Nomenclature of random variables $\{X=0, Y=0\}$ same as $\{X=0\}\cap \{Y=0\}$? just a small doubt.
My exercises keep oscillating their nomenclature on this small detail and I always have the other version.
Let $X,Y$ be random variables.
Is $\{X=0, Y=0\}$ the same as $\{X=0\}\cap \{Y=0\}$?
Another example. 
Let $N$ be the number of Users on a webpage. Two files are available for download, one with 200 kb and another with 400 kb size.
$$
\begin{align}
X_n(w) := w_n = \{ & 0:=\text{user downloads no file}, \\
& 1:=\text{user downloads the first file (200 kb)}, \\
& 2 :=\text{user downloads the second file (400 kb)}, \\
& 3:=\text{user downloads both files (600 kb)}\}
\end{align}
$$
I want to express, at least one user downloaded the 200 kb file. Here's how I expressed it
$\{X_1 + X_2 + \cdots + X_n \geq 1\}$. Would this be ok? The book expressed it as $\{X_1=1\}\cup\{X_1=3\}\cup \cdots \cup\{X_n=1\}\cup\{X_n=3\}$. 
Another thing to express: no user downloaded the 200 kb file. I expressed it as
$|\{X_k=1, 1 \leq k \leq N\}|=0$. The book as $\{X_1 \neq 1\}\cap \cdots \cap \{X_n \neq 1\}$. Would my solution be ok?
I'm always in doubt when I'm allowed to use symbols like $+$ and $|\mathrm{modulo}|$ (to get the number of elements). Is this generally always allowed? Many thanks in advance!
Thanks in advance guys!
 A: $\{X=0,Y=0\}$ and $\{X=0\}\cap\{Y=0\}$ are the same thing.  Both notations refer to
$$
\{\omega\in\Omega : X(\omega)=0\ \  \&\ \  Y(\omega)=0\} = \{\omega\in\Omega : X(\omega)=0\}\cap\{\omega\in\Omega : Y(\omega)=0\}.
$$
Your notation saying
$$
\begin{align}
X_n(w) := w_n = \{ & 0:=\text{user downloads no file}, \\
& 1:=\text{user downloads the first file (200 kb)}, \\
& 2 :=\text{user downloads the second file (400 kb)}, \\
& 3:=\text{user downloads both files (600 kb)}\}
\end{align}
$$
seems confused.  I suspect maybe you meant
$$
\begin{align}
\Omega = \{ & 0:=\text{user downloads no file}, \\
& 1:=\text{user downloads the first file (200 kb)}, \\
& 2 :=\text{user downloads the second file (400 kb)}, \\
& 3:=\text{user downloads both files (600 kb)}\},
\end{align}
$$
although even that may differ from what's appropriate if you're bringing in $n$ different random variables.  Your later notation makes it look as if what the author of the  book had in mind is that $X_k$ is the number of kb downloaded by the $k$th user, for $k=1,\ldots,n$.  Just what $w$ is, you're not clear about, and at this point I'm wondering if you're confusing $w$ with $\omega$.  Probably what is needed is this:
$$
\begin{align}
\{ & 0:=\text{user downloads no file}, \\
& 1:=\text{user downloads the first file (200 kb)}, \\
& 2 :=\text{user downloads the second file (400 kb)}, \\
& 3:=\text{user downloads both files (600 kb)}\}^n
\end{align}
$$
i.e. then $n$th power of that set of four elements.  This is the set of all $n$-tuples where each component of an $n$-tuple is one of these four elements.  Then, when $\omega$ is any such $n$-tuple, $X_k(\omega)$ is its $k$th component, which is one of those four elements.
For example, if $n=3$, so there are three users, then
$$
\begin{align}
\Omega = \{ & (0,0,0), (0,0,1), (0,0,2), (0,0,3), (0,1,0), (0,1,1), (0,1,2), (0,1,3),\ldots\ldots\ldots \\  \\
& \ldots\ldots\ldots, (3,3,3) \},
\end{align}
$$
with $64$ elements.  If, for example, $\omega=(2,3,0,1)$, then $X_2(\omega)=3$.
A: Your second example is incorrect: if no user downloaded the 200K file, but at least one user downloaded the 400K file, we will still have $X_1 + \dots + X_n \ge 1$.
