Tagged Questions

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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This about the famous article Zermelo, E., Beweis, daß jede Menge wohlgeordnet werden kann, Math. Ann. 59 (4), 514–516 (1904), available here. Edit: Springer link to the ...
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Continuity and the Axiom of Choice

In my introductory Analysis course, we learned two definitions of continuity. $(1)$ A function $f:E \to \mathbb{C}$ is continuous at $a$ if every sequence $(z_n) \in E$ such that $z_n \to a$ ...
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Defining cardinality in the absence of choice

Under ZFC we can define cardinality $|A|$ for any set $A$ as $$|A|=\min\{\alpha\in \operatorname{Ord}: \exists\text{ bijection } A \to \alpha\}.$$ This is because the axiom of choice allows any ...
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For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice

How to prove the following conclusion: [For any infinite set $S$,there exists a bijection $f:S\to S \times S$] implies the Axiom of choice. Can you give a proof without the theory of ordinal numbers....
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Axiom of choice and automorphisms of vector spaces

A classical exercise in group theory is "Show that if a group has a trivial automorphism group, then it is of order 1 or 2." I think that the straightforward solution uses that a exponent two group is ...
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Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice?

I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of ...
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Non-aleph infinite cardinals

I'm now confused with a concept of $\aleph$. 1.$\aleph$ is a cardinal number that is well-ordered in ZF.(Defined as an initial ordinal that is equipotent with). Does that mean $\aleph_x$ in ZF may ...
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Can one construct a non-measurable set without Axiom of choice?

Is axiom of choice required to show the existence of non-measurable sets? Is there a Lebesgue non-measurable set that can be constructed without axiom of choice? Related question on MO says it is ...
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Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$?

$\mathbb R$ is an uncountably dimensional vector space over $\mathbb Q.$ We can define as many endomorphisms of this vector space as we want by picking their values on the elements of the basis. ...
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Existence in ZF of a set with countable power set

Is it consistent with ZF for there to exist a set $S$ such that the power set $P(S)$ is countable? If so, what is the weakest form of the axiom of choice needed to prove that no such set exists?
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Which set is unwell-orderable?

In textbook it says that every well-orderable set is equipotent to an initial ordinal number. However, of course unwell-orderable cannot equipotent to any ordinal number, but is there any set is ...
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Is Banach-Alaoglu equivalent to AC?

The Banach-Alaoglu theorem is well-known. It states that the closed unit ball in the dual space of a normed space is $\text{wk}^*$-compact. The proof relies heavily on Tychonoff's theorem. As I have ...
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Axiom of choice and calculus

I thought many results in calculus need axiom choice. For example, I thought one needs AC to prove that a bounded sequence in the real line has a convergent subsequence. Recently I was taught that one ...
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Is Zorn's lemma required to prove the existence of a maximal atlas on a manifold?

In the definition of smooth manifolds, complex manifolds, and similar constructions, one starts by defining a property on neighborhoods in the space, specifying how they relate on overlapping ...
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The cardinality of a countable union of countable sets, without the axiom of choice

One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does not have cardinality $\aleph_2$. My solution shows that it does not have ...
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The Continuum Hypothesis & The Axiom of Choice

Does anyone here know of a reference to an analysis on a proposed relationship between The Continuum Hypothesis and The Axiom of Choice?
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Is the axiom of choice needed to show that $a^2=a$?

A comment on this answer states that choice is needed for the statement that $a^2=a$ for all infinite cardinals $a$. In Thomas Jech's Set Theory (3rd edition), his theorem 3.5 proves this statement ...
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Axiom of Choice Examples

In the wikipedia article, two examples are given which use/ do not use the axiom of choice. They are: Given an infinite pair of socks, one needs AC to pick one sock out of each pair. Given an ...
I have this exercise involving the axiom of choice, but I don't understand where it's needed: Let $(X_i)_{i \in I}$ and $(Y_i)_{i \in I}$ be pairwise disjoint sets with $|X_i| = |Y_i|$. Prove, using ...