The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.
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Relationships between AC, Ultrafilter Lemma/BPIT, Non-measurable sets
How is it possible to reconcile the following...
In 1970, Solovay constructed Solovay's model, which shows that it is consistent with standard set theory, excluding uncountable choice, that all ...
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Bourbaki Proof of Zorn's Lemma in Lang's Algebra
Serge Lang in his book Algebra has a nice appendix on set theory at the end of the book. In particular, in paragraph 2, pp. 881-884 he provides a proof of Zorn's Lemma from "other properties of sets ...
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Prove that the statement implies the Axiom of Choice
Prove that the following statement implies the Axiom of Choice:
Let $ C $ is a set (of sets) and $ B $ is a set such that for all $ c \in C $, there exists a $ b \in B $ such that $ b \not\in c $. ...
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Defining cardinals without choice
According to Wikipedia if we assume AC we define a cardinals number to be an ordinal that is not in bijection with any smaller ordinal.
Without AC, one takes the cardinality of a set $X$ to be the ...
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2 questions about “$\mathbb R$ has a Hamel basis over $\mathbb Q$”
I would like to know if there are "interesting" equivalences to the statement in the title. I am not interested in more general statements, like "every vector space has a Hamel basis" or "every vector ...
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Axiom of Choice and Ascending Chain Condition
Matsumura, in his "Commutative Ring Theory" p. 14 proves that "A partially ordered set $\Gamma$ satisfies the ascending chain condition $\Leftrightarrow$ every nonempty subset of $\Gamma$ has a ...
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Axiom of Choice - Naive Counterexample [duplicate]
Possible Duplicate:
Axiom of choice question
I know there is a lot of discussion on the axiom of choice and, in fact, I attended once a lecture on it, but I still cannot understand the ...
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Cardinality of $\mathbb R\setminus\mathbb Q$ without AC [duplicate]
Possible Duplicate:
Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice?
Showing that $\mathbb{R}$ and $\mathbb{R}\backslash\mathbb{Q}$ are equinumerous using ...
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Problem 1.30 (The Maximum Principle is equivalent to the axiom of choice) (J.Bell, Boolean-valued Models and Independence Proofs, 3rd edition)
Problem 1.30 (The Maximum Principle is equivalent to the axiom of
choice)
(i) Let $\{a_i : i ∈ I\} ⊆ B$ satisfy $\bigvee_{i∈I} a_i = 1$. A partition of unity $\{b_i : i ∈ I\}$ in B is called a ...
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Does negation of Axiom of Choice imply symmetry?
It seems that every construction of a model in which the Axiom of Choice fails involves some kind of symmetry. Is there an example of a construction of a model where AC fails but no argument involving ...
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Theorem's relying on algebraic closures
When working with fields, it's a usual method to work on an algebraic closure of a field to obtain results about that field. In general (i. e. unless you're explicitly considering "well-behaved" ...
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What's the thorny issue on: “If all $S\in \ell $ are nonempty, does it follow that $\prod_{S\in \ell} S$ is nonempty? when $\ell$ is infinite?”
I'm reading Paolo Aluffi's ALGEBRA, Chapter 0.
Here he proposes that there's a thorny issue:
What is this thorny issue?
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Using Zorn's Lemma
Background: I am trying to use Zorn's lemma to show the existence of ultrafilters containing an arbitrary filter on a set $X$. My argument goes as follows:
Let $\mathcal{F}_0$ be a filter on $X$. If ...
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Simple and intuitive example for Zorns Lemma
Do you know any example that demonstrated Zorn's Lemma simple and intuitive? I know some applications of it, but in these proof I find the application of Zorn Lemma not very intuitive.
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What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that…
... $\aleph_1$ is the immediate successor of $\aleph_0$?
I was reading the wiki article on $\aleph_1$ where a distinction is made between the two. If there's isn't a cardinal between $\aleph_1$ and ...
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1answer
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Axiom of choice on function [duplicate]
Possible Duplicate:
Using a choice function to find an inverse for $F\colon A\to P(B)$
Let $F:A \rightarrow \mathcal P (B)$ be arbitary functions which covers $B$.
Use AC to show there is ...
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Question about trees and generalizing the Principle of Dependent Choices.
One form of the Principle of Dependent Choices is that for any tree $T$ of height $\omega$ such that every node of $T$ has a successor, there is a branch of $T$ of length $\omega$. In this post, I ...
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Questions about generalizations of the Principle of Dependent Choices
I've encountered two different generalizations of the Principle of Dependent Choices (DC) to initial ordinals $\kappa>\omega$. First, some notation. By $A\prec B$, I indicate that $A$ is injectable ...
