2
votes
2answers
23 views

Proof of: $X$ is finite $\iff X$ is Tarski-finite

I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876). In the fourth chapter, he deals with different definitions of finite set. Here is the classical one: ...
2
votes
2answers
102 views

Is the Axiom of Choice necessary to prove $\mathbb R \approx \mathcal P(\omega)$?

I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876), and I'm getting quite confused about cardinal arithmetic without AC. Here (Which sets are well-orderable ...
1
vote
1answer
36 views

Why do we need the axiom of choice in showing the non-emptiness of an infinite Cartesian product

Given $I$ a set of indexes and $X_i$ a set of topological spaces, define The Cartesian product: $\prod_{i \in I}X_i = \{ f:I \rightarrow \bigcup X_i | f(i) \in X_i \}$ I have read that we need ...
2
votes
0answers
32 views

Is the Axiom of Choice restricted to $\Bbb R$ independent of ZF? [duplicate]

This is really a yes-no question, and I am hundred percent certain that the answer is "yes". I simply have not found it written directly anywhere. True, the Axiom of Choice is independent of ZF if we ...
0
votes
2answers
41 views

Bernstein sets, Well-Ordering theorem vs Axiom of Choice

In the construction of Bernstein sets (see here), is it necessary to use the well-ordering theorem? Why can't you just use the Axiom of Choice to pick two points?
1
vote
1answer
50 views

Proving Equivalence of Two Version of Axiom of Choice

I am working on an assignment that requires proving the equivalence of two versions of the axiom of choice. (1st form): For any relation $R$, there is a function $H \subseteq R$ with dom $H =$ dom ...
0
votes
2answers
58 views

Theorem 4.15 page 22, Real and Abstract Analysis, Hewitt and Stromberg

Following is Theorem 4.15 from Real and Abstract Analysis, Hewitt and Stromberg Theorem: Every infinite set has a countably infinite subset. Proof: Let $A$ be a infinite set. We show by induction ...
1
vote
1answer
39 views

factorial of infinite Cardinals

Let $S_A$ be set of all bijections over $A$ such that $Card(A)=\kappa$. Define foctorial as $\kappa!:=Card(S_A)$. Show that if $\kappa$ is infinite, then : $\kappa!=2^\kappa$ First, I've ...
2
votes
1answer
36 views

If $|X|<|Y|$ then $|Y|=|Y-X|$ (with $Y$ infinite)

Like the title says, I would like to prove that if $|X|<|Y|$ then $|Y|=|Y-X|$. (with $Y$ infinite) I know I have to use the axiom of choice, but I've no idea about how to proceed. Any help is ...
2
votes
1answer
14 views

If $A$ is D-Infinite then $|P_{\infty}(A)|=|P(A)|$

I want to prove that if $A$ is a D-Infinite set (i.e. it contains a countable subset $X$), then the set of the infinite parts of $A$, $P_{\infty}(A)$ has the same cardinality of $P(A)$. I know that ...
2
votes
1answer
54 views

Zorich's misinterpretation of “Axiom of Choice”?

I'm reading Zorich'es "Mathematical Analysis I", Ed 4, 2004, and wonder if this is a trifle misinterpretation of "Axiom of Choice". Ch 1.4 "Supplementary Material" says: 8°. (A x i o m o f c h o i ...
1
vote
1answer
25 views

What is wrong with the following proof saying Zorn's lemma implies Hausdorff maximum principle?

I am reading 'Topology' by J.R. Munkres's first chapter on set theory. In the exercises 5-7 on page 72 he asks the reader to show that Zorn's lemma implies Hausdorff maximum principle via the ...
0
votes
2answers
35 views

Does defining the closure of a set as the intersection of all closed set that contain it requires the axiom of choice?

Given a set $S$, the closure of $S$ is sometimes defined as the intersection of all the closed sets that contain it. This type of argument is pervasive in mathematics when one want to construct the ...
1
vote
2answers
88 views

Confusion regarding one formulation of the Axiom of Choice.

One formulation of the Axiom of Choice is: The Cartesian product of non-empty sets is always non-empty. Cartesian product is defined as making "every possible pair" between elements of two sets. ...
59
votes
6answers
5k views

Why can't you pick socks using coin flips?

I'm teaching myself axiomatic set theory and I'm having some trouble getting my head around the axiom of choice. I (think I) understand what the axiom says, but I don't get why it is so 'contentious', ...
4
votes
1answer
31 views

Do we need AC to prove Principle of Dependent Choices

For any nonempty set $X$ and any entire binary relation $R$ on $X$, there is a sequence $(x_n)$ in $X$ such that $x_nRx_{n+1}$ for each $n \in \mathbb{N}$. (Here an entire binary relation on $X$ is ...
3
votes
2answers
62 views

Assuming the axiom of choice, how to find explicit group structure of a given set

Let us assume the axiom of choice. This is equivalent to every nonempty set having group structure. My question is, given some nonempty set, can we define the binary operator in a constructive way ...
2
votes
1answer
55 views

Cardinality of union of pairwise disjoint elements needs choice?

