Tagged Questions

10
votes
2answers
102 views

The relationship of ${\frak m+m=m}$ to AC

Two simple questions: (Of course ${\frak m}$ denotes a cardinal in the weak sense in the claims below.) Can we prove in ZF that $\aleph_0\le{\frak m\Rightarrow m+m=m}$? If not, what is the ...
3
votes
1answer
64 views

(Non) equivalence of regular cardinal definitions

The usual definition of a regular cardinal is "$\kappa$ is regular if $cf(\kappa) = \kappa$", which, assuming the axiom of choice, is equivalent to this definition: "$\kappa$ is regular iff it cannot ...
6
votes
2answers
57 views

Zorn's lemma and maximal linearly ordered subsets

Let $T$ be a partially ordered set. We say $T$ is a tree if $\forall t\in T$ $\{r\in T\mid r < t\}$ is linearly ordered (such orders can be considered on connected graphs without cycles, i.e. on ...
1
vote
1answer
57 views

Jech: Set Theory exercise 3.13, how do I avoid Choice?

In an effort to finally learn set theory rigourously, I've decided to start plowing through Jech's Set theory, making sure to do each of the exercises. Here is Jech's problem 3.13 (pg. 34 in the ...
2
votes
1answer
63 views

Can we write every uncountable set $U$ as $V∪W$, where $V$ and $W$ are disjoint uncountable subsets of $U$? [duplicate]

Is it true that for every uncountable set $U$, we can write $U=V∪W$, where $V$ and $W$ are disjoint uncountable subsets of $U$ ?
10
votes
6answers
203 views

Can every infinite set be divided into pairwise disjoint subsets of size $n\in\mathbb{N}$?

Let $S$ be an infinite set and $n$ be a natural number. Does there exist partition of $S$ in which each subset has size $n$? This is pretty easy to do for countable sets. Is it true for ...
3
votes
1answer
46 views

Dissecting a proof of the $\Delta$-system lemma (part II)

This is part II of this question I asked yesterday. In the link you can find a proof of the $\Delta$-system lemma. In case 1 it uses the axiom of choice (correct me if I'm wrong). Now one can also ...
2
votes
2answers
99 views

Isomorphic Free Groups and the Axiom of Choice

When I read about free group, the proof which concerns about two free groups $F(X)$ and $F(Y)$ are isomorphic only if $\operatorname{card}(X) = \operatorname{card}(Y)$ has a sentence going as follows: ...
4
votes
1answer
54 views

Dependent choice and Zorn's Lemma

How much of Zorn's lemma can be saved if we assume only ZF+DC without full choice? More precisely: assume we have a partially ordered (inductive) set which is of size continuum. Then can we apply ...
2
votes
2answers
39 views

Mapping on cardinal without Axiom of Choice

Define $|A|\le|B|$ iff there exists injective mapping $A \to B$. If Axiom of Choice is assumed then this is equivalent as $|A|\le|B|$ iff there exists surjective mapping $B \to A$. But: If Axiom of ...
9
votes
1answer
105 views

Does a nonlinear additive function on R imply a Hamel basis of R?

A function is additive if $f(x+y) = f(x) + f(y)$. Intuitively, it might seem that an additive function from R to R must be linear, specifically of the form $f(x) = kx$. But assuming the axiom of ...
5
votes
2answers
171 views

Intuition behind the Axiom of Choice

Why is it different to make one choice or many choices than to make infinite choices from a theoretical point of view in which indeed you are not going to do any? How could that be different from ...
4
votes
3answers
161 views

Why the need of Axiom of Countable Choice?

Two theorems: $(1)$ Countable Union of Countable Sets is Countable $(2)$ Cartesian Product of Countable Sets is Countable Linked are the formal proofs on Proofwiki. I do not understand why they ...
2
votes
3answers
81 views

$\mathbb{R^+}$ is the disjoint union of two nonempty sets, each closed under addition.

I saw Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition. and have a question related to the answer (I'm not sure if this is the right place to post ...
4
votes
2answers
59 views

Are there versions of the axiom of choice that restrict the size of the factors?

