Does this lead to a contradiction within ZF?

Mathematics Asked by quanticbolt on December 14, 2020

Let the following be an axiom (which I will denote P):

If $x,y$ are sets and $f:yto x$ is a surjection, then the existence of an injection $f:xto y$ guarantees a choice function exists.

Does P lead to contradiction within ZF in any obvious Russell’s paradox like sense?

One Answer

If I understand you correctly, you want to say that if there is a surjection from $y$ onto $x$, then if there is any injection, there is one splitting the surjection.

This cannot lead to contradiction, since it follows from the Axiom of Choice. Another question would be whether or not this implies the Axiom of Choice. This is a bit more complicated.

Let "The Partition Principle" denote the statement "if there is a surjection from $y$ to $x$, then there is an injection". And what we're asking is: Does the Partition Principle implies AC provable from $sf ZF$? And the answer is that we don't know.

Correct answer by Asaf Karagila on December 14, 2020

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