Proof verification for union/intersection of indexed set families Can someone tell me if this proof is correct?Thanks
Can I please get a proof verification or a showing of how to prove these set equalities correctly?
Let $B$ be an arbitrary set.$C_{b}=B-\{b\}$ is a family of sets
Prove:
(1)If $B=\emptyset$ or $|B|=1$,
$\bigcup\limits_{b\in B}C_{b}= \emptyset$,
(2)Prove if $|B|>1$,$\bigcup\limits_{b\in B}C_{b}=B$,
(3)Prove $\forall B$, $\bigcap\limits_{b\in B}C_{b}=\emptyset$
Proof(1):
case1:If $B=\emptyset$ 
Then $\bigcup\limits_{b\in B}\emptyset= \emptyset$
case2:If $|B|=1$,
$\bigcup\limits_{b\in B}C_{b}= \emptyset$
Assume By way of contradiction(BWOC) $y \in C_b$ for some $b\in B$
then $y \in B-\{y\}$ and $B=\{y\}$ Since $|B|=1$
this contradicts the assumption that $y \in C_b$ for some $b\in B$.
Proof(2):
case3:Assume $|B|>1$ then $\bigcup\limits_{b\in B}C_{b}= B$
Since $C_b \subset B$ for all $b$ this is proved.
Assume $y\in B$
let $b\neq y$ then $y \in C_b$ for $b\neq y$
thus $y\in \bigcup\limits_{b\in B}C_{b}$
Proof(3):
Prove $\bigcap\limits_{b\in B}C_{b}=\emptyset$
Assume BWOC $y \in \bigcap\limits_{b\in B}C_{b}$
since $y \in C_b$ for all $b$, and $y\in B$,
$y \in B-\{y\}$ this contradicts the assumption so $\bigcap\limits_{b\in B}C_{b}= \emptyset$

Basically it is correct. 
But not very well written.
at (1):
I do not understand why you do a proof by contradiction. It makes the proof a little bit 'confusing' in my opinion.
Alternative proof:
If $B=\emptyset$, then $C_b=\emptyset$. Hence $\bigcup_{b\in B} C_b=\emptyset$.
Note that this statement would always be true and does not depend on $C_b$. Because $B$ is empty the condition $b\in B$ never holds, so the union is empty.
If $|B|=1$, then $B=\{b\}$ and $C_b=\emptyset$. So $\bigcup_{b\in\{b\}} C_b=\emptyset$.
These proofs are trivial, but your proof by contradiction makes it more complicated as it is. At least in my opinion.
at (2):
You want to show that $\bigcup_{b\in B} C_b=B$.
The proof of "$\subseteq$" is good.
When you proof "$\supseteq$" it is again not so well written.
Let $y\in B$. We have to show $y\in\bigcup_{b\in B}$. Since $|B|>1$ there is some $x\in B$ with $x\neq y$. So $y\in C_x=B\setminus\{x\}$. Thus $y\in\bigcup_{b\in B} C_b$.
It is not clear where the fact $|B|>1$ is used, or at least not mentioned clearly.
at (3):
I like your proof. 
Maybe you should drop the BWOC and just write "Assume ...". This already indicates a proof by contradiction. 
