we know that exists. So we can see that for any , and if the is not a lower bound of ...now multiply 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 ? 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.

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.

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 instead of (sorry for the accident).

If we multiply by negative one we get some number , by the properties of ordered fields we then have that and with just one more observation we have by def. that is


Can you handle it from there?

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

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.