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5 votes
3 answers
487 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 ? ...
InTheSearchForKnowledge's user avatar
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 ...
Manuel Bonanno's user avatar
0 votes
1 answer
63 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 $X,Y\in F$...
eitan.sh21's user avatar
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 ...
boyler's user avatar
  • 375
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 ...
MB7800's user avatar
  • 83
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 ...
lelouch_l8r4's user avatar
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 ...
rea_burn42's user avatar
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-...
Transfinite Pyramid Scheme's user avatar
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,...
imida k's user avatar
  • 295
3 votes
1 answer
461 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"...
bad at math's user avatar
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 ...
Manuel Bonanno's user avatar
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 ...
Keshav Srinivasan's user avatar
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 $s(n) \in N$ ...
IssaRice's user avatar
  • 1,175
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 ...
wsz_fantasy's user avatar
  • 1,732
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 ...
Boccherini's user avatar

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