# Proving set identities

I am attempting to work on some proofs for my math assignment, but I'll be honest in that I am really struggling to understand them. I read through the power point given by my teacher; however, even after asking for help I'm not really understanding why the proofs work the way they do. I know this might be a lot to ask, but I would love if someone could maybe help me work out these problems or give me some guidance on how to approach them.

Let $A,B$ and $C$ be sets. Show that

a) $(A\cup B)\subseteq (A\cup B \cup C)$
b) $(A\cap B \cap C)\subseteq (A\cap B)$
c) $(A- B)-C\subseteq A-C$
d) $(A- C)\cap (C-B)=\emptyset$
e) $(B-A)\cup (C-A)=(B\cup C)-A$

I am on the first one still I currently have: $x \in A \cup B$, so $x \in A$ or $x \in B$.

• you can make a start by considering an arbitrary element of left set and show that it is or isn't contained in the right set – danimal May 21 '15 at 16:14
• To show two sets are equal: Prove $S \subseteq T$ and $T \subseteq S$. To show something is a subset: Assume $x \in S$ and then prove that $x \in T$ also. This is the standard way of proving subset relations. – user222031 May 21 '15 at 16:15
• Venn Diagrams can be helpful guides for figuring out how the proofs will work. With a little formalization, Venn Diagrams can essentially be proof! – Bob Krueger May 21 '15 at 17:07

(a): Suppose $x\in A\cup B$. Then $x\in A$ or $x\in B$. If $x\in A$, then $x\in A\cup B\cup C$. If, however, $x\in B$, then $x\in A\cup B\cup C$. Either way, $x\in A\cup B\cup C$ when $x\in A\cup B$. Thus, $A\cup B\subseteq A\cup B\cup C$.
(b): Suppose $x\in A\cap B\cap C$. Then $x\in A$ and $x\in B$ and $x\in C$. Hence, we have that $x\in A$ and $x\in B$; that is, $x\in A\cap B$. Thus, $A\cap B\cap C\subseteq A\cap B$.
(c): Suppose $x\in (A-B)-C$. Then $x\in A-B$ and $x\not\in C$. That is, $x\in A$ and $x\not\in B$ and $x\not\in C$. Hence, $x\in A$ and $x\not\in B$; that is $x\in A-C$. Thus, $(A-B)-C\subseteq A-C$.
(d): Using some set algebra (you can use element-chasing proofs as above, but I would advise against this when simple set algebra can take care of everything), where $S^C$ denotes the complement of the set $S$, we have the following: \begin{align} (A-C)\cap(C-B) &= (A\cap C^C)\cap(C\cap B^C)\tag{definition}\\[0.5em] &= (C^C\cap C)\cap(A\cap B^C)\tag{associativity}\\[0.5em] &= \varnothing\cap(A\cap B^C)\tag{$S-S=\varnothing$}\\[0.5em] &= \varnothing. \end{align} (e): \begin{align} (B-A)\cup(C-A) &= (B\cap A^C)\cup(C\cap A^C)\tag{definition}\\[0.5em] &= (B\cup C)\cap A^C\tag{distributivity}\\[0.5em] &= (B\cup C)-A\tag{definition} \end{align}