iid random variables visualization I am thinking $X_1, X_2, \cdots$ being iid random variables, that map $\omega \in \{0,1\}$ to $\{0,1\}$. If $\omega = 0$, then $X_i = 0$; if $\omega = 1$, then $X_i = 1$.
However, if we define this way, all the $X_i$'s are the same. Where is the randomness coming from? In this stage, we have not yet define any probability measure. How should we visualize iid random variables?
 A: If you use this definition for all $X_i$, then they are not independent because they will all have the same value. When you fix a single $\omega$, you automatically fix values (realizations) for every random variable in the universe. In your definition, choosing either value of $\omega$ fixes $X_1\left(\omega\right)=X_2\left(\omega\right)=...$ to be the same value.
When one works with random variables, one rarely defines them on the outcome space directly, nor does one attempt to precisely define what the outcome space is. Rather, one assumes that there exists some outcome space, but one works only with the probability measure induced by various distributions. So, when you say that $X_1$ and $X_2$ are independent Bernoulli random variables, you are implicitly assuming the existence of some suitable probability space $\left(\Omega,\mathcal{F},P\right)$, but it is not necessary to explicitly define all of these for virtually anything that you might want to do with the random variables. It is sufficient to know $P\left(A\right)$ for events $A$ involving the random variables.
However, if you really wanted to define an outcome space for two independent $Bernoulli\left(p\right)$ random variables, you would need four outcomes with
\begin{eqnarray*}
X_1\left(\omega_1\right) &=& 1, X_2\left(\omega_1\right) = 1\\
X_1\left(\omega_2\right) &=& 1, X_2\left(\omega_2\right) = 0\\
X_1\left(\omega_3\right) &=& 0, X_2\left(\omega_3\right) = 1\\
X_1\left(\omega_4\right) &=& 0, X_2\left(\omega_4\right) = 0\\
\end{eqnarray*}
with $P\left(\left\{\omega_1\right\}\right) = p^2$, $P\left(\left\{\omega_2\right\}\right) = P\left(\left\{\omega_3\right\}\right) = p\left(1-p\right)$, and $P\left(\left\{\omega_4\right\}\right) = \left(1-p\right)^2$.
