What is an example of a nonempty subset of $\mathbb{R}$ that is bounded above that does not contain its least upper bound? This is an on-a-review sheet for my final. I thought the completeness axiom stated that any nonempty subset of $\mathbb{R}$ that is bounded above has an upper bound. Can anyone explain?
2026-04-03 21:46:17.1775252777
What is an example of a nonempty subset of $\mathbb{R}$ that is bounded above that does not contain its least upper bound?
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The completeness axiom says that any nonempty real subset $S \subseteq \mathbb{R}$ that is bounded above has a least upper bound (supremum), called $\sup S$. However, the completeness axiom does NOT require that the supremum of $S$ actually be an element of $S$, it only requires the supremum to be a real number.
So, using the example from @David Mitra above, the set $S = (0, 1)$ consists of all real numbers greater than 0 and less than 1. $S$ therefore doesn't contain 1 itself, but $1 = \sup S$. To prove this, note that if $0 < x < 1$, then $\frac{1+x}2$ (the number halfway between x and 1) is a member of $S$ yet is greater than $x$. Therefore, no such $x$ can be an upper bound of $S$.