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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.

1 Answers 1

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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.)

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.

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.

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 $g(0) \prec g(1) \prec g(0)$.

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    Thanks for such a thorough answer; there's a lot there for me to digest and contemplate. The notation issues I can study alone. Re the maths itseIf, I may have accidently implied antisymmetry of $\succ$ in respect of minimal elements in statement 2. What I intended to permit is multiple minimal elements in any ordered subset, which are cyclic in $\succ$. – 2017-01-30
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    p.s. it's not customary to assume $a\succ b \implies a>b$ is it?$ – 2017-01-30
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    @RobertFrost: If you want a partial ordering (see Wikipedia) then there cannot even be any cycles. Do you really want that? If you're certain then I'll think again and edit my answer. Yes for your 2nd comment; "$\prec$" is merely a symbol and you've to define whatever meaning you want. – 2017-01-30
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    I'm trying to walk a very tricky line between having an ordering without implying antisymmetry within a minimal cycle. The challenge is; that elements of a cycle can be considered equal; that's fine. So we definitely have a partial ordering. But in reality elements can also be cyclic by $\succ$ but only among the minimal elements of each totally ordered subset. – 2017-01-30
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    If it helps explain what I mean; $f(x)=(3x+1)/2^m: m\in\mathbb{N}, O\to O$ (O=odd integers). $f(x)\prec x$ – 2017-01-30
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    @RobertFrost: Read the wikipedia definition. Either you have a strict partial order, which **will forbid** any cycles, or you have a non-strict partial order, which allows only 1-cycles. Decide whatever you want, but you **cannot** have a partial order if you want cycles. Since you didn't specify anti-symmetry in your question, my answer stands. – 2017-01-30
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    @RobertFrost: Your last comment above makes no sense. If $m$ is fixed, $f$ is not a map from $O$ to $O$. If you cannot express yourself in notation clearly, use English. Let $f(x)$ be the largest odd factor of $(3x+1)$. Also, you cannot just say $f(x) \prec x$ for every $x \in O$. To define $\prec$ on $O$ you **must** define what it does on every pair from $O$. You could for example say $x \prec y$ iff $x = f(y)$. – 2017-01-30
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    $y=f(x)\implies y\prec x$ – 2017-01-30
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    @RobertFrost: As I said, that's **not** a valid definition. To define any binary relation you **must** define it for **every pair**..... – 2017-01-30
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    Sorry it must be frustrating. It's like I'm learning to speak a foreign language. – 2017-01-30
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    In respect of strict versus non-strict partial order. The rule in this case is; strict for non-minimal elements. But minimal elements not strict. – 2017-01-30
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    $y\prec x \iff y=f^m(x)$ for some $m\in\mathbb{N}$ sorry corrected. – 2017-01-30
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    @RobertFrost: You cannot say "I want this building to be a school for students and a playground for children." If you want to define a binary relation, define it **from scratch**. If you want to ask whether there is a binary relation satisfying some properties, you must **precisely** state the properties. Your last comment can serve as a definition of $\prec$, which would not be anti-symmetric. The powers of two under your $\prec$ would look the same as the example in my answer. In any case, I've answered your question as stated, so is it not enough? – 2017-01-30
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    The only power of $2$ in the domain of $f$ is $1$, so obviously $1=1$. I don't understand. – 2017-01-30
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    @RobertFrost: Oh sorry for a moment I forgot and thought your $f$ was simply the Collatz map. In your case, your $\prec$ has rather unknown structure. Note that you would have $1 \prec 1$ under your definition. – 2017-01-30
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    That was my aim. Did I fail? – 2017-01-30
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    p.s. your point $1\succ 1$ is exactly what I was trying to draw to your attention when I pointed out that it had not been my intention to imply antisymmetry, among minimal elements. But it's a crucial part of the question that reflexivity is only permitted between minimal elements. – 2017-01-30
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    The Collatz map has nothing to do with your question. In particular, your $\prec$ fails to satisfy (2) because $1 \prec 1 \prec 1$ and $1$ is the only available candidate for predecessor of $1$. Even if you list a few properties of your $\prec$, it says nothing about Collatz's conjecture. And unless you want $\prec$ to be a partial order, I've already answered your question in full. – 2017-01-30
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    But $5$ is a successor of $1$ so it does have at least a little to do with it. I'm not sure if you're encouraging me to be more precise, or if it's not plain what I mean – 2017-01-30
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    @RobertFrost: (2) is about predecessors; successors are irrelevant. (2) fails because $1$ has no predecessor, whether or not your notion of "predecessor" includes the element itself. I find it very discourteous when you always reject my answer despite it being absolutely correct and a complete answer to your question as stated. This is not the first time you're doing this. I'm going to edit to address the question of a partial ordering and that's it. – 2017-01-30
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    @RobertFrost: Okay I've answered your question fully in any reasonable way. If you don't understand any specific point in my answer, ask. But I don't want to hear anything more about Collatz's conjecture since it's utterly irrelevant here. – 2017-01-30
