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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.)

Michal Dvořák
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  • 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. – Asaf Karagila Oct 16 '18 at 11:26
  • 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. – Carl Mummert Oct 16 '18 at 11:47
  • 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? – Michal Dvořák Oct 16 '18 at 11:49
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    Yes, completeness is necessary. For example, $\Bbb Q$ is an ordered field, but not a complete one. The LUBP is a completeness axiom. – Cameron Buie Oct 16 '18 at 11:57
  • 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. – Paul Frost Oct 16 '18 at 13:41
  • For a proof see https://math.stackexchange.com/q/612640. – Paul Frost Oct 16 '18 at 13:50
  • @CarlMummert If I look at the proof given in https://math.stackexchange.com/q/612640, then it seems to me that we need the axiom of dependent choice or perhaps the axiom of countable choice. Are there proofs avoiding their use? – Paul Frost Oct 16 '18 at 13:56
  • @Paul Frost: I suppose it depends on how you approach it, but with some work you should be able to choose rational numbers. For example, if $A$ is any nonempty bounded set of reals, let $B$ be the set of rationals that are upper bounds for $A$ and let $C$ be the set of rationals that are lower bounds for $B$. Then $A$ has a least upper bound if and only if $B$ has a greatest lower bound if and only if $C$ has a least upper bound, and in this case the three numbers are the same - no choice is needed to prove this. So we can prove $C$ has a least upper bound, which will not require choice. – Carl Mummert Oct 16 '18 at 14:11
  • Even more easily, if we need to choose an element from an interval, we can always choose a rational, which means we don't need the axiom of choice, because the rationals are countable. – Carl Mummert Oct 16 '18 at 14:14

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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?

José Carlos Santos
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  • 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? – Michal Dvořák Oct 16 '18 at 10:47
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    Please take a look at this answer. – José Carlos Santos Oct 16 '18 at 10:51
  • 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. – Asaf Karagila Oct 16 '18 at 11:17
  • @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? – José Carlos Santos Oct 16 '18 at 11:22
  • You are talking about defining the reals in a different way than the OP. I'm just saying that if you're going down that road, then you can go all in, and just define the reals to be the thing you want them to be in the first place. And this is in fact the more common definition of the axiom of completeness, rather than taking Cantor's lemma as an axiom. – Asaf Karagila Oct 16 '18 at 11:24
  • Well, $\mathcal{P}(\mathbb{N})$ contains countable sets, but set of all upper bounds for all bounded subsets of $\mathbb{R}$ is uncountable collection of uncountable sets, so your argument about $\mathcal{P}(\mathbb{N})$ is irrelevant I think, but still, that answer you linked. Anyways, we always need some kind of completeness, right? Because for example $\mathbb{Q}$ is also ordered field, but not complete. – Michal Dvořák Oct 16 '18 at 11:26
  • I'm afraid that linked answer relies on some Choice--Countable Choice for subsets of $\Bbb R,$ if I recall correctly, but maybe as much as Dependent Choice. – Cameron Buie Oct 16 '18 at 12:20
  • @Noah: "Otherwise, put $b_1:=b_0$ and choose a point $a_1\in A$ such that $a_1>m$." There's no guarantee that this eventuality must only occur finitely-many times, is there? – Cameron Buie Oct 16 '18 at 13:45
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    @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. – Noah Schweber Oct 16 '18 at 13:46
  • @Noah: Nice fix! I just realized what I had originally been thinking of: CC($\Bbb R$) is required to prove that every non-empty, bounded-above subset of the reals has a monotone non-decreasing sequence of elements that converges to its supremum. – Cameron Buie Oct 16 '18 at 13:51
  • @CameronBuie Yes, that's true, but the closure-downwards trick - when it's usable - gets rid of reliance on such arguments. – Noah Schweber Oct 16 '18 at 14:09