Questions tagged [well-orders]

For questions about well-orderings and well-ordered sets. Depending on the question, consider adding also some of the tags (elementary-set-theory), (set-theory), (order-theory), (ordinals).

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Why does proof of Zorn's lemma need to use the fact about ordinals being too large to be a set?

I'm not understanding why its necessary to invoke the knowledge about ordinals in order to prove Zorn's lemma. The Hypothesis in Zorn's lemma is Every chain in the set Z has an upper bound in Z Then ...
Pecan Lim's user avatar
7 votes
1 answer
98 views

What does the cardinality alone of a totally ordered set say about the ordinals that can be mapped strictly monotonically to it?

For any cardinal $\kappa$ and any totally ordered set $(S,\le)$ such that $|S| > 2^\kappa$, does $S$ necessarily have at least one subset $T$ such that either $\le$ or its opposite order $\ge$ well-...
Transfinite Pyramid Scheme's user avatar
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Using the Principle of Well Order, show that for all $n \in N$ it holds that $4^n -1$ is divisible by 3.

Using the Principle of Well Order, show that for all $n \in N$ it holds that $4^n -1$ is divisible by 3. I have already defined the set of counterexamples $C$, then I proved that for $n=1$ the ...
Marcos Daniel Castaeda Ramirez's user avatar
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1 answer
92 views

Well-ordering theorem and cardinality of real numbers

If we assume that the axiom of choice is right, the well-ordering theorem can be verified. So, the set of real numbers also can be constructed by well-ordering property. By the well-ordering property, ...
Eunseong So's user avatar
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Prove that if $A$ is any well-ordered set of real numbers and $B$ is a nonempty subset of $A$, then $B$ is well-ordered.

To prove: if $A$ is any well-ordered set of real numbers and $B$ is a nonempty subset of $A$, then $B$ is well-ordered. My attempt: Suppose towards a contradiction that given any well-ordered set of ...
lohg's user avatar
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1 vote
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Limit Countable Ordinal - is it a limit of a intuitive sequence of ordinals?

I am studying set theory, ordinal part. Set theory is new to me. I know that commutativity of addition and multiplication can be false in infinite ordinal world. $ \omega $ = limit of sequence $\, 1,2,...
imida k's user avatar
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3 votes
1 answer
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Difference between partially ordered, totally ordered, and well ordered sets.

I just started studying set theory and I'm a bit confused with some of these relation properties. Given a set A = {8,4,2}, and a relation of order R such that aRb means "a is a multiple of b"...
bad at math's user avatar
1 vote
1 answer
101 views

a proof idea: Every well-ordered set has an order-preserving bijection to exactly one ordinal.

I have seen a proof of the statement, and its usually by transfinite induction. And I'm trying to find out why my proof doesn't work, it seems too simple: Let $X$ be a well-ordered set. Define $X^{<...
hteica's user avatar
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2 votes
1 answer
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Intuitionistic well-orderings of uncountable sets

The well-ordering principle has always been considered to be highly unconstructive, as far as I know. However, I think intuitionistic mathematics can be compatible with the existence of a well-...
Keplerto's user avatar
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If i have an a well ordered set in which every chain admits an upper bound then the maximal element is unique

It is clear that the Zorn lemma guarantees the existence. I prove that the minimal element is unique, and obviously the set is totally ordered. So because the uniqueness of the successor of all ...
Manuel Bonanno's user avatar
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1 answer
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Using WOP to prove certain naturals can be written a certain way.

$\newcommand{\naturals}{\mathbb{N}}$ Use the Well-Ordering Principle to prove that every natural number greater than or equal to 11 can be written in the form $2s+5t$, for some natural numbers $s$ and ...
Mave's user avatar
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0 votes
1 answer
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A couple of well ordering proofs.

I'm having trouble understanding a couple of things when studying well orderings and ordinals. I know that given a well ordering $(A,<)$ there is no $a\in A$ s.t. $(A,<)\cong (A_a,<)$ where $...
cento18's user avatar
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2 votes
1 answer
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What is the generalization of well-orders to partial orders?