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    Thanks. I will accept, I'm grateful for your help and I apologise this has been frustrating. I feel really bad that my imprecision of notation has meant we've spent so long discussing a range of matters without getting close to the actual intent of the question. I will re-post it when I can express it more precisely and I will understand entirely if you ignore it when I do! But the question, written more skilfully, IS equivalent to the Collatz Conjecture. – 2017-01-30
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    Ok I did my best to make it more precise: http://math.stackexchange.com/questions/2121344/does-this-set-exist I can do no more than my best. – 2017-01-30
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    @RobertFrost: You're wrong about both the Collatz conjecture and that you are doing your best. Until you actually learn logic, you will be unable to understand that your attempts are just doomed to failure because you cannot grasp the logical structure of basic things like partial orderings, not to say the Collatz map. So many experts have already pointed out on your previous questions that anything that is equivalent to the Collatz conjecture **must ** involve the **specific** (3n+1) operation. Since your questions don't, they can **never** be equivalent. – 2017-01-31
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    @RobertFrost: As for doing your best, I can believe that you **sincerely believe** that you're doing your best. However, the actual fact is that you're not. The best that you can do right now is to learn logic; read through the first two references that I recommend under "Introduction" in my profile. **After that** you will be perfectly equipped to come up with **truly equivalent** formulations of Collatz conjecture, and will **also fully grasp** why doing so in any simple 1-page way will not solve it. – 2017-01-31
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    @RobertFrost: Finally, sorry for scolding you, but I feel you really need to re-evaluate your mathematical goals and current path. You will appreciate my advice only after you have benefited from it, so I don't expect you to thank me right now. – 2017-01-31
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    Thanks for your advice. I appreciate what you're saying, although you're commenting with limited knowledge of my understanding of Collatz, nor of the analysis which I'm working towards. But I will try to improve my notation to be clearer. With respect to the precise $(3n+1)$ operation I'm analysing a certain distillation of its properties which, to be a proof of Collatz would have to be accompanied by a proof that it is equivalent. What I study at the moment may at times be, or not be, equivalent, as I progress towards an attempt. – 2017-01-31
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    p.s. I would strongly advise you to let go of any emotional attachment you may have, to me ceasing to work on the Collatz conjecture as I tend to be quite resolute in my undertakings. But you are free of course to do as you will. – 2017-01-31
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    @RobertFrost: This shall be my final response to you on the Collatz conjecture, until you show a proper attitude towards learning logic. You are the one with **fundamental incompetence** with basic logic, as very clearly shown in all of your questions, and every field of mathematics requires a full grasp of first-order logic, which you do not have at all. I spend my time responding to you only because I hope you won't waste your time deluding yourself into believing you can make mathematical progress without grasp of basic logic. If you were a crackpot or troll I wouldn't bother. All the best. – 2017-01-31
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    @RobertFrost: Maybe, but we have to always sit down and think rationally what is best for ourselves and others. Dreams $\ne$ reality. I can fly in my dreams sometimes, by sheer will alone. Okay enough chit-chat. I'll leave you to think about what I've said. I hope to see you genuinely progress in mathematics a year from now. Bye! – 2017-01-31
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    "You cannot use variable subscripts in a variable that is quantified" - Why that? How then would you write a statement with more than 26 nested quantifiers? At least in the statement criticized, the indices seem to do no harm ... – 2017-02-09
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    @HagenvonEitzen: As I stated, you cannot use **variable** subscripts in a variable that is quantified, in first-order logic. If you want to go to higher-order logic, you'd have to precisely specify which system you wish to use. Standard presentations of higher-order logic don't do it like this either. You say "seem to do no harm", but people like Robert can't even see clearly what will do harm. Formally speaking, first-order logic has a single symbol for each of countably many variables. How we encode first-order sentences in ASCII is irrelevant to what is valid. – 2017-02-10
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    @HagenvonEitzen: To give you an analogy, to say that his notation is valid is like saying that "$\exists x^2\ ( x^2 = -1 )$" is valid. Or "$\forall f(n)\ ( f(n) < f(n+1) )$". Or "$\forall ! f(x)\ ( g(x) = 0 )$"... Those who are serious about doing and communicating mathematics to others must either use conventionally accepted notation or clearly define whatever idiosyncratic notation they wish to use. Robert did neither. According to modern conventions, we can use **textual** subscripts in quantified variables. "$min_x$" is not textual but clearly denotes a variable related to $N_x$. – 2017-02-10
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    @HagenvonEitzen: Finally, it is conventional in logic to use **meta natural numbers** for variable subscripts when we need an arbitrary number of them when working **in the meta-system**. With respect to the formal system we are studying they are **still textual** and merely serve as one possible encoding of the countably many variables as objects in the meta-system. In the meta-system, on the other hand, we can use those natural number subscripts for reasoning about arbitrary formulae. Those who do not grasp the distinction between the formal system and meta-system had better avoid this. – 2017-02-10