If there is an indexed family $(i\mapsto A_i)_{i\in I}$ of pairwise disjoint sets $A_i$, why do we need choice to show that $$ \left|\textstyle{\bigcup_{i\in I}A_i}\right| = \sum_{i\in I}|A_i|? $$ It ...
1
vote
1answer
40 views

Zorn's lemma problem

Let $n$ be a positive natural number. Prove using Zorn's lemma that there is a set A of points in the plane that satisfies: 1. Any line in the plane does not contain $n+1$ points of A. 2. For every ...
3
votes
1answer
68 views

Jech's axiom of choice, problem 2 first chapter

I have a question stemming from Jech's book on the axiom of choice., chapter 1 exercise 2. We are asked to show that a family of sets of natural numbers has a choice function. Now the version of the ...
7
votes
2answers
171 views

Plausibility argument for Zorn's Lemma

In "Mathematical Physics" by Robert Geroch, the following 'plausibility argument' is given for Zorn's Lemma [If every totally ordered subset of a partially ordered set $S$ is bounded above, $S$ has a ...
8
votes
2answers
86 views

How should one think about results that depend on AC?

I just encountered this: "(Theorem of A. H. Stone) Every metric space is paracompact... Existing proofs of this require the axiom of choice... It has been shown that neither ZF theory nor ZF ...
1
vote
2answers
149 views

The Axiom of choice

I'm a little lost with this proof: If every set is equipotent to an ordinal, then we have the axiom of choice And I want to know if someone can help or maybe give me a hint of how to proceed.
1
vote
1answer
71 views

$\aleph(X) < \aleph(\mathcal{P}(\mathcal{P}(\mathcal{P}(X))))$

Prove: $\aleph(X) < \aleph(\mathcal{P}(\mathcal{P}(\mathcal{P}(X))))$ With $W(X)=\{\langle A,R\rangle: A \in \mathcal{P} (X),R \in \mathcal{P}(X \times X)$ and $ R $ wellorders $ A \}$ And ...
1
vote
1answer
48 views

What does the Axiom of Choice have to do with right inversibility?

I have encountered an exercise that asks to prove that, these two statements are equivalent: every surjective function has a right inverse. Axiom of choice. Definition: Given a function $f$, we ...
2
votes
1answer
48 views

well ordering principle implies Zorn's Lemma

In here: Proving that well ordering principle implies Zorn's Lemma. I asked how to finish a proof of this statement. After a few helpful remarks, I think I have managed to finish it. What do you ...
4
votes
1answer
104 views

Proving that well ordering principle implies Zorn's Lemma.

I am trying to prove that well ordering principle implies Zorn's Lemma. I think that I'm close but don't quite know to make the last step of my proof. Here is what I wrote so far: Given that on every ...
3
votes
1answer
59 views

The union of a countable set of countable sets is countable

Here is the proof provided in my lecture notes: Let $A = \{B_n | n < \omega =\mathbb{N}\}.$ Assume each $B_n$ is countable. For each $n < \omega,$ let $E_n$ be set of all bijections between ...
0
votes
2answers
72 views

Do we need the axiom of choice in here?

Axiom of choice: Given $\mathbb{F}$ is a set of non-empty sets. Then, there is a function $f$ with $\text{Dom}(f)=\mathbb{F}$ such that, for every $A \in \mathbb{F}$, $f(A) \in A$. The ...
3
votes
2answers
98 views

Proving Zermelo's Theorem implies the Axiom of Choice

I thought it would be fun to try and prove this. It turned out to be pretty simply and maybe too simple so I was wondering if anybody could verify if this proof is correct. Suppose we have a family ...
1
vote
2answers
61 views

How do i formally write down a countable choice function?

Let $A$ be an infinite set. Then, we can construct an injective function $f:\omega \rightarrow A$. But how do i construct this via orginal statement of $AC_\omega$? (i.e. $\forall countable X, ...
2
votes
1answer
124 views

Logic: Teichmüller-Tukey Lemma and the Axiom of Choice

How can you proof that the Teichmüller-Tukey Lemma (which says that if $S$ is nonempty and of finite character, $S$ contains a maximal element with respect to the subset ordering), implies the Axiom ...
1
vote
2answers
89 views

Logic: on the Axiom of Choice

Let $X,Y,Z$ be infinite sets and $f:X \rightarrow Y$ be a surjective function. How can I prove that if $|Y| \le |Z|$ and for every $y \in Y$ is $|f^{-1}(y)| \le |Z|$, the following inequality holds: ...
2
votes
1answer
30 views

How can one tell if a sequence is well-defined; is the axiom of choice needed?

