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A friend of mine asked me why $\aleph_0$ is the smallest infinity and if we must use axiom of choice to say that it is the smallest infinity.

Do you know if Axiom of Choice is neccesary?

HeMan
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Let's define what "smallest" means; since cardinality is a description of bijections, i.e. $\mbox{card}A=\mbox{card}B$ iff there is a bijection between $A$ and $B$, then we can define an ordering on cardinals by saying: cardinal $c_1$ is smaller than cardinal $c_2$ if there exists a surjection from some set with cardinality $c_2$ onto some set with cardinality $c_1$.

The definition of infinite is generally taken to be "not finite", i.e. not in bijective correspondence with any finite set. Using only this definition we can say that for any infinite set $A$, there exists a surjection $f:A\rightarrow\Bbb{N}$, so that $\mbox{card}\Bbb{N} \leq \mbox{card}A$ for every infinite set $A$.

Disclaimer: I don't work in foundations or set theory, so this may not be rigorous, but I'm confident that it's the right idea.

Sort of Damocles
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    It's weaker (in ZF without choice) for a set $S$ to have a surjection onto $\mathbb{N}$ than for $\mathbb{N}$ to have an injection into $S.$ And it is consistent with ZF that there are infinite sets with neither of these properties, e.g. amorphous sets. However, ZF + countable choice is enough to show both of these properties are equivalent to a set being infinite in the standard sense. – Elliot Glazer Oct 03 '17 at 02:54
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    You neec the axiom of choice to prove that for any infinite set $A$ there is a surjection $f:A\to\mathbb N.$ – bof Oct 03 '17 at 02:55