My question is: why does $$A_\infty=\bigcup_{n\geq 5}^{\infty} \,A_n \,\,?$$
Doesn't $A_{\infty}$ contain (an isomorphic copy of) $A_4$, which is not a simple group? I'm sorry if this is a stupid question, I just haven't been able to see it. This comes from here: Is the question phrased properly? and is my proof correct? (An infinite alternating group is simple).