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Proof that recursive construction is equivalent to dependent choice
I am trying to prove the following: In Zf, dependent choice (DC) and recursive construction (RCW) are equivalent where
and
(RCW):
Can you tell me if my proof in one direction is correct and ...
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Why can countable recursive constructions not be done using countable choice?
Why can countable recursive constructions not be done using countable choice? For example, replace $Z$ by $\omega$ in the following theorem
Why can't one prove it using countable choice? (The proof ...
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Proving Dedekind finite implies finite assuming countable choice
I'd like to show that if a set $X$ is Dedekind finite then is is finite if we assume $(AC)_{\aleph_0}$. As set $X$ is called Dedekind finite if the following equivalent conditions are satisfied: (a) ...
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Follow up on “Proof of $X \times X \hookrightarrow X$ implies $[X]^2 \hookrightarrow X$”
This is a follow up on this earlier question of mine.
We have the following statements:
(HSO) For every infinite set $X$ there exists an injection $f: X \times X \hookrightarrow X$
(HSU) For every ...
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Countable unions of countable sets
It seems the axiom of choice is needed to prove, over ZF set theory, that a countable union of countable sets is countable.
Suppose we don't assume any form of choice, and stick to ZF. What are the ...
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Is $\kappa^\lambda=2^\lambda$($2 \le \kappa<\lambda$,$\lambda$ infinite) valid in set models of ZF?
Let $2 \le \kappa<\lambda$(both cardinal numbers), in which $\lambda$ is infinite. Then these formula as follows hold where in ZFC:
$\lambda+\kappa=\lambda$
$\lambda\cdot\kappa=\lambda$
...
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1answer
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$A+\alpha\sim A$ when $\omega\le\alpha<h(A)$
I have been considering about $A+\alpha\sim A$ when $\omega\le\alpha<h(A)$, in which $h(A)$ is the Hartogs number of $A$.
If there is $\alpha$ s.t. $\omega\le\alpha<h(A)$, we can get $\alpha ...
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Is there any Dedekind-infinite set can be split to two smaller Dedekind-infinite sets?
I have been considering about $A+\alpha\sim A$ when $\omega\le\alpha<h(A)$, in which $h(A)$ is the Hartogs number of $A$.
If there is $\alpha$ s.t. $\omega\le\alpha<h(A)$, we can get $\alpha ...
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1answer
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Proof of $X \times X \hookrightarrow X$ implies $[X]^2 \hookrightarrow X$
Consider the following two statements (where $[X]^2$ denotes the set of all unordered two-element subsets of $X$):
(HSO) For every infinite set $X$ there exists an injection $f: X \times X ...
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Proving equivalences of statements equivalent to AC
I'm doing the following exercise from Just/Weese:
Show in ZF that (WO) implies (IC) and that (IC) implies (SC).
where
(WO) Every set can be well-ordered.
(IC) For any two sets $X,Y$ either there ...
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Why is AC needed for $|\bigcup X_i|=|\bigcup Y_i|$, $\forall i$ $|X_i|=|Y_i|$, $\{X_i\}_{ i\in I}$, $\{Y_i\}_{i\in I}$ pairwise disjoint?
On Page 60, Set Theory, Jech(2006),
5.9 If $\{X_i : i \in I\}$ and $\{Y_i : i \in I\}$ are two disjoint families such that $|X_i| = |Y_i|$ for each $i \in I$, then $|\cup_{i \in I}X_i| = |\cup_{i ...
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Statement not provable from ZF
I'm doing the following exercise from Just/Weese:
Some thoughts: To show that the statement is not provable from $ZF$ I could either show that it implies the axiom of choice or Tychonoff or I could ...
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Using a choice function to find an inverse for $F\colon A\to P(B)$
Let $A$ and $B$ be arbitrary non-empty sets and let $F\colon A\to P(B)$, be an arbitrary function which covers $B$ in the sense that $\forall b \in B$, $\exists a \in A$ such that $b \in F(a)$ holds.
...
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1answer
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Is $\aleph_0$ the minimum infinite cardinal number in $ZF$?
$\aleph_0$ is the least infinite cardinal number in ZFC. However, without AC, not every set is well-ordered.
So is it consistent that a set is infinite but not $\ge \aleph_0$? In other words, is it ...
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What fragment of ZFC do we need to prove Zorn's lemma?
It is extremely well-known that Zorn's lemma is a theorem of ZFC. My interest is in a certain finitely-axiomatisable fragment of ZFC, sometimes called RZC (restricted Zermelo with choice) or ZBQC. The ...
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Is this theory $\kappa$-categorical when $\kappa$ is infinite and not $\le \aleph_0$?
