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An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty sets. The axiom of choice is related to the first of Hilbert's problems.

46 votes
Accepted

Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein?

Without the axiom of choice, it is possible that there is a vector space $U\neq 0$ over a field $k$ with no nonzero linear functionals. Let $V$ be the direct sum of countably many copies of $U$, and $ …
Jeremy Rickard's user avatar
5 votes
Accepted

Exterior powers and choice

As YCor notes in comments, (2) is a special case of (1), so I'll only address (1). Suppose $\Lambda^k\varphi:\Lambda^kV\to\Lambda^kW$ is not injective, and let $x\neq0$ be in the kernel. Then $x$ ca …
Jeremy Rickard's user avatar
13 votes
Accepted

Axiom of choice and algebraic tensor product

I think both can be proved without choice, essentially because, in both cases, whenever you're tempted to choose a basis, you can manage with a little care to get by with a basis of a finite dimension …
Jeremy Rickard's user avatar
12 votes
5 answers
1k views

Does k(X) have a k-basis for every set X, without AC?

This question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?. For any field $k$, the field $k(x)$ of rational functions in one variable has an explic …
Jeremy Rickard's user avatar
11 votes

Properties of vector spaces without AC

I think vector spaces must still be flat (tensor product is exact). I don't think any of the steps in the following proof use choice, although it's quite possible I'm mistaken: Finite dimensional vec …
Jeremy Rickard's user avatar