What does $\{x|x = 7\} = \emptyset$ mean? (Probability) To my understanding $\{x|x = 7\} = \emptyset$  means that the number seven is a not allowed value. But I do not understand the meaning of "$x|x$". Can anybody please explain $\{x|x = 7\} = \emptyset$ more in detail.
It's from an introduction book into probability theory. It says:" ... Not possible events are for example $\{x|x = 7\} = \emptyset$ , $\{x|x = 0\} = \emptyset$ , ...". Basis is the Throwing Dice example as e. g. also available on the page http://www.mathsisfun.com/data/probability.html
 A: I guess the bar $|$ means "such that". Therefore "the set of possible outcomes of $x$ such that $x=7$ is an empty set"
A: The formalization of the "Throwing a Dice" example involves the construction of a set $\Omega$, known as the sample space, whose elements $\omega$ correspond to each possible result of the experiment. Hence, you can take as your sample space the six elements set
$$
  \Omega =\{ \bullet, \bullet\bullet, \bullet\bullet\bullet, \bullet\bullet\bullet\bullet, \bullet\bullet\bullet\bullet\bullet, \bullet\bullet\bullet\bullet\bullet\bullet \} \, .
$$
A random variable $X$ is a function which maps $\Omega$ to the set of real numbers. For example, you can construct a random variable $X$ which gives you the number of dots in the dice's face which turned up. Here is the function:
$$
  X(\bullet) = 1 \qquad X(\bullet\bullet) = 2 \qquad X(\bullet\bullet\bullet) = 3
$$
$$
  X(\bullet\bullet\bullet\bullet) = 4 \qquad X(\bullet\bullet\bullet\bullet\bullet) = 5 \qquad X(\bullet\bullet\bullet\bullet\bullet\bullet) = 6 \, .
$$
For some subset $A$ of the real numbers, the inverse image $X^{-1}(A)$ is the set of elements $\omega\in\Omega$ such that $X(\omega)$ belongs to $A$. Symbolically,
$$
  X^{-1}(A) = \{ \omega\in\Omega : X(\omega)\in A\} \, .
$$
Note that you can consider the inverse image of some set by any function. The function doesn't need to have an inverse.
Consider these examples:
$$
  X^{-1}(\{5\}) = \{ \bullet\bullet\bullet\bullet\bullet \}
$$
$$
  X^{-1}(\{1,6\}) = \{ \bullet, \bullet\bullet\bullet\bullet\bullet\bullet \}
$$
$$
  X^{-1}((-\infty,3]) = \{ \bullet, \bullet\bullet,\bullet\bullet\bullet \}
$$
Now consider $X^{-1}(\{7\})$. First, write the formal definition
$$
  X^{-1}(\{7\}) = \{ \omega\in\Omega : X(\omega)= 7\} \, .
$$
Is there any $\omega\in\Omega$ such that $X(\omega)=7$? No. Hence,
$$
  X^{-1}(\{7\}) = \emptyset.
$$
