Why can positive rationals be not well ordered? If we define the relation to be greater than(>), then every subset will have a least element. Or why are positive or even integers not well ordered? By the same logic we can always find a least element in any subset. I know I am wrong at some very fundamental point, but please explain it to me.
Elementary Set Theory – Well Ordering Principle for Rationals
elementary-set-theorywell-orders
Related Solutions
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.
A. is correct; it's one of the paradigm wellordered sets.
C. is not correct. This has no least member, since for any $x>0$ one can take $x/2$ and get a smaller number greater than $0$, and hence a positive rational.
I'm not sure what you mean by "inversing", but the inverse order has no direct bearing on whether something is wellordered. The only thing that matters is if there are any subsets without a least member. For instance, the integers are not wellordered because $\mathbb{Z}$ itself has no least member; the negative numbers keep going down forever. Likewise the positive rationals keep getting smaller and smaller forever.
To show that a set is not wellordered, you just need to show a non-empty subset such that for every $x$, there's a $y$ with $y<x$. To show that it is wellordered, you need to show that there are no such sets.
Best Answer
Your claim isn't true.
The positive rationals can be well-ordered
Since $\mathbb{Q}$ bijects with $\mathbb{N}$, the well-ordering on $\mathbb{N}$ will induce a well-ordering on $\mathbb{Q}$ and hence on the positive rationals.
However,
The usual ordering of positive rationals is not a well-ordering
The usual ordering is, of course, $\frac{a}{b}>\frac{c}{d}$ if and only if $ad>bc$ (where $a,b,c,d$ are positive integers).
If it is a well-ordering, then there is a least positive rational $p/q$. But halving it gives a smaller positive rational $p/(2q)$, so $p/q$ can't be the least, contradiction.