Intersection Notation and its Function in Probability Suppose event $A$ is flipping a coin, $A = \{H, T\}$
Suppose Event $B$ is rolling a die, $B = \{1, 2, 3, 4, 5, 6\}$. 
Next, suppose we flip the coin and roll the die. If we want to find the probability of getting, say, a head $H$ (denoted by $A_{H}$) and a $5$ (denoted by $B_{5}$), we write, 
$Pr(A_{H} \cap B_{5})$. 
What's troubling me is: 

What does $A_{H} \cap B_{5}$ mean? To me at least, the intersection of the two events is the empty set $\{\}$. 

I know the it used for a substitute for 'AND' in plain English, which makes sense, it just doesn't make sense from its traditional sense where the intersection means 'the set of shared elements'. 
 A: Let's pause time before we flip the coin and roll the die, and take the set
$\Omega$ of 'possible continuations' (or 'possible worlds').
Then $A_H\subseteq\Omega$ is the set of those 'worlds' where the coin is flipped to the head.
A: You have written down the sample space for the coin flip $A= \{H,T\}$ and the die roll $B=\{1,2,3,4,5,6\}.$ When you consider a situation where both a coin is flipped and a die is tossed, the natural sample space to consider is the cartesian product $\Omega = A\times B$ where the outcomes consist of a pair $(a,b)\in \Omega$ of outcomes, one from the coin flip and one from the die roll. So if you flip heads and roll a $2$ then that is the outcome $\omega = (H,2).$ 
The event that you get a head (without regard to how the dice roll comes out) is just the the set $$A_H = \{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)\} \subset \Omega$$ the event that you get a $5$ for the die roll can be written $$B_5 = \{(H,5),(T,5)\}.$$
The interection, which consists of all the outcomes where you both flip a head and roll a $5$ is then $$ A_H\cap B_5 = \{(H,5)\}.$$
Say you also spin a spinner that can come out red blue or green $C=\{R,B,G\}$ and you want to add it to the mix. Then you'd have $ \Omega = A\times B\times C$ and writing down everything would be more tedious, but you would end up with $$ A_H\cap B_5 = \{(H,5,R),(H,5,B),(H,5,G)\}$$ as the set of outcomes where you flip a head and roll a 5.
A: The key here is to recognize that a "plain-English" definition of the event can only be made precise once you specify a sample space.  So, with respect to the sample space $A$, the "coin flip is heads" event, $A_H$, is $\{H\}$.  But wait!  This sample space is clearly insufficient for your purposes: $B_5$ cannot be expressed in terms of the elements of that sample space.  Similarly, $A_H$ cannot reify w.r.t. $B$, but $B_5=\{5\}$ in the sample space $B$.  
To ask the question you want to ask, you need to use the sample space $A\times B$.  With respect to $A\times B$, the event  $A_H=\{H\}\times B$, because those are the outcomes in $A\times B$ for which the coin flip was heads.  Similarly, $B_5$ reifies to $A\times\{5\}$, and hence $$A_H\cap B_5=(\{H\}\times B)\cap(A\times\{5\})=\{(H,5)\}$$ as you expect.  
