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Is the statement "any nonempty subset of R that is bounded above has a least upper bound" an axiom or there is a way to prove it?

I am asking because this statement not immediately obvious to me to proclaim it an axiom.

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    $\begingroup$ This is often used as an postulate for the real numbers. E.g., in Spivak's Calculus. And then later in the book it is shown to be a property of a good construction of the reals. $\endgroup$
    – Simon S
    Mar 14, 2015 at 19:14
  • $\begingroup$ I think you can prove it when you construct the real numbers, although it is taken for granted in most courses. But we should wait for more answers. $\endgroup$
    – Asinomás
    Mar 14, 2015 at 19:14

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The answers to your two questions are "yes" and "yes".

First, this statement is one of the standard axioms for real numbers, called the "completeness axiom".

Second, in the standard "Dedekind cut" construction of the real numbers, one starts from an axiomatic description of the rational numbers, and then one constructs the real numbers and proves that all of their axioms hold, including the completeness axiom.

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It's an axiom or a DEFINITION of R. Real numbers are defined by Dedekind cuts (or equivalence classes of limits of Cauchy sequences). Dedekind cuts are essentially adding in least upper bounds. So R, by definition, satisfies the completeness axiom.

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Let $X\subset\mathbb{R}$, such that $X$ is bounded above. Then the closure of $X$ has a maximal element in $\mathbb{R}$, and this element is the least upper bound.

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    $\begingroup$ "If we continue this process" it is not obvious at all that we will "end up" after a finite number of steps. $\endgroup$
    – zesy
    Mar 14, 2015 at 20:06
  • $\begingroup$ As @SergeyZykov says, it is not obvious that a finite number of steps will get us to the process--it might not get us there at all. However, if no finite number of steps give us a least upper bound, we can construct a sequence of decreasing upper bounds this way, which is also bounded below by any element of X. It remains to show that this limit is, indeed, a least upper bound for X (although not any such decreasing sequence will do). How sound or non-circular this argument is might depend on how axioms are introduced in a certain course, and what the student is meant to assume. $\endgroup$ Aug 25, 2022 at 2:47
  • $\begingroup$ @JorgeZazueta; I've edited my answer. $\endgroup$
    – JMP
    Aug 25, 2022 at 8:37

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