One formulation of the axiom of choice is that an arbitrary product of nonempty sets must be nonempty. The axiom of countable choice AC$_\omega$ is known to be strictly weaker than AC, but still ...
8
votes
1answer
137 views

Stone's Representation Theorem and The Compactness Theorem

If you're working on $ZF$ and you assume the compactness theorem for propositional logic, then you have the prime ideal theorem, and thus you can show that the dual of the category of Boolean algebras ...
3
votes
1answer
60 views

well-ordering principle

I'm confused by the well-ordering principle . The proof is clear but I don't have any idea why is it true . It says that every non-empty set can be well-ordered but $C^{1}$ is a non-empty set but ...
6
votes
3answers
101 views

Proof of $CFE \implies BPI$

(CFE): Every filter of closed sets can be extended to a maximal one. (BPI): Every Boolean algebra contains a prime ideal. I am reading Herrlich's and Stepran's paper "Maximal filters, continuity and ...
1
vote
2answers
83 views

Question about proof in Jech's The axiom of choice ($AC_\omega$)

I'm looking at the following in Jech's The Axiom of Choice on page 20: 2.4.1. Example: The Countable Axiom of Choice implies that every infinite set has a countable subset. Proof. Let $S$ be ...
6
votes
1answer
85 views

How to prove that a regular cardinal cannot be expressed as a union of sets with less cardinality?

My question is about the following: Using the Axiom of Choice show that: If $\kappa\ge\omega$ is a regular cardinal, $\gamma\le\kappa$, and $\langle A_\alpha\mid\alpha\lt\gamma\rangle$ is a ...
9
votes
1answer
81 views

Proof of a basic $AC_\omega$ equivalence

On Wikipedia it is mentioned that "... in order to prove that every accumulation point $x$ of a set $S\subseteq \mathbf R$ is the limit of some sequence of elements of $S\setminus \{x\}$, one uses (a ...
12
votes
3answers
156 views

Axiom of Choice and Determinacy

In my set theory course we have been talking about the axiom of determinacy. One of the first things we showed was that $AD$ and $AC$ are incompatible. We later showed that $ZF+AD$ implies the ...
8
votes
1answer
106 views

Failure of Choice only for sets above a certain rank

Let $\alpha$ be an ordinal. How can we show that the following theory is consistent? $\mathrm{ZF}$ + "there exists a set with rank greater than $\alpha$ that is not well ordered" + "every set of rank ...
1
vote
1answer
65 views

Axiom of choice , Hartogs ordinals, well-ordering principle

I'm trying to prove the following: If it holds that if for any two sets $A$ and $B$, $A$ can be injected into $B$ or $B$ can be injected into $A$, then every set can be well-ordered (axiom of choice ...
3
votes
1answer
67 views

A union represented as a disjoint union: weaker than choice?

While looking at this problem, I was thinking about the more general statement: For all sets $I,X$ and $U:I\to \mathcal P(X)$, the power set of $X$, there exists a $V:I\to \mathcal P(X)$ with the ...
2
votes
1answer
45 views

Proving that a dense set in a poset by Zorn's lemma have a maximal antichain as a subset

So I am little confused here; Zorn's lemma says that if every chain has a supremum in a poset one defines, then the poset must have a maximal element. My question is, how does this lead to the proof ...
8
votes
1answer
99 views

Injection of union into disjoint union

Given a family of sets $(A_i : i \in I)$, we define the disjoint union: $$\sum_{i \in I} A_i = \bigcup_{i \in I} (\{i\} \times A_i).$$ There is a surjection $\sum_{i \in I}A_i \to \bigcup_{i \in I} ...
4
votes
1answer
71 views

Clarification of a proof in Herrlich

In Herrlich on page 5 he gives a proof of $\textbf{AC} \implies \textbf{WOT}$: He does not give a definition of cardinality $|X|$ before this proof and I searched the index for a definition but ...
12
votes
0answers
101 views

Is Dover publishing Moore's book on the Axiom of Choice? [closed]

Dover is publishing a paperback edition of Gregory H. Moore's Zermelo's Axiom of Choice: Its Origins, Development, and Influence. It's supposed to come out March 20th and is available for pre-order at ...
3
votes
2answers
79 views

Application of Zorn's Lemma

I have the following problem: Given $\{A_{\lambda}\}_{\lambda \in \Lambda}$ for some index set $\Lambda$. Prove that there exists $\{B_{\lambda}\}_{\lambda \in \Lambda}$ such that: 1) $(\forall ...
3
votes
1answer
83 views

Is every discrete topological space orderable?

I apologize for asking a question in topological terms when it's not really about topology, but here goes: If $S$ is a set, is it always possible to totally order $S$ so that every non-minimal ...
6
votes
1answer
76 views

Prove the existence of a set in the Euclidean plane

I got stuck on the following problem. Prove that there exists a subset $A$ of $\mathbb{R}^2$ such that every line in $\mathbb{R}^2$ goes exactly through two points in $A$. I know that I should apply ...
5
votes
2answers
138 views

Does linear ordering need the Axiom of Choice?

