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

I read in a paper on set theory that the supremum and the infimum of the empty set are defined as $\sup(\{\})=-\infty$ and $\inf(\{\})=\infty$. But intuitively I can't figure out why that is the case. Is there a reason why the supremum and infimum of the empty set are defined this way?

-

## marked as duplicate by Cameron Buie, Asaf Karagila, Amzoti, Nate Eldredge, RahulAug 21 '13 at 6:09

@CameronBuie I think it's related, but the focus of that question seems to be more on why the infimum and supremum exist for the empty set, rather than why they are what they are. The answers primarily answer that, and some indirectly imply why they are what they are, but I feel like the answers here are more explicit. Just my opinion though. – Ataraxia Aug 21 '13 at 4:50
I think that metacompactness' answer on the suggested duplicate explains the intuition quite well in its last paragraph. I also think that my answer can help to develop the intuition as to why this is the case, even if not addressing this problem directly. – Asaf Karagila Aug 21 '13 at 4:53
@AsafKaragila Yes, your answer do a very good job at explaining it now that I read it more carefully. +1 – Ataraxia Aug 21 '13 at 4:57
@AsafKaragila: whether your answer there would be a good answer for this question or not does not make this question a duplicate of that one. That question assumes that, if meaningful, $\sup(\{\})=-\infty$ and $\inf(\{\})=\infty$, and asks whether "it is only meaningful to talk about $\inf(E)$ and $\sup(E)$ if $E$ is non-empty and bounded". This question assumes $\inf(E)$ and $\sup(E)$ are meaningful on $\{\}$, and asks why $\sup(\{\})=-\infty$ and $\inf(\{\})=\infty$. Although they deal with the same ideas, these questions seem distinct to me. – robjohn Aug 21 '13 at 7:11
@robjohn: I recall (but cannot find at the moment) meta discussions in which it was agreed that when a question is answered by an answer of another question, even if they are not the same, it can be reasonable to close as a duplicate. – Asaf Karagila Aug 21 '13 at 7:31

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 $\infty$.

-

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.

-

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 $\infty$.

-

In addition to the reasons mentioned above about it making things easier, it also makes sense intuitively if you think about how each is defined. The supremum is the lowest upper bound on a set, so, since any real number is an upper bound on the empty set, no real number can be the lowest such bound (If x is that bound, then x - 1 is a lower upper bound.) Thus, it is defined to be $-\infty$ -- less than any real number, and similarly for the infimum.

-

As robjohn says, for sets $A$ and $B$, we want $\sup (A\cup B) = \max(\sup (A), \sup (B))$. If $B=\emptyset$, then we have $$\sup (A) = \max(\sup (A), \sup (\emptyset)).$$ In other words, $\sup (\emptyset)$ must be the identity element for the binary operation $\max$, which is $-\infty$. Dually, $\inf (\emptyset)$ must be $\infty$.

The nice thing about this viewpoint is that it generalizes immediately to complete lattices! The empty join is the least element and the empty meet is the greatest element.

-

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 $\infty$.

-