All Questions
Tagged with well-orders order-theory
154
questions
5
votes
3
answers
488
views
Motivation of inventing concept of well-ordered set?
I've started studying set theory for a while. I understand what is an ordered sets but i still fail to see what motivated mathematicians to invent these concept.
Could you please enlightment me ?
...
1
vote
1
answer
78
views
Complete iff Compact in Well-Ordered space
Let $T=(S, \leq, \tau)$ a well-ordered set equipped with order topology (defined here).
Definition 1: $T$ is called complete iff every non-empty subset of $T$ has a greatest lower bound (inferior) and ...
0
votes
1
answer
64
views
Partial order on sets and application of Zorn's Lemma to construct well-ordered subset
I would appreciate help with the following question:
Let $(A,<)$ a linear ordered set.
a. Let $F\subseteq P(A)$. Prove that the following relation is a partial order in $F$: $X\lhd Y$ for $X,Y\in F$...
1
vote
1
answer
81
views
Any subset of a well-ordered set is isomorphic to an initial segment of this well-ordered set.
I wanted to prove the fact for which I have a sketch of proof: Let $(W,\leq )$ be a well-ordered set and $U$ be a subset of $W$. Then considering the restriction of $\leq $ to $U\times U$, we have ...
0
votes
0
answers
30
views
Well-founded Relation on infinite DAGs
A well-founded relation on set $X$ is a binary relation $R$ such that for all non-empty $S \subseteq X$
$$\exists m \in S\colon \forall s \in S\colon \neg(s\;R\;m).$$
A relation is well-founded when ...
1
vote
1
answer
121
views
Partition of $\mathbb R$ in convex subsets/badly ordered sets
Background: These questions come from two different exercises, but since the first is much shorter and of the same kind of one of the others, I preferred to put everything in only one thread. (We work ...
2
votes
0
answers
75
views
Law of Trichotomy for Well-Orderings
Often in beginning set-theory courses, and in particular in Jech's book Set Theory, it is proved from scratch that given any two well-orderings, they are isomorphic or one is isomorphic to an initial ...
8
votes
1
answer
109
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-...
1
vote
0
answers
46
views
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,...
3
votes
1
answer
467
views
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"...
0
votes
1
answer
58
views
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 ...
1
vote
1
answer
95
views
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 ...
2
votes
2
answers
80
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 $s(n) \in N$
...
0
votes
0
answers
66
views
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 ...
0
votes
1
answer
172
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 ...