In textbooks, supremum is often defined as follows:
Let be a partially ordered set, and let .
(1) is a least element of if for all .
(2) is an upper bound of if for all .
(3) is called a supremum of if it is the least element of all upper bounds of . Denote it as .
If we express these definitions in wffs of first-order logic, then
(1) is a least element of if the following wff is true: .
(2) is an upper bound of if the following wff is true: .
Now if is an empty set , it follows that any element of is an upper bound of , for in (2) the left part of is always false, so the implication is vacuously true, and the whole quantifier expression is true. This is like what we do in proving that the empty set is contained in any set. Therefore, by definition, is the least element of the set consisting of all upper bounds of , that is, just the least element of . I'm not sure if it is right or not, does anybody know? Thanks.
I know that W. Rudin's book "Principles of Mathematical Analysis" discusses the extended real number system in P11-12, where two symbols and are introduced. But I think to extend the set to include something standing for infinity, only one symbol need to be added, that is, the symbol in the one-point compactification of (See, e.g. 29 of Munkres' "Topology", this proccess makes the extended system compact, the added symbol is a limit point of , that's why you add a bar above ). So, both and are actually the same; there is no difference between them; we write + or - in front of just in order to indicate how sequences in converge to -- whether the sequence is becoming larger and larger, or conversely. Am I right?
The other, two-point compactification, adds two points and , which at the same time provide with a smallest and a largest element. In the two-point compactification, and are not at all the same but are quite distinct (otherwise the order-structure of would get out of joint). So, two-point compactification is used if you are not satisfied with just compactifying but want a certain kind of closure of its order structure at the same time that makes, among other things, that and both exist.
In the one-point compactificaion you are not allowed to just posit that is at the same time equal to and , because, as I wrote already, this would make at the same time the largest and the smalles element, thus completely upsetting the order-structure because of transitivity.
It is just a fact of life that a single non-compact space can have different compactifications.
Yes! it can be readily proved that in order topology is really a compact space having space as its subspace , and that and are limit points of , hence a two-point compactification of which can preserve the order structure is proven.
Thank you Failure, your post deepened my understanding of the mysterious symbol since the study of calculus, thank you!
Interesting math! :-)
Rudin's book has shown that this "two-point" extension has the least-upper-bound property, and it can be expressed as closed interval . Th27.1 gives us the compactness.
In this ordering structure, is a convex subset, so the order topology is identical to the subspace topology(Th 16.4).
it is clear that any basis element of , intersects at some sufficiently large real number. Similar arguments apply to .