# Any set can be well-ordered; doesn't that imply $(0,1)$ has a smallest real number?

I came across the above theorem ( also known as Zermelo's theorem) in real analysis. Since any set can be well-ordered so can the set (0,1) which means there must exist a least element of the set.

That number, say 'p' is either rational or irrational ( I don't think we have a third option here). If 'p' rational, then there must be an irrational number 'q' such that 0< q <p (The proof is trivial) Likewise, if 'p' is irrational then there must exist a 'q' such that '0< q <p'

Thus a contradiction and no such 'p' exists and thus (0,1) is not well-ordered.

What am I missing here?

• "Since any set can be well-ordered so can the set (0,1) which means there must exist a least element of the set." -- With respect to which ordering, though? It isn't always the usual $\le$ ordering. Commented Dec 25, 2020 at 22:47
• Yeah, but saying that a set is well-ordered means more than that. Real numbers are not well ordered. Saying well-ordered means there exists a least element for all subsets of the set. Isn't it? Commented Dec 25, 2020 at 22:49
• Real numbers can be well ordered if you assume the Axiom of Choice. Commented Dec 25, 2020 at 22:53
• Yes, a least element with respect the order in question - an element $\ell \in S$ such that, for all $s \in S$, $S$ with order $\preccurlyeq$, that $\ell \preccurlyeq s$. However, in the case of $(0,1)$, $\preccurlyeq$ is not the usual $\le$. Commented Dec 25, 2020 at 22:54
• Carefully write out what you mean by "any set can be well-ordered". Either you have your definition wrong or you are misunderstanding the definition; writing it out will clarify the problem. Commented Dec 25, 2020 at 22:54

The well ordering states:

For every set $$S$$ there exists some total order $$\tilde{<}$$ with the property that every non-empty set has a least element.

where I write $$\tilde{<}$$ to distinguish this order from any more common one. Note that the order is not intrinsic to the set - one can well order the rationals, for instance, by saying that, for rationals in lowest terms (with positive denominators), we define $$\frac{a}b\,\tilde{<}\,\frac{c}d$$ if $$|a|+|b| < |c|+|d|$$ or if $$|a|+|b|=|c|+|d|$$ and $$a. We could get loads of other orders by choosing any bijection $$f:\mathbb N\rightarrow\mathbb Q$$ and saying that $$p_1\,\tilde{<}\,p_2$$ whenever $$f^{-1}(p_1) < f^{-1}(p_2)$$ in the natural numbers - or, in fact, we could do this with any bijection from a set with a known well-order.

None of these would be the usual ordering on the rational numbers - since the usual order is not a well-order, as you prove. The confusion is that your proof regards $$(\mathbb Q,<)$$ as an ordered set - that is, a set with additional structure on it - whereas the well-ordering theorem regards $$\mathbb Q$$ itself - a set with no extra structure.

Said otherwise, you proved that $$((0,1), <)$$ is not a well order for a specific $$<$$. All the well-ordering theorem says is that there is some $$((0,1),\tilde <)$$ that is a well-order - so your proof really just rules out one possible order, but doesn't contradict the existence of some other order.

• Thanks, I wasn't aware of the necessary distinction between the usual order (<) and any other custom defined order that would be required to create a well-ordered set. Commented Dec 25, 2020 at 23:01
• In fact it need not even be a bijection. Any injection $f : \mathbb Q \to S$ into a well-ordered set $S$ will do, by saying $p_1 < p_2$ whenever $f(p_1) < f(p_2)$. Commented Dec 26, 2020 at 7:32

The key word there is "can". The statement "S can be well-ordered" means "There is an ordering on S that is a well-ordering". It doesn't mean "Every ordering of S is a well-ordering". The standard ordering of $$\mathbb R$$ is not a well ordering. Any well ordering of $$\mathbb R$$ will not correspond to anything resembling a "normal" ordering (and, in fact, claiming that $$\mathbb R$$ can be well ordered requires the axiom of choice, which means that no ordering can be explicitly stated).

There is a distinction between $$\mathbb R$$ as a set and $$\mathbb R$$ as a mathematical object. When we speak of the real numbers as a mathematical object, we are speaking of not only the set of real numbers, but also the structure associated with $$\mathbb R$$, such as addition, multiplication, and ordering. Zermelo's theorem says that sets can be well ordered. That means that if we take just the space of $$\mathbb R$$, and ignore the normal structure, we can apply another structure such that the resulting object is well ordered.

• What do you mean by "no ordering can be explicitly stated"? How does that follow from the axiom of choice? Commented Dec 25, 2020 at 23:03
• @UtkarshRaj This is because a weak form of the Axiom of Choice is actually implied by the existence of a well-ordering of the reals. Commented Dec 25, 2020 at 23:05
• It's a little handwavy, but if you could explicitly state an ordering, then you wouldn't need the axiom of choice to do it, because "and now we use Choice to claim that there is an ordering" doesn't actually produce the ordering for you. But it's provable that you do need Choice to state an ordering; so you can't do so explicitly. Commented Dec 25, 2020 at 23:07
• Though there is a limit to how "explicit" an ordering could possibly be even in principle, given that equality of reals isn't even computable. Commented Dec 25, 2020 at 23:09
• @UtkarshRaj If you want a full explanation, you might want to ask another question on that. Basically, it's possible to prove the statement "it's not possible to prove that the real numbers can be well ordered without using the axiom of choice". But given any explicit ordering, proving that it is a well ordering would constitute proving that a well ordering exists. Commented Dec 25, 2020 at 23:16