# Tagged Questions

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### What do you need to know about ordinals to understand Zorn's Lemma's proof?

I'm trying to understand the proof of Zorn's Lemma but the one which does not use ordinals (Halmos' proof) is extremely long and I really feel I get lost somewhere along the way. On the other hand, ...
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### Proof that $\omega_1$ is countable (where is the flaw?)

I have discovered a "proof" of the fact, that $\omega_1$ can be countable, contradicting its definition. It mustn't assume axiom of choice, so I guess there are some parts of this construction which ...
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### AC iff $P(\delta)$ can be well-ordered for all $\delta\in{\bf On}$ [duplicate]

I'm trying to do the following exercise: Exercise. Assume ZF. Then AC is equivalent to the statement that $P(\delta)$ can be well-ordered for all $\delta\in{\bf On}$. I'm struggling with the ...
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### 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 ...
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### Does the set of all ordinals strictly dominated by a given set exist in ZF?

How do I prove that $$\{\alpha\in\mathsf{On}\,|\,\alpha\prec A\}\in V,$$ assuming $A\in V$? I know that if AC is assumed, this set is equal to $\mbox{card}(A):=\mu_\alpha(\alpha\approx A)$, and ...
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### Possible inaccuracy in Wikipedia article about initial ordinals

I quote from the Wikipedia article: "So (assuming the axiom of choice) we identify $\omega_\alpha$ with $\aleph_\alpha$, except that the notation $\aleph_\alpha$ is used for writing cardinals, and ...
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### Which set is unwell-orderable?

In textbook it says that every well-orderable set is equipotent to an initial ordinal number. However, of course unwell-orderable cannot equipotent to any ordinal number, but is there any set is ...
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### Why would the axiom of choice be needed if ordinals are well-ordered without AC?

Will ordinals be well-ordered without AC? This seems to be obviously true, as they are by definition well-ordered. Why would we then need the axiom of choice. We can just form a bijection function to ...
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### If $\kappa$ is a cardinal, $\aleph$ is any $\aleph$-number, and if $\kappa\leq\aleph$ then $\kappa$ can be well ordered as well.

I'm having trouble understanding the statement: If $\kappa$ is a cardinal, $\aleph$ is any $\aleph$-number, and if $\kappa\leq\aleph$ then $\kappa$ can be well ordered as well. I understand the ...
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### $\varepsilon$-number countability without choice

Let $\alpha\mapsto\varepsilon_\alpha$ be the enumeration of the $\varepsilon$-numbers--that is, those $\alpha$ such that $\omega^\alpha=\alpha$--by the ordinals. If we know that countable unions of ...
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### What is the smallest possible value of $\omega_1$ in $\mathrm{ZF}$?

It is consistent with $\mathrm{ZF}$ that a countable union of countable sets may be uncountable. As far as I understand it, this is because in absence of $\mathrm{AC}$ we cannot necessarily choose a ...
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### 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. ...
One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does not have cardinality $\aleph_2$. My solution shows that it does not have ...
### Non-existence of a surjection $\aleph_n \to \aleph_{n+1}$, without the axiom of choice
Firstly, let's establish what exactly I mean by these symbols. Let $\omega_0 = \{ 0, 1, 2, \ldots \}$, where $0, 1, 2, \ldots$ are the usual von Neumann representations of the natural numbers. Let $n$ ...