A union of n sets with diferent order, how can I prove? Let$k_1,k_2,...,k_n$ be any ordination of the indexes $1,2,...,n$then $\bigcup_{i=1}^{n}A_{k_i}=\bigcup_{i=1}^{n}A_i$ I tried the double contention technique with no good results, is there a handy theorem I can use?

Let $x\in \bigcup_{i=1}^{n}A_{k_i}$. Then for some m, $x\in A_{k_m}$. But $ A_{k_m}$ is one of $A_1, \dots A_n$. So $x$ is in on of $A_1, \dots A_n. $ Hence $x \in\bigcup_{i=1}^{n}A_i$. We proved so far: $\bigcup_{i=1}^{n}A_{k_i}\subseteq\bigcup_{i=1}^{n}A_i$.

Similarly: $\bigcup_{i=1}^{n}A_i\subseteq \bigcup_{i=1}^{n}A_{k_i}$