Least upper bound axiom

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The least upper bound axiom, also abbreviated as the LUB axiom, is an axiom of real analysis stating that if a nonempty subset of the real numbers has an upper bound, then it has a least upper bound. It is an axiom in the sense that it cannot be proven by the other axioms within the system of real analysis. This axiom is very useful since it is essential to the proof that the real number line is a complete metric space. The rational number line does not satisfy the LUB axiom and hence is not complete.

An example is $S = \{ x\in \mathbb{Q}|x^2 < 2\}$ . 2 is certainly an upper bound for the set. However, this set has no least upper bound — for any upper bound $x \in \mathbb{Q}$ , we can find another upper bound $y \in \mathbb{Q}$ with $y < x$ .

[edit] Proof that the real number line is complete

Let $\{ s_n\}_{n\in\N}$ be a Cauchy sequence. Let S be the set of real numbers that are bigger than $s n$ for only finitely many $n\in\N$ . Let $\varepsilon\in\R ^+$ . Let $N\in\N$ be such that $\forall n,m\ge N,$ $|s_n-s_m|<\varepsilon$ . So, the sequence passes through the interval $(s_N-\varepsilon ,s_N+\varepsilon )$ infinitely many times and through its complement at most a finite number of times. That means that $s_N-\varepsilon\in S$ and hence $S\not=\emptyset$ . Clearly, $s_N+\varepsilon$ is an upper bound for S. By the LUB Axiom, let b be the least upper bound. $s_N-\varepsilon\le b\le s_N+\varepsilon$ . By the triangle inequality, $\forall n\ge N,$ $d(s_n,b)\le d(s_n,s_N)+d(s_N,b)\le\varepsilon +\varepsilon =2\varepsilon.$ Therefore, $s_n\longrightarrow b$ and so $\R$ is complete. Q.E.D.