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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Is there a known well ordering of the reals?
So, from what I understand, the axiom of choice is equivalent to the claim that every set can be well ordered. A set is well ordered by a relation, $R$ , if every subset has a least element. My ...
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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 well-...
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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 ...
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A way to well-order real line
How is well-ordering in real line possible?
I know that the axiom of choice provides possible well-ordering, but intuitively, this does not seem to make sense.
How can you compare 1.111111.... and 1....
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Totally ordering the power set of a well ordered set.
Let's say I take a set $S$, where $S$ can be well ordered. From what I understand, one can use that well ordering to totally order $\mathscr{P}(S)$.
How does a body actually use the well ordering of $...
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Well-ordering theorem and second-order logic
I am confused by this sentence in the Wikipedia article for "Well-ordering theorem":
...the well-ordering theorem is equivalent to the axiom of choice, in the sense that either one together with ...
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There is a well ordering of the class of all finite sequences of ordinals
I am trying to solve this exercise from Jech's book on set theory:
Ex. 3.6: There is a well ordering of the class of all finite sequences of ordinals such that for each $\alpha$, the set of all ...
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Why isn't this a well ordering of $\{A\subseteq\mathbb N\mid A\text{ is infinite}\}$?
So, to explain the title, I'm referring to the necessity of the axiom of choice in the existence of a well ordering on reals, or any uncountable set.
Now, while tweaking some sets, I came across this ...
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Hessenberg power of ordinals
According to these notes on ordinal arithmetic:
The Hessenberg sum $\alpha + \beta$ is the supremum of ordinals that are isomorphic to some well-order on $\alpha \sqcup \beta = (\{0\} \times \alpha) \...
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Well-orderings of $\mathbb R$ without Choice
The question is about well-ordering $\mathbb R$ in ZF. Without the Axiom of Choice (AC) there exists a set that is not well-ordered. This could occur two ways: a) there are models of ZF in which $\...
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Prove that no ordinal is an element of itself
I would like to show that the successor set of an ordinal: $s(\alpha) =\alpha \cup\{\alpha\}$ is an ordinal. To do so, I need to show that that $S(\alpha)$ is strictly well-ordered by $\in$.
I can't ...
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Given $S \hookrightarrow T$ construct $U ≈ T$ disjoint from $S$ in Z set theory?
I was recently thinking about the fact (in ZFC) that, given a (first-order) structure $A$ that embeds into another structure $B$, there is some structure $C$ isomorphic to $B$ such that the domain of $...
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Prove the comparability theorem for well ordered sets using transfinite induction
My question is about the same proof discussed here, but I am confused about more than this question addresses.
In Naive Set Theory, Halmos phrases the comparability theorem as follows:
The ...
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Explicit well-ordering of $\mathbb{N}^{\mathbb{N}}$
Is there an explicit well-ordering of $\mathbb{N}^{\mathbb{N}}:=\{g:\mathbb{N}\rightarrow \mathbb{N}\}$?
I've been thinking about that for awhile but nothing is coming to my mind. My best idea is ...
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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 ...
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