# Tag Info

4

No, this is really just induction with finitely many choices. The axioms of choice would come in if you would have wanted to say there is a countably infinite set of algebraically independent elements. With this proof, you would invoke Dependent Choice, which is stronger than countable choice. But I reckon the proof can be slightly modified to use just ...

4

The axiom of choice is absolutely not needed here. We only need to choose one element from $b$, since it's not empty we can do it. And the axiom of choice is not needed. Then simply map everything not in $B$ to $b$, while keeping the rest in place. This is similar to the problem of inversing functions. The axiom of choice is needed in order to construct an ...

4

Given any family of sets $A_i$ indexed by $i\in I$, consider the map $\phi:\mathcal{A}\to I$ from the disjoint union of the family to $I$. The existence of a one-sided inverse to $\phi$ is precisely a choice function. Hence the converse is immediate.

3

Once you have a choice function from $\{X_i\times\{i\}\mid i\in I\}$, meaning some $f$ whose domain is $I$ and $f(i)=\langle x,i\rangle$ where $x\in X_i$, simply take the projection onto the left coordinate, and show that this is a choice function for the original family.

2

In fact, the following are equivalent (in ZF): The axiom of choice. Every coequaliser in $\mathbf{Set}$ is absolute. Indeed, one can prove (in ZF) that every surjection is the coequaliser of its kernel pair, so it suffices to prove the following assertion: The coequaliser diagram $$R \rightrightarrows A \rightarrow A / R$$ is absolute if and only ...

2

For the purpose of applying set theory as a foundational subject, one must be able to express typical mathematical reasoning, such as Let $r$ be a real number... Let $n$ be an integer... Let $a$ be a sequence of real numbers... Let $f$ be a real-valued function of the reals... Let $S$ be a subset of the reals... And in general, if you have a kind of ...

2

Axiom of choice implies König's theorem which estimates cardinality of the product from below, namely $|\prod_{i\in I}A_i|\geq|\sum_{i\in I}A_i|$, the sum stands for the disjoint union. In particular, the product has at least as many elements as any of the factors, which is stronger than just non-empty. However, if $I=\mathbb{N}$ and $A_i=A$ "every ...

2

Yes, currently the problem is open whether or not the Partition Principle and the Axiom of Choice are equivalent. There are two major factors for this (in my opinion): Many people become less interested in choiceless results. So while they might be very happy to hear about them, they prefer to put their research efforts towards other directions. The ...

2

If $X$ is non-empty, then there is no dependence on the axiom of choice. To see this, note that $X_\alpha$ is a continuous map of $X$ with the projection map $\pi=\pi_\alpha(x)=x_\alpha$. This follows from the fact that if a product $X=\prod_{i\in I}X_i$ is non-empty, then for each $x\in X_i$ there is a function $f\in X$ with $f(i)=x$. Simply pick one ...

1

Your definition of $R$ is extremely cumbersome. You don't have to mention the set over which you apply separation if the definition is obviously bounded. Writing $R=\{(i,h)\mid h\in H(i)\}$ is far clearer. And if that is the relation that you were defining, then the proof is fine.

1

Rubin and Rubin has the wonderful Equivalents of the Axiom of Choice II which includes hundreds of equivalents and proofs of the equivalence. For simpler equivalents, like the one you are interested in, you don't need to go to something like that. Most basic books about set theory will cover the axiom of choice and include a proof like that, and it appears ...

Only top voted, non community-wiki answers of a minimum length are eligible