This question is in the context of the following exercise: Let $X$ be a first countable topological space, let $A \subseteq X$, and let $x \in X$ with $x \in \overline{A}$. Then there exists a ...
2
votes
1answer
105 views

(Revisited$_2$) Injectivity Relies on The Existence of an Onto Function Mapping Back to Its Preimage

QUEST: For any sets $X$ and $Y$, there exists an injective function $f:X\rightarrow Y$ if and only if there exists a surjective function $g:Y\rightarrow X$. QUESTION$_1$: How do you people ...
-2
votes
1answer
90 views

Axiom of Choice: An Invocation Necessary for A Proof on Surjectivity [closed]

Prove that if $f\colon X\rightarrow Y$ is surjective, then there must exist a function $g\colon Y\rightarrow X$ such that $f\circ g=1_Y$, where $1_Y$ is the identity map on $Y$.
-1
votes
2answers
183 views

The intersection of an infinite descending chain of non-empty sets [closed]

I am trying to prove something by contradiction and I am stuck as described below: From the (false) assumption, I have shown that for any $i \in \mathbf{N}$, $A_i \neq \emptyset$ and that $A_1 ...
2
votes
1answer
179 views

How do we know we need the axiom of choice for some theorem?

I have been working through Munkres Topology book and in an exercise he says that there was a theorem he proved in a previous section that relied on the axiom of choice and the task is to find it. I ...
1
vote
1answer
58 views

Cardinality and surjective functions

Let $A$ denote a set and $P(A)$ be the power set. By definition for cardinalities $|A|\le|B|$ iff there exists an injection $A \hookrightarrow B$. Note that there is an obvious surjection $P(A) \to ...
2
votes
1answer
148 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 $U$ ?
6
votes
2answers
192 views

The Axiom of Choice and definability

I've seen a lot of relations between the notion of the existence of a definable set with a given property and the necessity of AC is proving that there is a set with the property. For example: Under ...
4
votes
2answers
67 views

Does the existence of this quotient set depend on the Axiom of Choice?

We know the familiar equivalent relation on $\mathbb{R}$, which is $$ x\sim y\Leftrightarrow x-y\in\mathbb{Q} $$ After quoting this relation, we have the quotient set $$ \mathbb{R}/_\sim = \{x + ...
4
votes
4answers
180 views

Do we always use the Axiom of Choice when picking from uncountable number of sets?

I know there are uncountable number of equivalence classes defined by the relation defined here: Equivalence classes of "$x \sim y \Longleftrightarrow x -y $ is rational". (the relation is ...
3
votes
3answers
136 views

Equivalence with the axiom of choice

This is a problem from Tao's Analysis I. We are asked to show that the axiom of choice is equivalent to the statement that for any sets $A$ and $B$ for which a surjection $g:B\to A$ exists, an ...
2
votes
3answers
88 views

Cardinals of set operations without AC

Given info: $|A|=\mathfrak{c}$ , $|B|=\aleph_0$ in ZF (no axiom of choice). Prove: $|A\cup B|=\mathfrak{c}$ If $B \subset A\implies|A \backslash B|=\mathfrak{c}$? I have found several places ...
3
votes
2answers
158 views

Wellordering vs. Zorn's lemma

Many mathematicians outside mathematical logic dislike wellorderings, ordinals and corresponding transfinite arguments. They use zorn's lemma instead and claim one does not need ordinals at all. ...
4
votes
2answers
462 views

Second-countable implies separable/Axiom countable choice

Let $(X,\mathscr T)$ be a topological space, and $(B_n)_{n\ge1}$ a countable basis for X. Under this assumptions, X is separable. The proof of this assertion is as follows: We can assume without ...
4
votes
2answers
89 views

Existence of an injection from $\Bbb N$ without the axiom of choice

If you have a set $A$, and it satisfies that for some $x\in A$, there is a bijection between $A$ and $A\setminus\{x\}$. Does that imply that there is an injection from $\Bbb N$ to $A$? It is clearly ...
4
votes
2answers
266 views

Choice function for a collection of nonempty subsets of $\{0,1\}^\omega$ [duplicate]

Possible Duplicate: Finding a choice function without the choice axiom Is possible to construct a Choice function for a collection of nonempty subsets of $X=\{0,1\}^\omega$? (Without AC) ...
3
votes
2answers
77 views

Is it correct, or legitimate to write $P(\prod_{i \in I}{X_i}) = \prod_{i \in I}{P(X_i)}$?

I'm trying to formulate Axiom of Choice in terms of $P(\prod_{i \in I}{X_i})$, but end up with the following questions. Let $P(X)$ be the power set of $X$.In general, is it correct, or legitimate to ...