For context, I have no set theory background and my understanding of sets is naive. I had a discussion with a friend that I thought for any well-ordered subset of the reals, I could simply take the least element, second least, etc. to show it is countable. It is the same procedure described in Is a well-ordered subset of $(\mathbb{R},<)$ countable?. However he told me I had to make use of ordinals (which I don't know about) to index into the set. Is there an intuitive (or at least relatively elementary) reason why I can't count up using my naive procedure to conclude the subset is countable? The main concern I see is about if I ever "finish counting" which I don't know how to reason about and seems subtle.
What’s wrong with the naive “take successive smallest” procedure in proving well-ordered subsets of $\mathbb R$ are countable
elementary-set-theoryreal numberswell-orders
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I assume you know the general theorem that, using the axiom of choice, every set can be well ordered. Given that, I think you're asking how hard it is to actually define the well ordering. This is a natural question but it turns out that the answer may be unsatisfying.
First, of course, without the axiom of choice it's consistent with ZF set theory that there is no well ordering of the reals. So you can't just write down a formula of set theory akin to the quadratic formula that will "obviously" define a well ordering. Any formula that does define a well-ordering of the reals is going to require a nontrivial proof to verify that it's correct.
However, there is not even a formula that unequivocally defines a well ordering of the reals in ZFC.
The theorem of "Borel determinacy" implies that there is no well ordering of the reals whose graph is a Borel set. This is provable in ZFC. The stronger hypothesis of "projective determinacy" implies there is no well ordering of the reals definable by a formula in the projective hierarchy. This is consistent with ZFC but not provable in ZFC.
Worse, it's even consistent with ZFC that no formula in the language of set theory defines a well ordering of the reals (even though one exists). That is, there is a model of ZFC in which no formula defines a well ordering of the reals.
A set theorist could tell you more about these results. They are in the set theoretic literature but not in the undergraduate literature.
Here is a positive result. If you work in $L$ (that is, you assume the axiom of constructibility) then a specific formula is known that defines a well ordering of the reals in that context. However, the axiom of constructibility is not provable in ZFC (although it is consistent with ZFC), and the formula in question does not define a well ordering of the reals in arbitrary models of ZFC.
A second positive result, for relative definability. By looking at the standard proof of the well ordering principle (Zermelo's proof), we see that there is a single formula $\phi(x,y,z)$ in the language of set theory such that if we have any choice function $F$ on the powerset of the reals then the formula $\psi(x,y) = \phi(x,y,F)$ defines a well ordering of the reals, in any model of ZF that happens to have such a choice function. Informally, this says that the reason the usual proof can't explicitly construct a well ordering is because we can't explicitly construct the choice function that the proof takes as an input.
We would like, in general, for properties of well-orders to be preserved under order-preserving maps - that is, we want it to be the order that matters, not the elements. The two sets you name, $\{z \in \mathbb{Z} : z \leq 100\}$ and $\{z \in \mathbb{Z} : z \leq 1\}$, are order-isomorphic; they differ only in elements, not in structure.
The property "$A$ is a continuation of $B$" is a property of the order - if $A$ is a continuation of $B$ and both are well-orders, then $A$ and $B$ are both isomorphic to ordinals and the ordinal of $B$ is greater than the ordinal of $A$. In your proposed change, "continuation" would be a property of the particular set; possibly an interesting notion anyway, but not what we usually care about when talking about well-orders.
Best Answer
Given any infinite well-ordered set $X$, you can define an injection $f:\mathbb{N}\to X$ by repeatedly taking the least remaining element as you describe. To be precise, you can recursively define $f(n)$ to be the least element of $X\setminus\{f(m):m<n\}$.
However, this map may not be surjective, and so it may not show that $X$ is countable. For a very simple example, consider $X=\{1\}\cup\{1-1/n:n\in\mathbb{Z}_+\}\subset\mathbb{R}$. This is well-ordered with the usual ordering inherited from $\mathbb{R}$. The least element is $1/2$, then the second least element is $2/3$, then the third least element is $3/4$, and so on. However, repeatedly taking least elements like this does not exhaust all of $X$: the map $f:\mathbb{N}\to X$ hits each element of the form $1-1/n$ but does not hit $1$, so it fails to be surjective.