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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.2011-04-04
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    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).2011-04-04

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