Random Vector IID It is my understanding that random vectors are deterministic that is
$$X:\Omega\longrightarrow \mathcal{R}^2$$
where
$$X(\omega)=\begin{bmatrix}X_{1}(\omega)\\X_{2}(\omega)\end{bmatrix}$$
is completely determined once $\omega$ is selected. The only randomness occurs from not knowing which $\omega$ is selected. If I am wrong please correct me. My question is if a random vector has iid component functions do those functions have to be identical,i.e. $X_1(\omega)=X_2(\omega)$? I believe so and I came up with this logic to validate my belief.
If the two functions are not identical then if I consider a partition of the real number line using elements of the borel space  $B_i\in (\mathcal{R}, \mathcal{B})$  where $\cup_{i}^{\infty}  B_i=\mathcal{R}$  and $B_i\cap B_j=\emptyset$ then  if we consider $B_i$ we have
$$A_{1i}=X_1^{-1}(B_i)\neq X_2^{-1}(B_i)=A_{2i}$$
$$P(A_{1i})= P(A_{2i})$$
where the first follows from non-equal functions and the second from identically distributed.
Then we have two partitions of $\Omega=\cup A_{1i}=\cup A_{2i}$ and $P(A_{1i})=P(A_{2i})$ and $A_{1i}\neq   A_{2i}$ which cannot happen. If I am not correct can someone provide an example where a random vector has iid components but the functions  are not identical or if I am correct provide a better explanation then the one I have provided.

Suppose the probabilistic experiment is tossing a coin 2 times; $X_1$ is Bernoulli random variable equal to $1$ if we have heads in the first toss, $X_2$ is Bernoulli random variable equal to $1$ if we have heads in the second toss. Evidently $X_1$ and $X_2$ are different iid random variables.
