Tagged Questions
2
votes
2answers
131 views
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 ...
2
votes
2answers
104 views
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 ...
3
votes
3answers
215 views
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 ...
2
votes
2answers
140 views
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 ...
0
votes
1answer
133 views
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 ...
7
votes
2answers
182 views
$\varepsilon$-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 ...
4
votes
1answer
138 views
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 ...
8
votes
4answers
197 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.
...
12
votes
1answer
2k views
The cardinality of a countable union of countable sets, without the axiom of choice
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 ...
5
votes
2answers
722 views
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 $n$ ...
