Find $\left\{ x \in U: X_{A}=1 \right\}$ Let $U$ be the universe. Suppose $A \subseteq U$ with $A \ne \emptyset$, and $A \ne U$. Let $X_{A}$ be the characteristic function of A. Find
(a) $\left\{ x \in U: X_{A}(x)=1 \right\}$
(b) $\left\{ x \in U: X_{A}(x)=0 \right\}$
(c) $\left\{ x \in U: X_{A}(x)=2 \right\}$
(d) $\left\{ x \in U: X_{A}(x) \leq 1 \right\}$
From the definition of a characteristic function of $A$ is $1$ iff $x \in A$ or $0$ iff $x \notin A$.
I found that the answer to (a) is $A$, but I don't understand how to come to that answer. I really do not understand how to approach parts (c) or (d).
How do I get on the right track here?
 A: It's helpful to read out the definitions of the sets in English. Let's look at the example (a).

The set $S_a=\{x\in U: X_A(x)=1\}$ is:

The set of all values $x$ such that $X_A(x)=1$.

while the function $X_A$ is equal to

The function that is equal to $1$ precisely for elements of $A$.

Clearly, the two definitions hint that $S_a=A$, but now, you have to prove it strictly. To do that, to prove that two sets are equal, you need to prove:


*

*That $S_a\subseteq A$. 


So, you take $x\in S_a$. By the definition of $S_a$, this means $X_A(x)=1$. Since we know that $X_A(x)=1$ if and only if $x\in A$, we can therefore conclude that $x\in A$, therefore $S_a\subseteq A$.


*That $A\subseteq S_a$.


Let $x\in A$. Therefore, by definition of $X_A$, we have $X_A(x)=1$. This means that, by the definition of $S_a$, that $x\in S_a$.

For (b), (c) and (d), take the same approach:


*

*Write down what the set is by definition.

*Try to "guess" what the set would be.

*Prove, rigorously, that what you guessed is true.