Consider this statement: "every set can be linearly ordered." Can we prove it without AC?
2
votes
1answer
66 views

From Hausdorff maximal principle to well-ordering principle

How to prove the well-ordering principle if we have known the Hausdorff maximal principle?
7
votes
1answer
170 views

Pathological linear functionals and ZF

Let $S$ be an infinite set. Let $C(S)$ be the vector space of all functions $S \to \mathbb{R}$, and let $C_c(S)$ be the subspace of functions of finite support. Is the existence of a nonzero linear ...
2
votes
1answer
64 views

Where in this argument ultrafilter is used?

http://en.m.wikipedia.org/wiki/Dimension_theorem#section_1 Let's first not assume any choice principle. Let $V$ be a vector space over a field $F$ and $\beta_1,\beta_2$ be bases for $V$. Suppose ...
3
votes
1answer
84 views

Tychonoff Theorem in the Realm of $\neg AC$

It's widely know that the Tychonoff Theorem is equivalent to the Axiom of Choice; thus, assuming the negation of the axiom of choice, I'd like to know if there is a canonical example of a collection ...
8
votes
4answers
406 views

Why isn't this a valid argument to the “proof” of the Axiom of Countable Choice?

I am having a little trouble identifying the problem with this argument: Let $\{A_1, A_2, \ldots, A_n, \ldots\}$ be a sequence of sets. Let $X:= \{n \in \mathbb{N} : $ there is an element of the set ...
1
vote
1answer
55 views

Finite character, monotonic functions, partial functions

Could you explain to me why family of partial functions and family of monotonic functions are of finite character? I'm asking this because I'm currently reading a proof of a theorem concerning a ...
8
votes
1answer
243 views

Do you need the Axiom of Choice to accept Cantor's Diagonal Proof?

Math people: It is my understanding that set theorists/logicians compare cardinalities of sets using injections rather than surjections. Wikipedia defines countable sets in terms of injections. ...
1
vote
2answers
138 views

Mathematical questions whose answer depends on the Axiom of Choice

Question inspired by the following surprising claim: The chromatic number of the $R^n$ hyperplane may depend on whether the Axiom of Choice is available or not. http://shelah.logic.at/files/E33.pdf ...
4
votes
2answers
98 views

Martin's Axiom and Choice principles

My main questions are: (1) Does MA have any relation to choice principles such as AC, PIT/UL, DC, AD, etc.? (2) In particular, does MA($\kappa$) it imply AC for collections of cardinality $\kappa$? ...
6
votes
1answer
126 views

Is axiom of choice required for there to be an infinite linearly independent set in a (non-finite-dimensional) vector space?

In discussing this answer, I noted that while the statement: Any vector space has a basis is equivalent to the axiom of choice, I wondered if the statement that: Any vector space either has ...
4
votes
1answer
116 views

$\bf AD$ implies $\bf AC_{\omega}(\mathbb{R})$?

How to show that $\bf AD$ implies $\bf AC_{\omega}(\mathbb{R})$? $\bf AD$ is abbreviated for axiom of determinacy. $\bf AC_{\omega}(\mathbb{R})$ states that for each family $(X_i)_{i∈\omega}$, in ...
3
votes
2answers
118 views

is equipotence between Cantor set and $[0,1]$ due to AC?

I only known a surjection from Cantor to $[0,1]$, but this thing only means that Cantor is equipotent whit [0,1] when AC is true. What if ¬AC?
2
votes
1answer
81 views

Axiom of choice and the number of choice functions

The axiom of choice guarantees that every collection of disjoint non-empty sets has at least one choice function. Intuitively, a stronger result holds. We expect that the number of choice functions ...
3
votes
1answer
91 views

Help with a proof of Zorn's lemma

I am working on a proof of Zorn's lemma using AC, and need some help. I'll skectch main steps, and describe where I am stuck. Let $(A,<)$ be partially ordered set without any maximal elements, ...
5
votes
1answer
83 views

Cardinality of set of well-orderable subsets of a non-well-orderable set

Suppose, with the negation of Axiom of Choice, we have a non-well-orderable set $A$, and its power set $P(A)$, let $P'(A)$ be $\{x \in P(A): x \text{ is well orderable}\}$ Is there an injection from ...
5
votes
1answer
89 views

Equivalent of the countable axiom of choice?

Given $\{A_n\}_{n \in \omega}$ a countable collection of nonempty sets, there is a function $f$ with domain $\omega$ and $f(n) \in A_n$ for each $n \in \omega$. Is it the case that it is equivalent to ...
4
votes
3answers
135 views

Equivalence relations and the axiom of choice

Let $X,Y$ be two sets and $f:X\rightarrow Y$ a function. If we define an equivalence relation $\sim$ on $X$ and we assume that if $x_1\sim x_2$ then $f(x_1)=f(x_2)$, we can define a function ...

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