In my analysis course, we are considering $(\mathbb{R},+,\cdot,\leq)$ as axiomatically constructed ordered field. Now, together with that, we added a completness axiom stated as follows:

Axiom: Let $(I_n)$ be a sequence of intervals in $\mathbb{R}$ such that for all $n\in\mathbb{N}$ $I_{n+1}\subset I_n$. Also let $\lim_{n\to\infty} |I_n|=0$. Then $$\bigcap_{n=1}^\infty I_n\neq \emptyset$$

Now, I wish to use the least upper bound property (LUBP), that is, any bounded (from above) subset of $\mathbb{R}$ has a least upper bound. My lecturer proposed that this is equivallent to the Axiom of Choice, because we have all subsets of $\mathbb{R}$ (an uncountable set) and for each of these sets $X \in \mathcal{P}(\mathbb{R})$, there is a set of all upper bounds of $X$. Which is also uncountable. So in order to select the $\min{X}$, we need AC, right In this construction, do we need AC to get to LUBP? In addition, what would we need to be able to prove LUBP as a theorem? (Also, if I am wrong, I would love to see a proof that LUBP can be proven from the axiom i gave.)

  • $\begingroup$ There is absolutely no statement about the existence of a choice function from sets of real numbers which is equivalent to AC over ZF. The axiom of choice can fail in all kind of ways, while still keeping a choice function from the sets of reals. $\endgroup$ – Asaf Karagila Oct 16 '18 at 11:26
  • $\begingroup$ The least upper bound property for the reals is provable in ZF set theory without the axiom of choice. Actually there are several equivalent definitions for the reals, but the least upper bound property is provable for each of them without the axiom of choice. $\endgroup$ – Carl Mummert Oct 16 '18 at 11:47
  • $\begingroup$ Just to assure myself, when taking $\mathbb{R}$ as an ordered field (by axioms), then I also have to include some kind of completeness axiom, right? Else, if I constructed $\mathbb{R}$ all the way from Peano axioms, $\mathbb{Z},\mathbb{Q}$, cauchy sequences... then completeness would become a theorem, right? $\endgroup$ – Michal Dvořák Oct 16 '18 at 11:49
  • 1
    $\begingroup$ Yes, completeness is necessary. For example, $\Bbb Q$ is an ordered field, but not a complete one. The LUBP is a completeness axiom. $\endgroup$ – Cameron Buie Oct 16 '18 at 11:57
  • $\begingroup$ Your axiom does not make sense in the above form (consider $I_n = (0,1/n)$). You need the requirement that the $I_n$ are closed intervals. $\endgroup$ – Paul Frost Oct 16 '18 at 13:41

No, you don't need the axim of choice here. What we are stating here, is that a certain family of sets (namely, the family of all sets of upper bounds of the non-empty upper bounded real numbers) is such that each element of that family has a minimum. Besides, if we define the real field as an ordered field such that the Archimedian property holds and that the axiom that you mentioned holds too, then we can prove that property, and we don't need the axiom of choice to do that.

By the way: let $\mathscr N=\mathcal{P}(\mathbb{N})\setminus\{\emptyset\}$. The set $\mathscr N$ is uncountable and each of its elements has a minimum. Do you need the axiom of choice to prove that?

  • $\begingroup$ Well, yeah, good point here, actually $\mathbb{N}$ is well ordered, so in each subset, there is a minimum, but $\leq$ is not a well ordering on $\mathbb{R}$ (or, if we use AC, anything can be well ordered, but we don't want to use it). So, how do i go about the proof of the least upper bound property from the axiom i stated? $\endgroup$ – Michal Dvořák Oct 16 '18 at 10:47
  • 1
    $\begingroup$ Please take a look at this answer. $\endgroup$ – José Carlos Santos Oct 16 '18 at 10:51
  • $\begingroup$ If we define the real numbers to be an ordered field satisfying the least upper bound property then we don't need anything to prove that. The remark you make is irrelevant here. $\endgroup$ – Asaf Karagila Oct 16 '18 at 11:17
  • $\begingroup$ @AsafKaragila I don't understand your remark. Who talked about defining the real numbers to be an ordered field satisfying the least upper bound property? $\endgroup$ – José Carlos Santos Oct 16 '18 at 11:22
  • 1
    $\begingroup$ @CameronBuie Ah, sorry, I typed too fast. Here's the right thing: first, we close $A$ downwards. Now we can restrict attention to rationals for choosing $a_i$ and $b_i$. OK, but since the rationals are well-orderable we're golden. $\endgroup$ – Noah Schweber Oct 16 '18 at 13:46

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.