Is $\emptyset$ bounded? Why then $\inf \emptyset = \infty$ is reasonable?

Question

I know the proof that $\emptyset$ (empty set) has $\infty$ as infimum. This seems to imply to me that $\emptyset$ is unbounded, I mean a set with $\infty$ as lower bound: how can this be bounded?! But if $\emptyset$ is unbounded then $\forall M>0$ $\exists x \in \emptyset $ s.t. $x > M$, which is impossible because there can't exist an element in $\emptyset$. How to resolve this paradox?

thanks.

Answer 1

A subset $S$, of the real numbers, is bounded when there exists some real number $M$ such that $|x| \le M$ for all $x \in S$.

(The same for bounded from above and bounded from below separately.)

According to this definition the empty set is bounded. You can chose $M$ whatever you like. The assertion is true, it is vacuously true.

Further more, every real number is an upper bound an every real number is a lower bound, e.g., $26$ is a lower bound and $-45$ an upper bound of the empty set.

To resolve the paradox one needs to note that for $m$ a lower and $M$ and upper bound of some set $S$, one can conclude that $m \le M$ provided that $S$ is non-empty.

One needs some element in $S$ to make the obvious argument work. (For any $s \in S$, we have $m \le s$ and $s \le M$. Thus $m \le M$.)

A takeway is that vacuous truth can have some counterintuitive consequence.

Answer 2

As it's been pointed out in the comments, infimum is such that

$$\inf X \le x \forall x \in X \land (y \le x \forall x \in X \implies y \le \inf X) $$

by definition. Of course $k \le x \forall x \in \emptyset$ for every finite $k$ — then $\inf \emptyset = \infty$!

Conversely, and even more surprisingly, you can state that $\sup \emptyset = - \infty$ by a similar argument, so that nice relationship $\inf X \le \sup X$ doesn't even hold for $\emptyset $

You might want to check the concept of vacuous truth.

Answer 3

I love to think of the infimum of a set $S\subset \Bbb R$ as an imaginary point $p$ moving from $-\infty$ toward $\infty$. The point $p$ stops when it hits the lowest point of $S$ then $p=\inf S$.

When $S=\emptyset$, since $p$ never stops because it doesn't hit any thing at all, so it moves toward $\infty$. This justifies $\inf \emptyset=\infty$ for me.

Answer 4

Something more abstract for intuition:

Let $M$ be a monoid (group if you like). Then we usually define: $\prod_{i=1}^0 a_i$, i.e. the empty product to be the identity of that monoid.

The case of infima is completely analogous. Let: $$a \wedge b:= \min(a,b) = \inf(a,b)$$

Then $\mathbb{R}$ equipped with this operation is a monoid. The identity is of course $\infty$, hence:

$$\inf \emptyset = \bigwedge_{i=1}^0 a_i = \infty$$

Of course $\inf$ is defined for infinite sets too, but the reasoning still holds.