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Math Help - What's the supremum of an empty set?

  1. #1
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    What's the supremum of an empty set?

    In textbooks, supremum is often defined as follows:
    Let (A,\leq) be a partially ordered set, and let B\subseteq A.
    (1) a\in B is a least element of B if a\leq x for all x\in B.
    (2) a\in A is an upper bound of B if x\leq a for all x\in B.
    (3) a\in A is called a supremum of B if it is the least element of all upper bounds of B. Denote it as \sup B.
    If we express these definitions in wffs of first-order logic, then
    (1) a is a least element of B if the following wff is true: a\in B\wedge \forall x(x\in B\rightarrow a\leq x).
    (2) a is an upper bound of B if the following wff is true: a\in A\wedge \forall x(x\in B\rightarrow x\leq a).
    Now if B\subseteq A is an empty set \emptyset, it follows that any element of A is an upper bound of B, for in (2) the left part of \rightarrow 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, \sup B is the least element of the set consisting of all upper bounds of B, that is, just the least element of A. I'm not sure if it is right or not, does anybody know? Thanks.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by zzzhhh View Post
    In textbooks, supremum is often defined as follows:
    Let (A,\leq) be a partially ordered set, and let B\subseteq A.

    (3) a\in A is called a supremum of B if it is the least element of all upper bounds of B. Denote it as \sup B.

    Therefore, by definition, \sup B is the least element of the set consisting of all upper bounds of B, that is, just the least element of A. I'm not sure if it is right or not, does anybody know? Thanks.
    I think you have argued your case correctly: \sup \emptyset is the least element of A (if it exists). If A does not have a least element, then \sup \emptyset does not exist, of course.
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  3. #3
    Super Member Gamma's Avatar
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    You are correct zzzhhh. For example in the real numbers generally we define sup \{\emptyset\} = - \infty and inf\{ \emptyset\} = \infty
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    Super Member Failure's Avatar
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    Quote Originally Posted by Gamma View Post
    You are correct zzzhhh. For example in the real numbers generally we define sup \{\emptyset\} = - \infty and inf\{ \emptyset\} = \infty
    Actually no: in the real numbers \mathbb{R} we have that \sup\emptyset and \inf\emptyset do not exist: because in \mathbb{R} there is neither a smallest nor a largest element. In particular: \pm\infty\notin\mathbb{R}. But in the extended real numbers \overline{\mathbb{R}}:=\mathbb{R}\cup\{-\infty,+\infty\} you are right.
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    I know that W. Rudin's book "Principles of Mathematical Analysis" discusses the extended real number system in P11-12, where two symbols +\infty and -\infty are introduced. But I think to extend the set \mathbb{R} to include something standing for infinity, only one symbol need to be added, that is, the symbol \infty in the one-point compactification of \mathbb{R}(See, e.g. \S29 of Munkres' "Topology", this proccess makes the extended system compact, the added symbol \infty is a limit point of \mathbb{R}, that's why you add a bar above \mathbb{R}). So, both +\infty and -\infty are actually the same; there is no difference between them; we write + or - in front of \infty just in order to indicate how sequences in \mathbb{R} converge to \infty -- whether the sequence is becoming larger and larger, or conversely. Am I right?
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  6. #6
    Super Member Failure's Avatar
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    Quote Originally Posted by zzzhhh View Post
    I know that W. Rudin's book "Principles of Mathematical Analysis" discusses the extended real number system in P11-12, where two symbols +\infty and -\infty are introduced. But I think to extend the set \mathbb{R} to include something standing for infinity, only one symbol need to be added, that is, the symbol \infty in the one-point compactification of \mathbb{R}(See, e.g. \S29 of Munkres' "Topology", this proccess makes the extended system compact, the added symbol \infty is a limit point of \mathbb{R}, that's why you add a bar above \mathbb{R}). So, both +\infty and -\infty are actually the same; there is no difference between them; we write + or - in front of \infty just in order to indicate how sequences in \mathbb{R} converge to \infty -- whether the sequence is becoming larger and larger, or conversely. Am I right?
    Actually no, I don't think so. You are talking of two different types of compactification of the reals. The one-point compactification basically identifies the reals with a circle and adds only one singe new point \infty. The one-point compactification does not give you a smallest -\infty and largest +\infty element, however.
    The other, two-point compactification, adds two points -\infty and +\infty, which at the same time provide \mathbb{R} with a smallest -\infty and a largest +\infty element. In the two-point compactification, -\infty and +\infty are not at all the same but are quite distinct (otherwise the order-structure of \mathbb{R} would get out of joint). So, two-point compactification is used if you are not satisfied with just compactifying \mathbb{R} but want a certain kind of closure of its order structure at the same time that makes, among other things, that \inf \emptyset and \sup \emptyset both exist.
    In the one-point compactificaion you are not allowed to just posit that \infty is at the same time equal to \inf\emptyset and \sup\emptyset, because, as I wrote already, this would make \infty 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.
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    Yes! it can be readily proved that \mathbb{R}\cup\{-\infty,+\infty\} in order topology is really a compact space ^{[1]} having space \mathbb{R} as its subspace ^{[2]}, and that +\infty and -\infty are limit points of \mathbb{R}\;^{[3]}, hence a two-point compactification of \mathbb{R} which can preserve the order structure is proven.
    Thank you Failure, your post deepened my understanding of the mysterious symbol \infty since the study of calculus, thank you!
    Interesting math! :-)

    [1]Rudin's book has shown that this "two-point" extension has the least-upper-bound property, and it can be expressed as closed interval [-\infty,+\infty]. Th27.1 gives us the compactness.
    [2]In this ordering structure, \mathbb{R} is a convex subset, so the order topology is identical to the subspace topology(Th 16.4).
    [3]it is clear that any basis element of (x,+\infty], x\in\mathbb{R}, intersects \mathbb{R} at some sufficiently large real number. Similar arguments apply to -\infty.
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