How many infima and suprema can there be?

Question

The Wikipedia definitions of infimum and supremum include the words "greatest" and "lowest", implying that there's at most one infimum and one supremum for any given subset.

But in the empty set article, it says that any element of a set is the infimum and at the same time the supremum of the empty set.

Is the empty set a special case, or can any subset have an infinite number of suprema and infima?

Answer 1

Let $(A,<)$ be a partially ordered set. For $B\subseteq A$ we say that $x=\sup(B)$ if two conditions occur:

1. $\forall y(y\in B\rightarrow y\le x)$, that is $x$ is an upper bound of $B$;
2. $\forall y(y\text{ an upper bound of } B\rightarrow x\le y)$. That is if $y$ also satisfies the first clause, then $x\le y$.

(We say that $x=\inf(B)$ if the same clauses occur with $\ge$ instead of $\le$)

From this definition it is obvious that every element is vacuously an upper bound for the empty set, as well a lower bound. As there are no elements in the empty set, the first clause is true for every $x\in A$.

This definition also implies that only the minimum element, if it exists, is a supremum for $\varnothing$, and only the maximal element, if it exists, is an infimum for the empty set. For example, if $x$ is not the minimum of $A$ then for some $y$ we have $x\nleq y$, and since $y$ is an upper bound of $\varnothing$ we have that $x$ is not the supremum.

In particular this is why in the context of the real numbers we have that $\sup(\varnothing)=-\infty$ and $\inf(\varnothing)=\infty$.

Answer 2

There is nothing in the empty set article that says every element is the supremum and at the same time the infimum of the empty set. It doesn't say that. What it says is every element is an upper bound (not the upper bound but an upper bound) and at the same time a lower bound of the empty set. You shouldn't start a posting like this by misquoting the article.

Answer 3

If a subset of the real numbers has a lowest upper bound, then that lowest upper bound in unique. (The same is, of course, also true for greatest lower bounds.) This is because the real numbers are totally ordered, so that, given any two distinct real numbers, one of them has to be greater than the other. Thus, in particular, there can be only (at most) one lowest or greatest anything in the real numbers.

The empty set has no lowest upper bound in the real numbers: every real number is, trivially, an upper bound of the empty set, and there is no lowest real number. However, on the extended real number line, $-\infty$ is the lowest of all numbers, and thus the lowest upper bound of the empty set.

On a partially ordered set, it can happen that some subsets might have several lowest upper bounds (or greatest lower bounds). For example, consider the set $\lbrace \mathrm A, \mathrm B, \mathrm X, \mathrm Y \rbrace$, where $\mathrm A$ and $\mathrm B$ are both less than $\mathrm X$ and $\mathrm Y$ (that is, $\mathrm A < \mathrm X$, $\mathrm A < \mathrm Y$, $\mathrm B < \mathrm X$ and $\mathrm B < \mathrm Y$), but $\mathrm A$ and $\mathrm B$ are incomparable, and so are $\mathrm X$ and $\mathrm Y$ (i.e. $\mathrm A \nleq \mathrm B$, $\mathrm B \nleq \mathrm A$, $\mathrm X \nleq \mathrm Y$ and $\mathrm Y \nleq \mathrm X$). Then $\mathrm X$ and $\mathrm Y$ are both lowest upper bounds of the subset $\lbrace \mathrm A, \mathrm B \rbrace$ (which has no greatest lower bound), and $\mathrm A$ and $\mathrm B$ are both greatest lower bounds of $\lbrace \mathrm X, \mathrm Y \rbrace$ (which has no lowest upper bound).