# Completeness Axiom

Completeness axiom: a non-empty subset of $\mathbb{R}$ which is bounded above, has a supremum in $\mathbb{R}$.

Natural question is, what is $\sup(\emptyset)$?

In many books, the concept of supremum is stated for non-empty subsets of $\mathbb{R}$, and do not mention anything about $\emptyset$. In some books, it is stated that $\emptyset$ has no supremum, since every real number is an upper bound for $\emptyset$. This is the point, about which I want to get clear explanation. The books give following argument to prove this:

For exery $x\in \mathbb{R}$, since $\emptyset \subseteq \{x\}$ hence $\sup(\emptyset) \leq x$ for every $x\in \mathbb{R}$.

In this reasoning, they have used the fact that $A\subseteq B\Rightarrow \sup(A)\leq \sup(B)$; but this is proved for non-empty sets $A, B$; how can we use it for empty set?

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Phillip, $x$ is an upper bound of a set $A$ iff for all $y$, if it happens that $y\in A$, then $y\le x$. Recall that false implies anything. If $A$ is empty then "$y\in A$" is false, and therefore " if $y\in A$, then $y\le x$" is true. This holds for all $y$, and therefore, by definition of upper bound, $x$ is indeed an upper bound of the empty set. This is an example of a vacuous condition: We do not prove that $y\le x$, rather, than an implication is true. –  Andres Caicedo Apr 4 '11 at 6:08
Phillip, while many people think that the empty set is denoted by the Greek letter phi, it is actually a Danish letter (whose name I never learned). –  Asaf Karagila Apr 4 '11 at 7:35

A real number $x\in\mathbb{R}$ is an upper bound for $S\subseteq\mathbb{R}$ when $$\forall y\in S, y\leq x.$$ But for every real number $x\in\mathbb{R}$, the statement $$\forall y\in\emptyset, y\leq x$$ is vacuously true - there are no elements $y\in\emptyset$. Thus every $x\in\mathbb{R}$ is an upper bound for $\emptyset$, and because there is no least real number, there is no least upper bound (i.e., supremum) of $\emptyset$.
@Zev: We say that $x$ is an upper bound of $\phi$ if $\forall y\in \phi, y\leq x$; but there is no such $y$ in $\phi$ satisfying this condition; hence $x$ is not an upper bound....is this correct? if not, what's wrong? And, the fact $A\subseteq B \Rightarrow sup(A)\leq sup(B)$ is true for non-empty sets; how can we use it for empty set? –  user8186 Apr 4 '11 at 6:10
If it helps, you can think of $sup(\emptyset)$ as being $-\infty$. Note that $\infty$ and $-\infty$ are not real numbers, so that this is technically a false statement. But, one can think of $-\infty$ as being a symbol with the property that, for every real number $x\in\mathbb{R}$, we have that $-\infty\leq x$. Thus, $-\infty$ would fit the definition of being the least of all the upper bounds of $\emptyset$. This can be made rigorous with the extended real number system. –  Zev Chonoles Apr 4 '11 at 6:17
@Phillip: you are unnecessarily focused on the fact that $A\subseteq B \Rightarrow sup(A)\leq sup(B)$. However, using the (technically meaningless) answer $sup(\emptyset)=-\infty$, this would continue to be true; for any $B$, we have $\emptyset\subseteq B$, and we also have $-\infty=sup(\emptyset)\leq sup(B)$ regardless of what real number $sup(B)$ ends up being. –  Zev Chonoles Apr 4 '11 at 6:27
Putting $\sup(\emptyset):=-\infty$ ensures the permanence of the following principle: For every $c<\sup(A)$ there is an $x\in A$ with $x>c$. –  Christian Blatter Apr 4 '11 at 9:56