What's the supremum of an empty set?

**zzzhhh** · August 4th 2009, 09:05 AM

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.

**Failure** · August 4th 2009, 09:20 AM

Originally Posted by zzzhhh

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.

**Gamma** · August 4th 2009, 10:18 AM

You are correct zzzhhh. For example in the real numbers generally we define $sup \{\emptyset\} = - \infty$ and $inf\{ \emptyset\} = \infty$

**Failure** · August 6th 2009, 10:11 AM

Originally Posted by Gamma

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.

**zzzhhh** · August 11th 2009, 05:27 AM

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. $\S$ 29 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?

**Failure** · August 11th 2009, 06:10 AM

Originally Posted by zzzhhh

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. $\S$ 29 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.

**zzzhhh** · August 11th 2009, 10:11 AM

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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