2
$\begingroup$

I actually dont need this question to be answered for university or something, it was just a discussion I had with a friend.

Does the empty set have an upper or lower bound?

Since the supremum is the minimal bound it must be $-\infty$ since there is no element in the empty set such that it is bigger than $-\infty$. It somehow makes sense. On the other hand the infimum is $\infty$.

Can one define something like that for the empty set? Or is it just heuristic...?

$\endgroup$
4
$\begingroup$

Let $X$ be a set equipped with an order relation $\le$. The supremum of a subset $A$ is computed (if it exists) in two steps.

  1. Consider the set $UB(A)$ of upper bounds of $A$: $$ UB(A)=\{x\in X: a\le x,\text{ for all }a\in A\} $$

  2. Take the minimum $m$ of $UB(A)$ (if it exists).

When both steps succeed, $m$ is the supremum of $A$.

Now, if $A$ is the empty set, what's the set of upper bounds? Let $x\in X$; may $x$ fail to be in $UB(\emptyset)$? In order for this to happen there should be $a\in\emptyset$ such that it is false that $a\le x$. But this isn't possible, because no element can be found in $\emptyset$. Therefore no element of $x$ fails to be in $UB(\emptyset)$, that is, $$ UB(\emptyset)=X $$ and so the supremum of $\emptyset$ is, if it exists, the minimum element of $X$.

If your set is $\mathbb{R}$ with the usual order relation, then the supremum of $\emptyset$ doesn't exist. If you're considering $\bar{\mathbb{R}}$, the set $\mathbb{R}$ with $-\infty$ and $\infty$ added in the usual fashion, then the supremum of the empty set is indeed $-\infty$.

$\endgroup$
  • $\begingroup$ Thank you very much for the answer! $\endgroup$ – Thorben Apr 11 '14 at 20:49
  • $\begingroup$ But can't we also say that since there is no $a \in \phi$, the condition that for any $x \in X$ to be in $UB(\phi)$, $a \leq x$ for all $a \in \phi$ fails? That way we can say that no element of $X$ satisfies the condition to be in $UB(\phi)$. $\endgroup$ – Sidd Jul 17 '16 at 8:43
  • $\begingroup$ @Sidd No. An element $x$ fails to belong to $UB(A)$ if there exists $a\in A$ with $a\not\le x$. If $A=\emptyset$, you cannot find such an $a$. $\endgroup$ – egreg Jul 17 '16 at 9:13
  • 1
    $\begingroup$ @Sidd An element $x\in X$ does not belong to $UB(\emptyset)$ if and only if there exists $a\in\emptyset$ such that $a\not\leq x$. Can you find $a\in\emptyset$? No. So we conclude that, for every $x$, it is false that $x\notin UB(\emptyset)$. Hence, for every $x$, it is true that $x\in UB(\emptyset)$. Note that the condition for belonging to $UB(A)$ has “for all $a\in A$”, not an existential quantifier. $\endgroup$ – egreg Jul 19 '16 at 10:22
  • $\begingroup$ @egreg I totally get what you are saying, but looking from the other side, what if we can prove that $x \notin UB(A)$. I mean an element $x$ belongs in $UB(A)$ if and only if $a \leq x$ for all $a \in A$, and since there is no element in $A$, that fails. Won't that mean that $x \notin UB(A)$? $\endgroup$ – Sidd Jul 20 '16 at 13:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.