All Questions
Tagged with cardinals axiom-of-choice
183
questions
2
votes
1
answer
308
views
Dependence of the equation $a+1=a$ for infinite cardinals $a$ on the axiom of choice
let $A$ be a set such that for all $n \in $ N $ A ≉ N_n$
where $N_n = \{ 0 ,1 ,2 ...... n-1\} $
and $a$ be the Cardinality of $A$ meaning ($|A| = a$)
is it possible to prove that $a+1=a$ without ...
- 81
0
votes
1
answer
87
views
Is there a way to construct larger cardinals without choice axiom?
From Cantor's Theorem, we know that $|\mathcal{P}(X)| > |X|$. So, we can define inductively a set with cardinality $\aleph_n, \forall n \in \mathbb{N}$. Let $\lbrace A_i\rbrace_{i \in \mathbb{N}}$ ...
- 31
17
votes
2
answers
402
views
Combinatorial proof, without axiom of choice, that for any set $A$, there is no surjection from $A^2$ to $3^A$
The well known proof of Cantor's theorem (stating that $A<2^A$ for any set $A$) does not make any use of the axiom of choice. I have now spent some time wondering if the analogous result $A^2<3^...
- 2,243
0
votes
0
answers
71
views
Does the proof that "a proper ascending chain of subsets of the naturals is countable" necessitate axiom of choice?
Let $(A_x)_{x\in X}$ be a collection of disjoint subsets of $\mathbb{N}$. Using the Axiom of Choice, i.e. assuming there exists a function $f:\{A_x\}_{x\in X}\to \bigcup_{x\in X} A_x$, with $f(A_x)\in ...
4
votes
1
answer
153
views
There is no surjective function from a set in the Hartogs number of its power set
I'm trying to prove an equivalent state of the Axiom of Choice :
Given two non-empty sets, there is a surjective function from one of them into the other one.
If we prove that for any non-empty set $X$...
- 191
5
votes
2
answers
216
views
When can you subtract cardinals? Does $|A|+|C|=|B|+|C|>|C|$ imply $|A|=|B|$ in ZF?
Suppose that $|A|+|C|=|B|+|C|$, and that both sides are strictly larger than $|C|$. Does this imply $|A|=|B|$ in the absence of the axiom of choice?
With choice, cardinals are totally ordered, and $\...
- 23.5k
0
votes
1
answer
32
views
Reordering a Sequence of Sets Whose Union is the Whole Set
I have a set $ B $ that can be written as $ B = \cup_{\nu < \lambda} B_{\nu} $ where $ \kappa $ is the cardinality of $ B $, that is uncountable, and $ \vert B_{\nu} \vert < \kappa $ with $ \...
- 51
2
votes
1
answer
156
views
Does Cantor's paradox require the axiom of choice?
Wikipedia's article about Cantor's paradox claims that it requires that the cardinals are linearly ordered. But can't we show that the cardinality of the universal set is the greatest cardinality ...
- 1,049
0
votes
0
answers
41
views
Are cardinals always partial-well-ordered without AC? [duplicate]
With AC, we have that cardinals are well-ordered. Without AC, we can have two sets which are incomparable, meaning that, at best, cardinals are only partially ordered.
But do we still have, without AC,...
- 8,000
1
vote
1
answer
116
views
Showing cardinality inequalities involving equivalence relations without the Axiom of Choice
I am studying a course on ZF Set Theory and am currently looking at the cardinalities of infinite sets.
Consider any equivalence relation $\equiv$ on any set $X$. Show that $$2 ^{ |X/{\equiv}|} \leq ...
- 4,331
13
votes
0
answers
256
views
Conditions needed to prove $|A|^{|P(A)|}=|P(P(A))|$
I'm trying to figure out if ${|A|}^{|P(A)|}=|P(P(A))|$ (where $A$ is infinite) is provable without the Axiom of Choice.
I know unconditionally we have the lower bound: $|A|^{|P(A)|}\geq 2^{|P(A)|}=|P(...
- 855
2
votes
2
answers
311
views
For infinite cardinal $\kappa$, $\kappa$ . $\kappa$ = $\kappa$ . Set Theory Enderton p-162, Lemma 6R
Enderton's proof goes like below;
Let B be any infinite set of cardinality $\kappa$.
Let $H=${$f|f=$$\emptyset$ or $f: A×A-A$ is a bijection for some A $\subseteq B$}
Then he showed that any chain in ...
user1072974
19
votes
1
answer
645
views
Can free groups on different cardinals be isomorphic in ZF?
For the free group on a finite number of generators, $F_n$, this is simple. It is enough to find a group that can be a quotient of one and not the other, so it is sufficient to find a group that can ...
- 5,633
21
votes
1
answer
529
views
When does $|X^n|\leq |k^X|$ without the axiom of Choice?
We're working in $\sf{ZF}$ set theory, we assume $n$ is a finite cardinal, and the set $X$ is not finite. My question is: what's the smallest cardinal $k$ for which $\sf{ZF}$ proves $|X^n| \leq |k^X|$?...
- 2,119
6
votes
1
answer
105
views
What's the groupification of the cardinal numbers (without using AC)?
The set-theoretic cardinal numbers form an abelian monoid under addition (ignoring size problems). Under the axiom of choice, given two cardinal numbers $c$ and $d$, there exists a cardinal number $e$ ...
- 8,215