2
votes
1answer
60 views

A well ordering on $\mathbb{R}$ and bigger sets

Consider the set of sequences $S = \{f:\mathbb{N}\to\mathbb{N}\}$, define an order on $S$ by the following: Based on the well-ordering of $\mathbb{N}$ and induction, either $f_1 = f_2$ or there is a ...
3
votes
1answer
103 views

Is there anything wrong with this use of the axiom of choice?

I have a proof of the following theorem: Let A be a finite set and X be a perfectly normal topological space, and let $\{(A,\succsim_x): x\in X\}$ be a family of binary relations on $A$ satisfying ...
7
votes
0answers
248 views

Proving equivalence of a tree-based version of Countable Choice for families of finite sets.

In this paper by Good and Tree, the following result is mentioned without proof as part of Proposition 6.5: Each of the following statements imply those beneath it. The countable union of ...
8
votes
3answers
332 views

Without appealing to choice, can we prove that if $X$ is well-orderable, then so too is $2^X$?

Without appealing to the axiom of choice, it can be shown that (Proposition:) if $X$ is well-orderable, then $2^X$ is totally-orderable. Question: can we show the stronger result that if $X$ is ...
4
votes
3answers
183 views

Countable ordinals are embeddable in the rationals $\Bbb Q$ — proofs and their use of AC

Yesterday, Asaf Karagila's answer to my question sparked an extensive discussion on ways of proving that all countable ordinals are embeddable in $\Bbb Q$, and whether particular solutions to this use ...
3
votes
1answer
117 views

Is every discrete topological space orderable?

I apologize for asking a question in topological terms when it's not really about topology, but here goes: If $S$ is a set, is it always possible to totally order $S$ so that every non-minimal ...
1
vote
3answers
105 views

Question regarding well-ordering theorem

The well-ordering theorem states that every set can be well-ordered. That is, there is a total order on the set in which every non-empty subset has a least element. I was wondering whether it's also ...
2
votes
1answer
145 views

Is the set $S$ of functions $S = \{ f : \mathbb R \to \mathbb R \}$ from the reals to the reals a totally ordered set?

Is the set $S = \{ f : \mathbb R \to \mathbb R \}$ of functions $f$ from the reals to the reals a totally ordered set ? I think the question is clear. Note that there is a bijection to $g(x)$ gives ...
6
votes
3answers
427 views

How to define a well-order on $\mathbb R$?

I would like to define a well-order on $\mathbb R$. My first thought was, of course, to use $\leq$. Unfortunately, the result isn't well-founded, since $(-\infty,0)$ is an example of a subset that ...
0
votes
1answer
123 views

Shortest proof of Kuratowski's lemma

I want a short proof of Kuratowski's lemma (about maximal chains). Does proving Kuratowski's lemma need first prove Zorn lemma? I am writing a book in which I claim that the reader needs to know ...
5
votes
2answers
333 views

What does a well ordering of $\mathbb{R}$ look like? [duplicate]

Possible Duplicate: Is there a known well ordering of the reals? I am having a hard time wrapping my head around what a well-ordering of $\mathbb{R}$ looks like. I have seen the ...
3
votes
1answer
255 views

Is a chain-complete lattice a complete lattice without the axiom of choice?

This question is inspired by this question. Consider the following result: Let $(L,\leq)$ be a chain-complete lattice. Then $(L,\leq)$ is a complete lattice. Can this result be proven without ...
2
votes
2answers
217 views

Bourbaki-Witt fixed point theorem: two questions

Consider the following theorem: Let $f\colon E \to E$ have the propery that $f(x)\geq x$, where $(E,\leq)$ is a non-void partially ordered set with the property that every totally ordered subset of ...
8
votes
4answers
227 views

Is there an element with no fixed point and of infinite order in $\operatorname{Sym}(X)$ for $X$ infinite?

Let $X$ be an infinite set. Let $\operatorname {Sym}(X)$ denote the group of all bijections from $X$ onto itself. I have been thinking about the existence of elements of infinite order in this group. ...
4
votes
4answers
607 views

Prove that for any infinite poset there is an infinite subset which is either linearly ordered or antichain.

Let $(X, \leq)$ be an infinite poset. Prove that there is an infinite $X' \subset X$ for which holds one of the following: 1) Induced order on $X'$ is linear. 2) $(X', ...
2
votes
1answer
149 views

How different are inductively ordered set from posets that satisfy the ascending chain condition?

I'd like to show that posets that satisfy the ascending chain condition (ACC) have maximal elements. I noticed this statement looks much like Zorn's lemma which holds for inductively ordered sets. ...
1
vote
2answers
155 views

How does this statement on posets follow from Zorn's lemma?

Let $P$ be a poset with a maximal element $r$, i.e. an element $r\in P$ such that $r\geq x$ for all $x\in P$. One can think of the poset as a tree with root $r$. I want to show that there exists a ...
8
votes
3answers
287 views

With Choice, is any linearly ordered set well-ordered if no subset has order type $\omega^*$?

I've been fumbling around with order types and ordinals these past few days. I read about partial, total, and well-ordered structures, and I'm curious to see if a linearly ordered set has no subset ...