Proving Set Identities

Proving Set Identities
$\begingroup$

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 $x \in B$.

$\endgroup$
3

1 Answer

$\begingroup$

(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}

$\endgroup$
3

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

James H. Sterling
Author

James H. Sterling

James Sterling reports on renewable energy developments, climate policy, ecological conservation, and green tech innovations around the globe.