Does this ordering of the positive, odd integers exist?

Question

Is there some ordering $\succ$ of the positive odd numbers $n$, satisfying:

Every element has an infinite number of greater elements:

$\forall n\in\mathbb{N} \quad\exists\quad X=\{x\succ n\}: \lvert X\rvert=\lvert \mathbb{N}\rvert $

Every element has a unique, immediately smaller element:

$\forall n_p\ \exists\ n_{p-1}: \nexists n_m \mid n_p\succ n_m \succ n_{p-1}$

Every totally ordered subset contains at least one minimal element:

$\forall\ n\neq n_{min_x}\exists\ n_{min_x}: n\succcurlyeq n_{min_x}$

Of all the possible totally ordered subsets $N_x$, there is exactly one having a unique minimal element satisfying: $n_{min_x}\succcurlyeq n_{min_x}$

And yet: $\succ$ does not totally order the positive odd numbers?

Where have I got with this? I am wondering if, by Zorn's Lemma, there must be a unique minimum element and therefore the set must be totally ordered by $\succcurlyeq$, meaning that the statement that $\succ$ does not totally order the positive odd numbers must be false.

I apologise in advance for any errors made herein.

Answer 1

First let me point out what is wrong with your question.

(1) Every element has an infinite number of greater elements:

$\forall n\in\mathbb{N} \quad\exists\quad X=\{x\succ n\}: \lvert X\rvert=\lvert \mathbb{N}\rvert $

This is not a valid translation into logic. It's never meaningful to write "$\exists X = \cdots$". My advice is to learn first-order logic properly.

(2) Every element has a unique, immediately smaller element:

$\forall n_p\ \exists\ n_{p-1}: \nexists n_m \mid n_p\succ n_m \succ n_{p-1}$

Your English sentence is okay, but the term for "immediately smaller neighbour" is "predecessor". Your logical translation is simply invalid. You cannot use variable subscripts in a variable that is quantified. Even if you fix that to "$\forall x\ \exists y\ ( \neg \exists z\ ( y \prec z \prec x ) )$", it totally does not say what you want to say. For example, it is true if $\prec$ is the empty ordering! Please go back and figure out how to state what you want in first-order logic. I'll assume the English sentence rather than the meaningless logic translation you gave.

(3) Every totally ordered subset contains at least one minimal element:

$\forall\ n\neq n_{min_x}\exists\ n_{min_x}: n\succcurlyeq n_{min_x}$

Again, the English is okay, but the translation is meaningless. If $n_{min_x}$ is some constant (which you did not even specify), then it is meaningless to use it in an inner quantifier. Secondly, why would you have the weird condition "$n = n_{min_x}$"? By standard definition "$n_{min_x} \preceq n$" is equivalent to "$n_{min_x} \prec n \lor n_{min_x} = n$". Again, I'll ignore your invalid logic translation.

(4) Of all the possible totally ordered subsets $N_x$, there is exactly one having a unique minimal element satisfying: $n_{min_x}\succcurlyeq n_{min_x}$

As above, "$n_{min_x}\succcurlyeq n_{min_x}$" is always true, so your condition is totally meaningless, even though it is precise. Specifically, this phrase can be totally ignored, and all we need to satisfy is that only one totally ordered subset has a unique minimal element.

(Notice also that it is useless to talk about "positive odd numbers" since any countable set will do. This is because you never used the values or ordering of the positive odd numbers anywhere in your desired properties.)

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There is no such ordering $\prec$. Property (4) is by itself already sufficient to make it impossible, because $\{1\}$ and $\{3\}$ are each a subset of the odd positive integers that is totally ordered under $\prec$ and have a unique minimal element.

Game over.

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If want $\prec$ to be a (strict) partial ordering, then (2) and (3) cannot be both satisfied. This is because there cannot be any cycles in the graph representing $\prec$. This is a basic fact provable by induction and will be left as an exercise. Now (2) would then imply that there is a descending chain $(a_n)_{n\in\mathbb{N}}$ where $a_{n+1} \prec a_n$ for every $n \in \mathbb{N}$, which we can construct by induction. Since $\prec$ is transitive, this chain forms a totally ordered subset, and hence has a minimal element by (3), which is impossible since that minimal element's predecessor is also part of the chain.

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If you throw away property (4), but do not require $\prec$ to be a partial ordering (namely it's just a binary relation), then there is a solution as follows:

Let $f$ be a bijection from the odd positive integers $O$ to the natural numbers, and let $g = f^{-1}$.

Define $\prec$ to be a relation on $O^2$ such that $x \prec y$ iff ( $f(x) < f(y)$ or $⟨f(x),f(y)⟩ = ⟨1,0⟩$ ) for every $x,y \in O$.

Then $\prec$ satisfies all the properties you asked for except (4).

It is trivial to see that (1) holds. (2) holds because the predecessor (under $\prec$) of $g(0)$ is $g(1)$ and the predecessor of $g(n+1)$ is $g(n)$ for any natural number $n$. (3) holds because any totally ordered subset $X$ of $O$ cannot have both $g(0)$ and $g(1)$ as members, and hence $x \prec y$ iff $f(x) < f(y)$ for every $x,y \in X$, which obviously implies that $X$ has a minimum under $\prec$. Finally $\prec$ is not a total ordering on $O$ since $g(0) \prec g(1) \prec g(0)$.