Random sequence is dense in (0,1) I have a question about probability and it has been confusing me for days..
Let $X_1,X_2,...$ be independent random variables with uniform(0,1) distribution from probability space ($\Omega$,$\cal F$,$P$).

*

*Show that:
$P(\omega\in\Omega: \{X_1(\omega),X_2(\omega),X_3(\omega)...\} \text{is dense in (0,1)})=1.$
I don't know how to interpret the question and have no idea where to start...
 A: Define $S:=\left\{ X_{n}\mid n=1,2\dots\right\} $ as random set.
Define $I:=\left\{ \left\langle r,s\right\rangle\in\mathbb{Q}\cap\left(0,1\right) \mid r<s\right\} $ and note that $I$ is countable. 
For $\left\langle r,s\right\rangle \in I$
let $E_{r,s}$ denote the event that $S\cap\left(r,s\right)=\emptyset$.
Then $P\left(E_{r,s}\right)=0$ and consequently $P\left(\bigcup_{\left\langle r,s\right\rangle \in I}E_{r,s}\right)=0$.
(This can be proved on base of $P\{X_1\notin (r,s)\}=1-(s-r)<1$ combined the fact that the $X_n$ are iid.)
Equivalently: $$P\left(\bigcap_{\left\langle r,s\right\rangle \in I}E_{r,s}^{c}\right)=1$$ 
and this
can be recognized as the event that $S$ is dense in $\left(0,1\right)$.
A: To interpret the question, first note that a set $A \subset [0,1]$ is dense in $[0,1]$ if for any $x \in [0,1]$ and $\epsilon > 0$, there exists $a \in A$ such that $|x-a| < \epsilon$. 
To show the set $S(\omega) \dot = \{X_1(\omega),X_2(\omega),\dots\}$ is dense in $I \dot = [0,1]$ with probability 1, we need to show that for any $x \in [0,1]$, $\epsilon > 0$ and almost every $\omega \in \Omega$, there is an $i$ such that $|X_i(\omega) - x| < \epsilon$. 
Let then $\epsilon > 0$ and $x$ such that $(x-\epsilon,x+\epsilon) \subset [0,1]$ be given. We have $P(|X_i - x| < \epsilon) = 2\epsilon$, since $P(X_i \in (a,b)) = b-a$ for any $0 < a < b < 1$. 
Hint: Use Borel-Cantelli on the event $\{X_i \in B(x,\epsilon)\}$. If you go back to the definitions we wrote out at the beginning, you'll see the conclusion follows immediately.
Remark: You have to make a minor modification of the proof to take into account all $x \in [0,1]$, even those at the endpoints, $x = 0$ and $x = 1$. 
Edit: As Did pointed out, the above will only lead you to show that for any $x \in \mathbb{R}$, $P(S\text{ not dense at } x) = 0$, while we need to show $P\left(\cup_{x \in I} \{S\text{ not dense at } x\}\right) = 0$. The union here is uncountable and so it is not even in our $\sigma$-algebra. However, it suffices to show
\begin{gather*}
\bigcup_{x \in I} \{S \text{ not dense at } x\} = \bigcup_{x \in I \cap \mathbb{Q}} \{S \text{ not dense at } x\}
\end{gather*}
and then use countable subadditivity of measures to conclude the probability of the right-hand set above is 0. 
The $\supset$ inclusion is immediate. If $x \in [0,1] \cap \mathbb{Q}^c$ and $\omega \in \{S \text{ not dense at } x\}$, then for this $\omega$ there is an $\epsilon > 0$ such that for any $i \geq 1$, $|X_i(\omega) - x| \geq \epsilon$. Choosing a $q \in I$ such that $|q - x| < \epsilon/2$, we see that $\omega \in \{S \text{ not dense at } q\}$. Otherwise, we'd have a $j$ such that $|X_j(\omega) - q| < \epsilon/2$, but then $|X_j(\omega)-x| < \epsilon$ by the triangle inequality, a contradiction. This yields the $\subset$ inclusion.
