Principles of Topology: Operations on Sets: Union, Intersection, and Difference I have been stuck on the following set of questions for some time now:


Let $X$ be a set with subsets $A$ and $B$. Prove the following: 
(a) $X\setminus(X\setminus A) = A$.
(b) If $A \subset B$, then $X\setminus B \subset X\setminus B$.
(c) $A \subset B$ if and only if $X\setminus B \subset X\setminus A$.
(d) $X\setminus A \subset B$ if and only if $A \cup B = X$.
(e) $A \subset X\setminus B$ if and only if $A \cap B = \emptyset$
(f) $A\setminus B = A\cap(X\setminus B)$
(g) $X\setminus(A\setminus B) = B \cup(X\setminus A)$.


I was wondering if anyone could give me an idea on how to properly approach these type of problems. 
Thank you for your time and Thanks in advance for your feedback.
 A: Hint:
To solve these kind of problems: $W= V$ you need to show that $W\subseteq V$ and $V\subseteq W$. In this case, you take an element $x$ that is in $W$ and show that it has to be in $V$, then you can conclude that $W\subseteq V$.
Example: a) Let $x\in X\setminus(X\setminus A)$, then $x\in X \wedge x\notin (X\setminus A)$, then $x\in X \wedge (x\in A \vee x\notin X)$, then $x\in A$. Therefore $X\setminus(X\setminus A)\subseteq A.$ Now you show the other way round.
e) If $A \subset X\setminus B$, then, for every $x\in A$, $x\in X\wedge x\notin B$. So there isn't a $x$ that's in $A$ and in $B$, therefore $A\cap B=\emptyset$. Now you have to show that if $A\cap B=\emptyset$ then $A \subset X\setminus B$.
A: You should use the very defining properties of the sets, trying to work with their elements: to get a feeling of what I mean, here's the solution of (b):
$A\subseteq B\iff (x\in A\Rightarrow x\in B) \iff  (x\not\in A\Leftarrow x\not\in B) \iff (x\in X\setminus A\Leftarrow x\in X\setminus B) \iff X\setminus B\subseteq X\setminus A$
A: The most straightforward way (not necessarily effective for simple statements) of dealing with set operations seems the use of indicator (or characteristic) function:
$$
I_A:X\to\mathbb Z_2,\quad I_A(x):=
\begin{cases}
1 & x\in A\\
0 & x\not\in A
\end{cases}
$$
Note that for $A,B\subset X$
$$
A=B\quad\text{iff}\quad I_A=I_B,\quad A\subset B\quad\text{iff}\quad I_A\leq I_B
$$
It is easily checked that the following rules apply to indicator functions (remember $\mod 2$):


*

*$I_{A^c}=1-I_A$ ($A^c:=X\backslash A$)

*$(I_{A})^2=I_A$

*$I_{A\cap B}=I_AI_B$

*$I_{A\cup B}=1-(1-I_A)(1-I_B)=I_A+I_B-I_AI_B$

*$I_{A-B}=I_A-I_AI_B$

*$I_{A\Delta B}=I_A+I_B-2I_AI_B=I_A+I_B$ 


($A\Delta B:=A\cup B-A\cap B=(A-B)\cup(B-A)$)
Everything else can be deduced from these simply verified rules.Therefore:


(a) $X\setminus(X\setminus A) = A\quad\Leftrightarrow\quad 1-(1-I_A)=I_A$.
(c) $A \subset B$ if and only if $X\setminus B \subset X\setminus A\quad\Leftrightarrow\quad $ $I_A\leq I_B$ iff $1-I_B\leq 1-I_A$.
(d) $X\setminus A \subset B$ if and only if $A \cup B = X\quad\Leftrightarrow\quad$ $1-I_A\leq I_B$ iff $I_A+I_B-I_AI_B=1$.
(e) $A \subset X\setminus B$ if and only if $A \cap B = \emptyset\quad\Leftrightarrow\quad$ $I_A\leq 1-I_B$ iff $I_AI_B=0$
(f) $A\setminus B = A\cap(X\setminus B)\quad\Leftrightarrow\quad$ $I_A-I_AI_B=I_A(1-I_B)$
(g) $X\setminus(A\setminus B) = B \cup(X\setminus A)\quad\Leftrightarrow\quad$ $1-(I_A-I_AI_B)=I_B+(1-I_A)-I_B(1-I_A)$. 


Only (d) and (e) is nontrivial (which is probably when the method is not most effective). But either case is immediate after further observations.  
It is also interesting to note that 3. and 6. together with $0:=I_\varnothing, 1:=I_X$ defines the so called Boolean ring (which in fact refers to the fact that any element is an idempotent by 2.).