Every well-ordered set is linearly ordered. However, is there a notion of well-orders that generalizes to partial orders? Maybe, the generalized definition will be that every non-empty subset has a ...
user107952's user avatar
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1 vote
2 answers
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Well-Ordering Principle From Recursion Theorem

As far as I understand, in intuitionistic logic we have neither (i) the well-ordering principle nor (ii) the recursion theorem. But can one deduce one from the other? I believe we cannot deduce (ii) ...
fweth's user avatar
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Order isomorphism on a subset and a segment

I am trying to understand Theorem 1.7.4 of Devlin's Joy of Sets. The Theorem states: There is no isomorphism of $X$ onto a segment of $X$ (supposing $X$ is a woset). The difficulty I am having is that,...
brocolliSally's user avatar
3 votes
2 answers
367 views

Construction of two uncountable sequences which are "interleaved"

I believe the answer to my following question is no, but some things about uncountable sets/sequences can be really counterintuitive so I wanted to double check: Does there exist a pair of uncountable ...
psychicmachinist's user avatar
0 votes
1 answer
117 views

Why is well-ordering needed to define the statement "$\forall i, \, P(i) \implies P(i + 1)$"?

I have learned a proof that the well ordering principle is equivalent to the inductive property for $\mathbb{N}$, and have understood it. However, I am confused as to the following statement my notes ...
Princess Mia's user avatar
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1 vote
1 answer
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What are the order types of computable pseudo-ordinals with no c.e. descending chains?

The notion of a “computable pseudo-ordinal”, i.e. a computable linearly ordered set with no hyperarithmetical descending chains, is an old one going back to Stephen Kleene. Joe Harrison wrote the ...
Keshav Srinivasan's user avatar
2 votes
2 answers
68 views

Does every well-ordered set obeying the non-induction Peano axioms have a well-ordering compatible with the successor operation?

Let $N$ be a well-ordered set together with a unary operation $s$ that obeys the following axioms (they are just the Peano axioms without induction): $0 \in N$ for each $n \in N$ we have $s(n) \in N$ ...
IssaRice's user avatar
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28 views

A class of linear orderings with a convexity property on ordinals

I am looking for a reference for the following class of linear orderings on a non-zero ordinal $\lambda$. Define $\lambda^-$ as the predecessor of $\lambda$ if $\lambda$ is a successor, and set $\...
nombre's user avatar
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Are there any guessing principles in set theory which guess well-orders of sets?

$\Diamond(\omega_1)$ was introduced by Jensen and is known to imply $\mathsf{CH}$ and the existence of a Souslin tree. It is the statement that there is a sequence (in particular a $\Diamond$-sequence)...
quanticbolt's user avatar
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Does a total or well ordering has a countable cofinal

I read a post sometime earlier asking about the cardinality of total ordering on a set, which I forgot about the detail but led me to this question. For a total/well-ordered set $A$ does there exist ...
wsz_fantasy's user avatar
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3 votes
4 answers
558 views

Why can't well ordered sets have infinite decreasing subsequences?

Chapter 2 of Mathematics for Computer Science presents the following: Define the set $\mathbb F$ of fractions that can be expressed in the form $\frac{n}{n+1}$. Define $\mathbb N$ as the set of ...
Dean DeRosa's user avatar
0 votes
1 answer
113 views

Countable versus uncountable dense linear orders

Let $(\Omega,\leq)$ be a dense linear order without endpoints. If $\Omega$ is countable, we know by Cantor that $(\Omega,\leq)$ is order-isomorphic to $(\mathbb{Q},\leq)$. Suppose that $\Omega$ is not ...
Boccherini's user avatar
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0 answers
23 views

Does this beginningless-past thought experiment result in several possible non-well-ordered sets?