Let $\mathcal L$ be a language with only a binary predicate $E$, and let $T$ be a theory of structures in which $E$ is an equivalence relation which partitions the structure into two infinite ...
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Confused about why “disjointifying” implies “AC”
Assume I have the following
(DIS) For every indexed family $\{A_i : i \in I \}$ there exists a family $\{B_i : i \in I \}$ of pairwise disjoint sets such that $B_i \subset A_i$ for all $i \in I$ and ...
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1answer
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Always win without a winning strategy
On Page 141, Axiom of Choice, Herrlich(2006)
Show that if in a game of the form $G(1, X_1, Y_1, A)$, the first player
has no winning strategy, then the second player can always win, even
...
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1answer
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Making a choice function if $A_i$ are well-ordered for each $i$
Let $A_i$ be a family of sets such that each $A_i$ is well-ordered. Let $\varphi(x,S,W)$ be the formula $$ \forall z ( (z,x) \in W \rightarrow z \notin S)$$
where $W$ is the well-order on $S$. Then ...
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2answers
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Is the negation of Gödel's completeness theorem consistent with $ZF$ without AC?
The proof of compactness and completeness of $\mathscr{FOL}$ (with Hilbert system) used Zorn's lemma. And Zorn's lemma is equivalent to the Axiom of choice in $ZF$.
So my question is can they be ...
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Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition.
Let $\Bbb R^+$ be the set of positive real numbers. Use Zorn's Lemma to show that $\Bbb R^+$ is the union of two disjoint, non-empty subsets, each closed under addition.
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Countable infinity and the axiom of choice
Many of the states of affairs about infinite cardinalities and their size, depend on the axiom of choice. What would a comprehensive list be of the properties of the cardinality of the natural numbers ...
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(ZF) Sequential Continuity of Derivatives and Darboux
Let $C$ be a connected set of reals and $f:C\rightarrow \mathbb{R}$ be differentiable on $C$.
It can be proven that $f$ is sequentially continuous on $C$ iff $f$ is continuous on $C$.
First, i want ...
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Convex and continuity (ZF)
Let $f:(a,b)\rightarrow \mathbb{R}$ be a continuous function.
Suppose $\exists k\in (0,1)$ such that $\forall x,y\in (a,b), f(kx+(1-k)y)≦kf(x)+(1-k)f(y)$.
Let $A=\{\lambda\in [0,1]|\forall x,y\in ...
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Infinite Dedekind Finite sets
I realized that i have used argument below many times before and I'm not sure if it is true.
Let $A=\{n\in \omega|\Phi(n)\}$.
Then $A\preceq \aleph_0$.
(i)Suppose $A$ is dedekind-infinite and find ...
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1answer
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Does dense subset in a polish space contain a countable subset which is dense in the space?
Let $X$ be a separable complete metric space.
Let $E$ be a dense subset of $X$ and fix $p\in X$.
I want to 'choose' an element for each $\overline{B(p,1/n)}\cap E$ in ZF. ($n\in \mathbb{Z}^+$)
Is it ...
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Need for axiom of choice?
When you invoke the axiom of choice, it is because whatever you need to choose consists of small parts you technically have to choose one by one. So if I need to use the axiom of choice, but ...
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Hall's theorem vs Axiom of Choice?
From Wikipedia
Let $S$ be a family of finite sets, where the family may contain an
infinite number of sets and the individual sets may be repeated
multiple times.
A transversal for $S$ is ...
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4answers
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Which kind product of non-zero number non-zero cardinal numbers yields zero?
Let $I$ be a non-empty set. $\kappa_i$ is non-zero cardinal number for all $i \in I$.
If without AC, then $\prod_{i \in I}\kappa_i=0$ seems can be true(despite I still cannot believe it).
But what ...
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1answer
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Can an infinite cardinal number be a sum of two smaller cardinal number?
Let $\kappa$ be an infinite cardinal number.
My question is whether there are $\lambda$ and $\mu$ such that both $<\kappa$ but $\lambda+\mu=\kappa$?
If AC holds, then the answer is definitely ...
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1answer
102 views
Does Axiom of Choice naturally hold?
Let $A$ be a set of non-empty sets, then $\bigcup A$ is a set. Furthermore, $(\bigcup A)^{A}$ is a non-empty set. Besides let $P$ be a binary predicate such that for all $X\in A$ there is a unique $x ...
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Connected subset of a separable metric space is separable?
Continuity and the Axiom of Choice
I have proved a small generalization of Brian's argument, that is, "If $f:X\rightarrow Y$ is sequentially continuous on $X$ and $X$ is separable, then $f$ is ...
