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From some reading, I've noticed that $\sup(\emptyset)=\min(S)$, but $\inf(\emptyset)=\max(S)$, given that $\min(S)$ and $\max(S)$ exist, where $S$ is the universe in which one is working. Is there some inherent reasoning/proof as to why this is? It seems strange to me that an upper bound of a set would be smaller than a lower bound of the same set.

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3 Answers

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The supremum is defined to be the least upper bound. An upper bound for a set $K$ is defined to be an element $b$ such that for all $k\in K$, $k\leq b$. If $K$ is empty, then the condition "for all $k\in K$, $k\leq b$" is true by vacuity (you have an implication of the form "if $k$ is an element of the empty set, then $\lt$something happens$\gt$", and the antecedent is always false so the implication is always true).

Therefore, every element of $S$ is an upper bound for $\emptyset$; since $\sup(\emptyset)$ is the least upper bound, that means that $\sup(\emptyset)=\min($upper bounds of $\emptyset) = \min(S)$.

Similarly, the infimum is the greatest lower bound, and every element of $S$ is a lower bound for $\emptyset$, so $\inf(\emptyset) = \max($lower bounds of $\emptyset) = \max(S)$.

Yes, it is somewhat counterintuitive that you have $\sup(\emptyset)\leq\inf(\emptyset)$, but in fact the empty set is the only one for which you can have $\sup(\emptyset)\lt \inf(\emptyset)$. The empty set often causes counterintuitive results.

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Interesting, thanks for that explanation. –  yunone Sep 1 '10 at 5:20
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A quick counterpoint to Arturo's answer: yes, the fact that $\operatorname{sup}(\emptyset) = -\infty < \infty = \operatorname{inf}(\emptyset)$ seems strange at first. But it is designed, among other things, to preserve an intuitive and useful property of infima and suprema:

Let $S$ and $T$ be subsets of $\mathbb{R}$ with $S \subset T$. Then

$\operatorname{inf}(T) \leq \operatorname{inf}(S)$

and

$\operatorname{sup}(T) \geq \operatorname{sup}(S)$.

Indeed, by adding elements to a set, its infimum can only get smaller and its supremum can only get larger. Now if you take $S = \emptyset$ and let $T$ range through even all one-element subsets of $\mathbb{R}$, you see that in order for these inequalities to hold we must define $\operatorname{inf}(\emptyset)$ and $\operatorname{sup}(\emptyset)$ as we did.

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A very natural explanation of the correct definition for the values of min and max on empty sets arises from their dual universal definitions - analogous to the universal GCD, LCM definitions that I presented in a post here.

First some notation. $\;$ Write $\ \ \rm x \le S \;\iff\; x \le s,\;\: \forall\: s \in S,\;$ and dually for $\rm\; x \ge S$

DEFINITION of $\:$ min $\quad$ $\quad\rm x \le S \;\iff\; x \le min\ S$

DEFINITION of max $\:\quad$ $\quad\rm x \ge S \;\iff\; x \ge max\ S$

For min, when $\;\rm S = \emptyset\;$ is empty, the first clause $\;\rm x \le S\;$ in the min definition is vacuously true, hence the definition reduces to $\;\rm x \le \min \emptyset\;$ for all $\;\rm x\;$. Hence $\;\rm \min \emptyset = \infty\;$. Dually $\;\rm \max\emptyset = -\infty\;$.

As I remarked in said GCD, LCM post, such universal definitions often facilitate slick proofs. For some nontrivial examples of min, max flavor consider the slick proofs of the integrality of various products of binomial coefficients by employing the floor function, e.g. see Joe Roberts: Elementary Number Theory. Instead I close with an analogous slick GCD, LCM proof from my mentioned post (see it for further details).

Generally, in any domain, we have the following dual universal definitions of LCM and GCD:

DEFINITION of LCM $\quad$ If $\quad\rm a,b\ |\ c \;\iff\; [a,b]\ |\ c \quad$ then $\quad\rm [a,b] \;\;$ is an LCM of $\;\rm a,b$

DEFINITION of GCD $\quad$ If $\quad\rm c\ |\ a,b \;\iff\; c\ |\ (a,b) \quad$ then $\quad\rm (a,b) \;$ is an GCD of $\;\;\rm a,b$

Note: that $\;\rm a,b\ |\ [a,b] \;$ follows by putting $\;\rm c = [a,b] \;$ in the definition. Dually $\;\rm (a,b)\ |\ a,b \;$

Such $\iff$ definitions provide slick unified proofs of both arrow directions, e.g. the fundamental

THEOREM $\rm\quad (a,b)\ =\ ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.

Proof: $\rm\quad\quad d\ |\ a,b \;\iff\; a,b\ |\ ab/d \;\iff\; [a,b]\ |\ ab/d \;\iff\; d\ |\ ab/[a,b] \quad\;\;$ QED

The conciseness of this proof arises by exploiting to the hilt the $\iff$ definition of LCM and GCD. Compare to less concise / general / illuminating proofs in many number theory textbooks.

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@Bill: I think where you say "min" and "max" in the first two paragraphs you mean "inf" and "sup"; the empty set has no minimum and no maximum. –  Arturo Magidin Feb 1 '11 at 16:33
    
@Arturo: above min,max are synonyms for inf,sup, cf. their universal definitions. –  Bill Dubuque Feb 1 '11 at 20:30
    
@Bill: That's nonstandard use of min and max, as far as I am aware; I don't see what I'm supposed to compare, though... –  Arturo Magidin Feb 1 '11 at 20:41
    
@Arturo: You're supposed to refer to the universal definitions of min, max that I am using above. –  Bill Dubuque Feb 1 '11 at 21:59
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