How does cardinality map a finite set to $(\{0\} \cup \mathbb{N})$? In reading Kevin Houston's "How to Think Like a Mathematician", there's a line stating the following:
Let $X$ be the set of finite sets. Then the cardinality of a set is a function on $X$, that is $|.|: X \to \{0\} \cup \mathbb{N}$. Note that we need $0$ in the co-domain as the set could be the empty set.
What does this mean exactly? From my understanding, this would imply that any finite set is mapped exactly to one natural number. For example, a finite set of $5$ elements would be mapped to the natural number $5$. However, is it possible to map the set itself to a value?
 A: It means what you think it means. 
The domain of definition for this function is the "set of all finite sets". 
Thus every element here is a "finite set".
Consider for a simpler example: $A=\{\{1,2\}, \{1,2,3\},\{5,7,8,12\}\}$ 
This is a set whose elements are the sets $\{1,2\}$ and $\{1,2,3\}$ and $\{5,7,8,12\}$. 
If you define a function on $A$ then you assign a value to each of $\{1,2\}$ and $\{1,2,3\}$ and $\{5,7,8,12\}$  yet not to $1,2,3, 5,7,8,12$. 
Tangential remark: there is a problem with the idea of a set of all finite sets, but this is not relevant in this context.  
A: You understood wrong. According to this definition, any finite set is mapped exactly to a set whose cardinality is a natural number or zero.
A: For every finite set, its function value is its cardinality. 
So $\{1,2,3\}$ is mapped to $3$ just as $\{2,4,8\}$ is, as both have the same cardinality $3$. $\emptyset$ is mapped to $0$ (as the only set with that property).
I don't quite see what the problem is: there is a class of all finite sets and we can define a function with those sets as input.
