1
$\begingroup$

Ghorpade-Limaye, A couse in Calculus and Real Analysis, p.5) says that $\emptyset$ does not have a supremum but there is not any explanation.

My question is: Why?

$\endgroup$
4
$\begingroup$

Well, a supremum of a set $A$ is a number $a$ such that

  • $a$ is $\ge$ every element of $A$ (such an $a$ is called an upper bound of $A$), and

  • $a$ is $\le$ any number $b$ with the previous property.

Now, here's the problem:

What isn't an upper bound of the empty set?

For example, $4$ is an upper bound of $\emptyset$. Why? Well, $4$ is $\ge$ every element of the emptyset. You don't believe me? OK, find me an element of the emptyset which is not $\le 4$.

(Similarly, every element of the emptyset is an elephant.)

So the problem is that every real number is an upper bound of $\emptyset$. But there is no least real number, so the emptyset doesn't have a least upper bound - that is, the emptyset has no supremum.

$\endgroup$
  • $\begingroup$ Yes. Thanks.... $\endgroup$ – James Ensor Oct 9 '16 at 17:25
4
$\begingroup$

Every real number is an upper bound for $\emptyset$. So there is no smallest real number. The same reasoning applies for the infimum. What is interesting, however, is that you can use this reasoning to find $\sup\emptyset = -\infty$ and $\inf\emptyset = \infty$ in the extended reals.

$\endgroup$
  • $\begingroup$ I could'nt believe that sup $\emptyset = -\infty$ and vice versa. $\endgroup$ – James Ensor Oct 9 '16 at 17:28
  • $\begingroup$ Yeah it is definitely counter-intuitive, but let's tackle the supremum. Every real number is an upper bound for $\emptyset$. This means $\sup\emptyset \le x$ for all $x \in\Bbb{R}$. But the only number that satisfies this is $-\infty$. $\endgroup$ – G. Snapsmath Oct 9 '16 at 17:35
  • $\begingroup$ Definition of SUP says that Let $S$ be a subset of $\mathbb{R}$. An element $M\in\mathbb{R}$ is called a supremem of the set $S$ if bla bla... So, $-\infty$ or $\infty$ is not a element. $\endgroup$ – James Ensor Oct 9 '16 at 17:49
  • $\begingroup$ i.e., $M$ could not equal to $-\infty$ or $\infty$. $\endgroup$ – James Ensor Oct 9 '16 at 17:50
  • 1
    $\begingroup$ $\overline{\mathbb{R}} = \mathbb{R}\cup\{-\infty, \infty\}$ So basically you just take $\mathbb{R}$ and give it a largest and smallest element. This is known as the extended reals $\endgroup$ – G. Snapsmath Oct 9 '16 at 18:16
3
$\begingroup$

Every real number is an upper bound of $\emptyset$. That is to say, the set of all upper bounds of $\emptyset$ is $\Bbb{R}$, which has no least element.

$\endgroup$
0
$\begingroup$

The supremum is the least upper bound. What's an upper bound of $\emptyset$? It is $a\in\mathbb R$ such that $x\leq a$ for all $x\in \emptyset$. Since no such $x$ exists, either you interpret that all $a$ are upper bounds, so there is no least upper bound, or that no $a$ is an upper bound, in which case there is no least upper bound either.

$\endgroup$

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