Ok, thank you, I understand the types of proof now. Going to also look into some books on set theory proofs and examples. Thanks again for all the help.
Ok, so now all I need to show is that all the other lower bounds are ≤ sup ##C##?
Your constructions are a lot more elegant than mine, I assume this comes with practice? I can also take what you said here, and then say, c ≤ α ≤ b must be true. Which to me means that any element from ##C##...
I was talking extraneously in the last part of my second sentence. But isn't this what I should be doing, I need to show that ##C## has a least upper bound, and by doing that, will show that the set of lower bounds has a supremum, and that would be by definition a greatest lower bound for ##B##...
Well, since C is bounded above by B, and since C is a subset of A, C has the least upper bound property, and that means that the inf B is greater than every element in C right? Then the inf B is the supremum of C? This should help cover the case of the sup C not being in C.
Now I'm really confused, you said...
Which is what I did. It makes sense with the example you gave, but I'm at a lose now I guess.
The supremum of ##C## is the greatest lower bound for ##B## it seems like, and that's what I said right? Maybe I worded it wrong?
Not sure why I do that, I think it's just instinct to want to jam algebra into it where ever I can. But I can see how it makes things harder to understand, I'll try to be more direct I guess, in the definitions when doing a proof.
Forgot that part I guess, but the use of ##B## as an upper...
Ok, I think I have it now.
Let ##B## be a non-empty subset of ##A## that's bounded below, such that b≤ any x∈##B##.
Let ##C## be the set of all lower bounds for ##B## such that for any y∈C, y≤b≤ any x∈##B##.
We know that ##A## has the least upper bound property, so ##C## also has a least upper...
Ok, first, thank you, this is exactly the type of help needed. I'm not very experienced with these types of proofs, I've done proofs in the past, but mostly number theory and typical geometry stuff, and the subject matter is much different, and using these concepts in the right order in the...
Took a break from the problem for a day, came back to it and this is what I have now.
Let A0 be a non-empty subset of A, A0 is bounded above and has the least upper bound property, such that b≥x for any x∈A0, and b is the least upper bound, sup A0=b. B is the set of all bounds y such that...
Ok, so I'm trying to prove that the set you call C has a property that implies the result?
This is a topology book, and it has a list of axioms, 8 total, and says the first 6 form a field, which is reminiscent of vector spaces that I studied in linear algebra, and then the last two axioms...
Ok, this makes more sense as a way to approach the proof. But as far as using the property in the proof, my construction was trying to show that any non-empty set of A will have the property, and in the book I'm working from it says that these properties can be generalized to any ordered set...
Homework Statement
Prove if an ordered set A has the least upper bound property, then it has the greatest lower bound property.
Homework Equations
Definition of the least upper bound property and greatest lower bound property, set theory.
The Attempt at a Solution
Ok, I think that my main...
Ok, I think I understand this now, so the first part shows injectivity and the second shows surjectivity, and the order of the composition changes to show that the sets your mapping actually allow for an inverse, and then bijectivity follows?
OK, I see where I went wrong, but how do I use the existence of g and h to show that the function is a bijection? It has to be injective and surjective, I know the definition of them but don't see how g and h show it's bijective. Can you point me in the right direction? Does 1 function show one...
Mod note: Moved from a technical section, so missing the homework template.
Here is what I'm trying to prove.
Let f:A->B. If there are two functions g:B->A and h:B->A such that g(f(a))=a for every a in A and f(h(b))=b for every b in B, then f is bijective and g=h=f^(-1).
I think I have most...