-1
votes
0answers
113 views

Construction of the field of real numbers within $ZF$ [duplicate]

I am interested in a problem whether the field of real numbers can be constructed within $ZF$. I will state the problem more precisely as follows. Definition 1 An ordered field $K$ is called ...
1
vote
1answer
17 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
31 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
76 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', ...
3
votes
1answer
29 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
55 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
47 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
37 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
59 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
164 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
84 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
137 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.
2
votes
1answer
67 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
46 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
46 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
93 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
56 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
71 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
77 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
58 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
112 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
85 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
29 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
98 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
85 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
168 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
153 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
55 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
144 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
181 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
63 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
168 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
126 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
150 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
379 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
85 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 ...
3
votes
2answers
241 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 ...
3
votes
3answers
762 views

Infinite set as union of disjoint countable sets.

Question: Prove, with the help of Zorn's lemma, that infinite set, $X$ can be represented union of disjoint countable sets. My attempt: I know that a countable union of countable sets must be ...
3
votes
2answers
127 views

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) ...
6
votes
0answers
118 views

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 ...
5
votes
2answers
948 views

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 ...
2
votes
1answer
81 views

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: $\lambda+\kappa=\lambda$ $\lambda\cdot\kappa=\lambda$ ...
1
vote
1answer
49 views

$A+\alpha\sim A$ 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 ...
1
vote
2answers
107 views

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 ...
3
votes
1answer
280 views

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 ...
3
votes
3answers
79 views

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 ...
5
votes
2answers
102 views

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 ...