All Questions
Tagged with axiom-of-choice well-orders
42
questions
2
votes
0
answers
106
views
How does one prove without the axiom of choice that the product of a collection of nonempty well-ordered sets is nonempty?
Suppose $\{X_{\alpha}\}_{\alpha\in\mathcal A}$ is an indexed family of nonempty well-ordered sets, where $X_{\alpha}=(E_{\alpha},\le_{\alpha})$ for each $\alpha$. It seems intuitively obvious that we ...
- 20.7k
4
votes
0
answers
105
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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 ...
- 41
0
votes
1
answer
130
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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, ...
0
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0
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66
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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 ...
- 1,732
2
votes
1
answer
186
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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 ...
- 4,331
2
votes
1
answer
64
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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 ...
- 5,633
5
votes
2
answers
578
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The Well-Ordering Theorem and Uncountably Infinite Sets
So, I've been doing some reading regarding the Well-Ordering Theorem, and I've come across something that doesn't quite make sense. Per Wikipedia, the Well-Ordering Theorem states that any set $X$ is ...
- 61
2
votes
1
answer
111
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Infinite Lexicographic Order on Bijections is Well-Order
Problem. Prove, without using $\mathsf{AC}$ if possible , that if $\alpha$ and $\beta$ are ordinals such that $\alpha$ is countable and $\beta>1$, then $\alpha^\beta$ is countable.
The induction ...
- 1,371
5
votes
2
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177
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Does ZF prove that the union of a $\subseteq$-chain of well-orderable sets is again well-orderable?
I'm interested in this mis-transcription of Folland: https://proofwiki.org/w/index.php?title=Zorn%27s_Lemma_Implies_Well_Ordering_Theorem&oldid=356679. It is dubiously stated that the union of a $\...
- 1,645
2
votes
1
answer
177
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How far are WO and AC equivalent?
In $\textsf{ZF}$ with first-order logic we can prove that the well-ordering theorem ($\textsf{WO}$) and the axiom of choice ($\textsf{AC}$) are equivalent. I was wondering how far (in a vague sense) ...
- 912
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1
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52
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Can someone explain these theorems to me (I have not been able to understand them very well).
Let $A$ be an arbitrary set. We will consider pairs $(B, G)$, where $B$ is a subset of $A$, and $G$ is an order relation in $B$ which well-orders $B$.
Let $\mathscr{A}$ be the family of all such ...
- 533
1
vote
1
answer
278
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Well ordering theorem and Zorn's lemma implies the axiom of choice.
I am curently studying the well ordering theorem's(WOL) and zorn's lemmas's(ZL) equivalence with the axiom of choice(AOC). I have constucted the proofs of WOL and ZL implying AOC as below:
...
- 183
3
votes
2
answers
361
views
Infinite subset of natural numbers
I want to show that if the set $A$ is an infinite subset of natural numbers $N$, then $\text{card}(A)=\text{card}(N)$. To do so, it suffices to find an injective function $f:N\rightarrow A$.
One ...
- 991
1
vote
1
answer
133
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Do I understand this proof for AoC $\implies$ well-ordering theorem correctly?
Am a real beginner in set-theory. I read that the Axiom of Choice (AoC) implies the Well-Ordering Theorem...
From what I understand, the gist of the proof is that given any set $X$, we can use AoC to ...
- 735
1
vote
1
answer
214
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Every partial order with well ordered chains is well founded is equivalent to the axiom of choice
The theorem that every partial order with well ordered chains must be well founded can be proved with the Hausdorff Maximal Principle which is equivalent to the axiom of choice. Does this theorem ...
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