The Dedekind completeness is defined in many ways. Let show one of them to discuss the question.

We define a Dedekind cut $R$ as a non-empty subset of $\Bbb Q$ with this structure:

  1. $R$ dont have a minimum

  2. If $q\in R$ then $r\in R$ for all $r>q$

  3. $\Bbb Q\setminus R\neq\emptyset$

We define $\Bbb R$ as the set of Dedekind cuts. Then we can check that $R\ge R'\iff R\subseteq R'$ define a total order in $\Bbb R$.

Then, from this definition of order, if we have some collection of Dedekind cuts $R_j$ bounded below by $A$ and bounded above by $B$, that is there is a Dedekind cut $A$ such that $R_j\subseteq A$ for all $j$ and a Dedekind cut $B$ such that $B\subseteq R_j$ for all $j$, then we can see that

$$I:=\bigcup_j R_j\quad\text{ and }\quad S:=\bigcap_j R_j$$

defines the infimum and the supremum of the collection of the $R_j$. As @AndreasBleas pointed in the commentaries for the case that $\min(S)=m$ exists, then we take $S\setminus\{m\}$ as the supremum of the collection of $R_j$.

Then my question: I cant see, for this definition, that we need some special axiom as Cauchy sequences or the nested intervals axiom to prove the existence of the infimum and supremum for bounded set of reals.

Then, my question, we need something more than the axioms of ZFC (by example in this case) to prove that $I$ and $S$ are the infimum or supremum respectively? The unique thing that we used here is the arbitrary union or intersection of sets, with nothing special in it. Thank you in advance.

  • 1
    $\begingroup$ Sure, with Dedekind cuts aproach you prove the completeness of the reals, you only need it as an axiom if you are making an axiomatic definition of the real numbers $\endgroup$ – la flaca Oct 19 '16 at 0:37
  • 1
    $\begingroup$ I guess you need to quote what you read more carefully, or we will not be able to make sense of it. $\endgroup$ – GEdgar Oct 19 '16 at 0:41
  • 1
    $\begingroup$ @Masacroso: As the quote says, one option is to prove it directly from the construction, which is what you have done. There's nothing more to it than that. $\endgroup$ – Eric Wofsey Oct 19 '16 at 1:11
  • 2
    $\begingroup$ Your construction of the infimum $I$ is fine, but your $S$ might have a smallest element and might thus fail to be a Dedekind cut because of clause 1 in the definition. The problem is easy to fix: If $S$ has a smallest element $m$, then $S-\{m\}$ is the supremum that you want. $\endgroup$ – Andreas Blass Oct 19 '16 at 1:17
  • 1
    $\begingroup$ You do need to know some basic facts about sets in order to say that the sets $S$ and $I$ exist, if that's all you're saying. $\endgroup$ – Eric Wofsey Oct 19 '16 at 1:29

The point here is that "a bounded collection of Dedekind cuts" is just a set of Dedekind cuts. The union of any set is guaranteed by the axiom of union, and since the set is not empty, the intersection of this set is guaranteed to exist by the axiom [schema] of separation.

So indeed, to prove that $\Bbb R$ [exists and that it] is complete we need far less than $\sf ZFC$. We needed extensionality, power set, infinity, union and separation.

This can be pushed even lower by restricting the power set axiom (to say something along the lines "the set of natural numbers has only two power sets", or some other finite number of iterations that is needed here), and to restrict which separation axioms we use. But let's not get ahead of ourselves.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.