# Why does $A_\infty=\bigcup_{n\geq 5}^{\infty} \,A_n$?

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).

• Well, $A_{5}$ already contains $A_{4}$. – Geoff Robinson Oct 20 '15 at 21:47
• I think I see. Does that mean that $A_\infty = \bigcup_{n\geq k}^{\infty} A_n$ for any $k\geq 5$? I feel my confusion came from not understanding why 5 was chosen in particular. – user282311 Oct 20 '15 at 21:51
• Yes because you can show that $A_{n-1}<A_n$. $A_5$ is the smallest simple alternating group (and also smallest simple non-abelian group). – Ali Caglayan Oct 20 '15 at 21:53
• Starting from 5 is to emphasize that $A_\infty$ is a union of finite simple groups. – Amin Nov 18 '15 at 9:54