Why is the supremum of the empty set $-\infty$ and the infimum $\infty$? [duplicate]

Solution 1:

This is commonly done so that $$\sup(A\cup B)=\max(\sup(A),\sup(B))$$ and $$\inf(A\cup B)=\min(\inf(A),\inf(B))$$

In other words, as a set decreases, its supremum should decrease. The smallest a set can be is the empty set and the smallest its supremum can be is $-\infty$. As a set decreases, its infimum should increase. The smallest a set can be is the empty set and the largest its infimum can be is $\infty$.

Solution 2:

Intuition

This goes against our intuition because we think of suprema as being 'big' and infima as being 'small'. The crux is that suprema are the 'least big big thing', and when anything is big $-$ as is the case compared to the empty set $-$ then the 'least big big thing' is the smallest thing you can have!

More precisely

Given a set $X$ equipped with some ordering $\le$, and some subset $Y \subseteq X$, we can define $\sup(Y)$, if it exists, to be an element $s \in X$ such that:

1. $y \le s$ for each $y \in Y$; and
2. If $y \le t$ for each $y \in Y$, then $s \le t$.

Why define it like this? Well (1) says it's an upper bound, and (2) says it's a least upper bound. (That's what a supremum is: a least upper bound.)

So what about when $Y = \varnothing$? Then (1) is satisfied trivially, since $\varnothing$ has no elements; and similarly (2) is only satisfied if $s$ is a minimal element of the set $X$. [Stare hard at (1) and (2) to see why these are the case.] So when $X$ is taken to be the (extended) real line, this gives $\sup(\varnothing)=-\infty$.

The same goes for $\inf(Y)$ and $\infty$.

Solution 3:

You want $\sup$ and $\inf$ to obey the following property for two sets $A$ and $B$ if $A \subset B$ then $\inf(B) \leq \inf(A)$ and $\sup(A) \leq \sup(B)$. These are very easy to prove for non-empty sets, in order to extend this property to empty sets you make the previously mentioned definition.

Solution 4:

This is an example of "vacuous implication", and as such it's sometimes easier to turn the question around and ask "what if it weren't so?" That is, what if $\sup \emptyset = s \in \mathbb{R}$ were not $-\infty$? This would mean that no $t < s$ could be an upper bound for $\emptyset$, which in turn implies that:

for any $t < s$, there exists an element $x \in \emptyset$ such that $t < x$.

But that property is impossible, as there is no element in $\emptyset$. This means that, should we desire to assign a "numerical" value to $\sup \emptyset$, it must be less than every element of $\mathbb{R}$, and that's what $-\infty$ is intended to mean. The same logic applies to $\inf \emptyset$, but with all ordering relations reversed so it must be larger than every real number, hence $\infty$.