As Terence tao explains we have the following conclusion that if we consider the empty set as subset of the extended reals. We have the following result:

We know that $sup(E) \geq inf(E)$. This would imply that $-\infty \geq +\infty$. This is non-sense based on the ordering on the extended reals. Why do we get this result ? Is there anything preventing us from getting this result ?

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    $\begingroup$ "We know that sup(E)≥inf(E)" No. You don't know this and this example shows that isn't true. It is always true if $E$ is not empty. But it is false if $E$ is empty. $\endgroup$ – fleablood Jun 21 '19 at 23:46
  • $\begingroup$ Look very closely at the definitions. They mention the elements of the set. This set has no elements. $\endgroup$ – The Count Jun 22 '19 at 2:00

Consider the nature of vacuously true statements.

Let $x \in \emptyset$. There is no such element so anything we say about is vacuously true. Is $x$ a green dragon? Yes. Since there are no $x \in \emptyset$ then every $x\in \emptyset$ (all zero of them) are green dragons. This is true because there aren't any $x \in \emptyset$ that aren't green dragons.

Suppose $r \in$ extended reals and $x\in \emptyset$. Is $r \le x$? Yes. Since there is not $x\in \emptyset$ that is vacuously true. $r$ is less than or equal to every element in $\emptyset$ because there aren't any elements in the $\emptyset$ were it is not true.

So the empty set is bounded above by $r$ and likewise for any $r\in$ extended reals that we choose. So what is the least upper bound. Well, as all extended reals are upper bounds that would be the same thing as asking what is the least extended real. And clearly that is $-\infty$.

So $\sup \emptyset = -\infty$.

If that's to breezy we have the definition of $\sup \emptyset = s$ is 1) it is an upper bound. and 2) the is no $r < s$ that is an upper bound.

Well is that true of $-\infty$?

1) if $x \in \emptyset$ then is $x \le -\infty$? Yes. That is vacuously true because there is no $x \in \emptyset$?

2) Is there any $r < -\infty$ that is an upper bound of $\emptyset$. Well, no, there isn't any $r < -\infty$ at all.

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    $\begingroup$ Perfect. That makes so much sense. $\endgroup$ – Learner Jun 21 '19 at 23:51

Vacuously, any $x$ has $x\geq y$ for all $y\in \varnothing$, so $\sup \varnothing = \inf \mathbb{R} = -\infty$; and $\inf \varnothing = \infty$ analogously.

For the second part, any $x\in E$ has $\inf(E) \leq x \leq \sup(E)$. If such an $x$ actually exists, then we can conclude $\inf(E)\leq \sup(E)$, but there's no such $x$ in the empty set.


The inequality $\sup(E)\ge\inf(E)$ is only true if $E$ is nonempty. Any element of $[-\infty,\infty]$ is an upper bound for $\emptyset$, so the least upper bound is $-\infty$.


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