Assume there is an infinitely large universe with a beginningless past in which two eternal particles move at a constant speed of 1 km per “day” with respect to each other. Intuitively (to me), it ...
Al260's user avatar
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Exercise 4 of Chapter 7 in Enderton's Elements of Set Theory

Let $<$ be the usual ordering on the set $P$ of positive integers. For $n$ in $P$, let $f(n)$ be the number of distinct prime factors of $n$. Define the binary relation $R$ on $P$ by $$ mRn \iff [f(...
blahblah's user avatar
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2 answers
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$\mathbb{N}\times\{0,1\}$ and $\{0,1\}\times \mathbb{N}$ not isomorphic

Show that the sets $W=\mathbb{N}\times\{0,1\}$ and $W'=\{0,1\}\times\mathbb{N}$, ordered lexicographically are non-isomorphic well-ordered sets. Any hints on how to prove this?
cut's user avatar
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0 answers
172 views

Prove: $X$ is well orderable $\implies$ $X \times X$ is well orderable [duplicate]

I am studying a course on ZF Set Theory and have recently been considering whether or not $X$ being well orderable implies that $X \times X$ is well orderable. More formally my question is the ...
FD_bfa's user avatar
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6 votes
1 answer
258 views

Does there exist an infinite set admitting precisely four linear orders?

I am studying a course on ZF Set Theory where I encountered the following question: Does there exist a set admitting precisely four linear orders? Since any set with $n$ elements exhibits $n!$ ...
FD_bfa's user avatar
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2 votes
1 answer
154 views

Is the set of all linear orders on $\mathbb{N}$ linearly orderable?

In studying the issue of linear orders and well ordering in the context of ZF Set Theory (without the Axiom of Choice), I have recently been thinking about the following question: Is the set of all ...
FD_bfa's user avatar
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1 vote
1 answer
69 views

Induction principle on well ordered set

I'm new and very grateful this site exists. If I do something wrong, feel free to tell me. I have to prove the following: Let $S$ be a subset of a well ordered set $L$ with the conditions: a) $0_L \in ...
anoniem's user avatar
  • 191
0 votes
1 answer
38 views

Well Ordering Principle Proof without mathematical induction with different approach

Denote $\Bbb Z_0$ be the set of all non-negative integers. Well Ordering Principle for $\Bbb Z_0$. Every non-empty subset $S$ of $\Bbb Z_0$ has a least element; that is, there exists $m \in S$ such ...
math404's user avatar
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6 votes
1 answer
114 views

Sequence of quadratic surds over nonnegative integers without having to delete or sort?

I am trying find an strictly increasing iterative sequence that gives this set sorted: $$[a+\sqrt{b}: a,b \in \mathbb{N_0}].$$ These are a subset of constructable numbers. When I look at it, there are ...
Teg Louis's user avatar
13 votes
1 answer
612 views

Why do we need "canonical" well-orders?

(I asked this question on MO, https://mathoverflow.net/questions/443117/why-do-we-need-canonical-well-orders) Von-Neumann ordinals can be thought of "canonical" well-orders, Indeed every ...
Vivaan Daga's user avatar
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1 vote
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Every Non empty subset has a least element implies linear order

Suppose $(A,R)$ be structure where R is a binary relation on $A$. Suppose $A$ has the property that every Non empty subset of $A$ has a least element w.r.t. the relation $R$. Then $R$ is a linear ...
user avatar
2 votes
2 answers
125 views

What is the meaning of "induction up to a given ordinal"?

Given an ordinal $\alpha$, what does it mean: "induction up to $\alpha$"? When $\alpha=\omega$, is this is ordinary mathematical induction? Also, Goodstein's Theorem is equivalent to "...
John's user avatar
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1 vote
1 answer
89 views

Question about a proof that any well-ordered set is isomorphic to a unique ordinal

I am studying a proof that every well-ordered set is isomorphic to a unique ordinal. However, I don't understand why $A = pred(\omega)$ (see yellow). One direction is clear: Let $x \in pred(\omega)$, ...
DerivativesGuy's user avatar
2 votes
1 answer
61 views

non-wellordered linear order that doesn't contain a copy of $\omega^*$ in ZF?

