# Do you need the Axiom of Choice to accept Cantor's Diagonal Proof?

Math people:

It is my understanding that set theorists/logicians compare cardinalities of sets using injections rather than surjections. Wikipedia defines countable sets in terms of injections. Cantor's diagonal proof that the real numbers are uncountable involves showing that there is no surjection from $\mathbb{N}$ to $(0,1)$. So do I need the Axiom of Choice to accept Cantor's Diagonal Proof?

I browsed the Similar Questions and I could not find an answer. I apologize if this is a duplicate.

StEFAN (Stack Exchange FAN)

1. If there is an injection from a non-empty set $$A$$ into $$B$$ then there is a surjection from $$B$$ onto $$A$$. This does not require the axiom of choice, although the inverse implication (that a surjection has an injective inverse) is in fact equivalent to the axiom of choice.
To add on this, $$\mathbb N$$ is well-ordered without the axiom of choice, so if there is a surjection from $$\mathbb N$$ onto a set $$A$$, then there is an injection from $$A$$ into $$\mathbb N$$ as well.
• Cantor's proof that the cardinality of $A$ is less than that of its powerset also uses the diagonal argument. Is the above still valid? Feb 14, 2013 at 22:48