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?
Best Answer
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 $\emptyset$.