Obviously a wellorder cannot contain a copy of $\omega^*$ (the dual order of $\omega$), and this can be proven in ZF. In ZFC, it is easy to prove any linear order which is not a wellorder does contain ...
Zoe Allen's user avatar
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2 votes
1 answer
39 views

Intersection of a chain of closed sets

I was recently attempting proving a conjecture of mine about the existence of certain minimal nonempty closed sets in a topological space. I opted to use Zorn's Lemma, and the proof would go through ...
Nick F's user avatar
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2 votes
2 answers
244 views

Product of Countable Well-Ordered Set with $[0,1)$ is Homeomorphic to $[0,1)$

As part of a proof that the long line is locally Euclidean, I'd like to prove the following: Proposition. If $A$ is a countable well-ordered set, then $A \times [0,1)$ with the dictionary order is ...
Nick A.'s user avatar
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0 votes
0 answers
73 views

Order type of this order over naturals

For no specific reason other than curiosity, I find myself wondering of a way to formalize a specific order of the natural numbers based on a sort of "recursive prime number factorization", ...
Uro's user avatar
  • 82
0 votes
1 answer
41 views

Initial segment of well ordered set

An initial segment (B) of an well-ordered set $(A,<)$ is a subset $X⊊A$ such that, for all $x \in X$ and for all $y∈A$ such that $y<x, y∈X$. I want to prove that this set is equal to $A_z =$ { $...
Mgh's user avatar
  • 3
1 vote
1 answer
73 views

How do you prove that there exists a highest element of any finite, nonempty subset of Natural Numbers? Is the following algorithmic proof valid?

Since the given set, $C \subset \mathbb{N}$ is non empty, hence by well ordering principle there exists $\alpha \in C$ which is the lowest element in C. Also, since the set $C$ is finite, $\quad \...
Pragnya Jha's user avatar
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0 answers
55 views

Partial order, well order and initial segment

I saw the following definition for initial segment in https://digitalcommons.kennesaw.edu/cgi/viewcontent.cgi?article=2161&context=facpubs If $\leq$ is a partial order in a set $X$, then a chain ...
Irene's user avatar
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6 votes
1 answer
209 views

Riddle: finite set that contains one of the three numbers

A colleague told me this two-part riddle: PART $\mathbb{N}$: Three players sit in a circle, each having a positive integer number $x_i$ written on their hat. Every player can see only the numbers of ...
Antoine's user avatar
  • 3,429
1 vote
1 answer
47 views

"Name" for this order relation of binary sequence

I am trying to define a Matrix $M_n$ of dimension $2^n \times n$ where each row corresponds to a element of the set $\{0,1\}^n$. To avoid ambiguity, I using an order relation over the set $\{0,1\}^n$ ...
Renato Fernandes's user avatar
0 votes
1 answer
42 views

How many automorphisms are there for $ \langle \omega , < \rangle $?

How many automorphisms are there for $ \langle \omega , < \rangle $? I'm not sure how to start this, although I expect there to be an upper bound of $2^{\aleph_0}$. ($\aleph_0^{\aleph_0} = 2^{\...
amlearn369's user avatar
1 vote
1 answer
44 views

Is the class of well-ordered sets a pseudo-elementary class? [closed]

I know that the class of well-ordered sets is not an elementary class. But, is it at least a pseudo-elementary class? If not, what is the proof?
user107952's user avatar
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1 vote
1 answer
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What is the class of models of the $\forall$-theory of the class of well-ordered sets?

I know that the class $W$ of well-ordered sets is not an axiomatizable class. However, it does have an associated first-order theory $Th(W)$. Now, consider the $\forall$-part of the theory $Th(W)$, ...
user107952's user avatar
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0 votes
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100 views

Cantor-Bendixson derivative sets

I try to show that a compact subset $A\subset \mathbb{C}$ is at most countable if and only if there exists a countable ordinal number $\alpha$ (i.e $\alpha <\omega_{1},$ where $\omega_{1}$ is ...
pospos's user avatar
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