Supremum and infimum

A set \(S\) is bounded above if there is \(M\) with \(s\le M\) for all \(s\in S\); each such \(M\) is an upper bound. The supremum \(\sup S\) is the least of them. If \(\sup S\in S\), it is the maximum; but \((0,1)\) has supremum \(1\) and no maximum.

A practical test: \(M=\sup S\) if and only if \(M\) is an upper bound and, for every \(\varepsilon>0\), there is \(s\in S\) with \(s>M-\varepsilon\). That is, no smaller bound works. The infimum case is symmetric, and \(\inf S=-\sup(-S)\).

Guiding examples: \(\sup\{1-1/n\}=1\) without belonging to the set, \(\inf\{1/n\}=0\), and \(\sup\{x:x^2<2\}=\sqrt2\). The last one already shows why \(\mathbb Q\) is not enough: in \(\mathbb Q\) that set has upper bounds but none is the least.

If \(S\subseteq T\), then \(\sup S\le\sup T\), and the supremum of a union is the larger of the suprema. That every nonempty set bounded above has a supremum is precisely the completeness axiom, the next concept.