Why all permutations of i.i.d. samples are equally likely?

Question

Suppose $X$ i.i.d

Why is$$P(X_{i_1}

I think we can reason this by probability measure is translational invariant? (But I don't think this is true in general)

Update: I know the intuition but please give a mathematical proof of this

Answer 1

Since $(X_1,\ldots,X_n)$ is i.i.d., the sequence $(X_{\pi_1,},\ldots,X_{\pi_n})$ is also i.i.d., for any permutation $\pi_1\pi_2\cdots\pi_n$ of the numbers $\{1,\ldots,n\}$. In particular, it follows that $(X_1,\ldots,X_n)$ is equal in distribution to $(X_{\pi_1,},\ldots,X_{\pi_n})$. That is, $\mathbb P((X_1,\ldots,X_n)\in A)=\mathbb P((X_{\pi_1},\ldots,X_{\pi_n})\in A)$ for all $A$. By choosing $A$ to be the set of all increasing tuples, it follows that $$ \mathbb P(X_1