Trouble with proof of arbitrary family of sets Having a lot of trouble stringing together a proof for this.$B$ be arbitrary.Given index sets by $C_{b}=B-\{b\}$
(a)$\bigcup\limits_{b\in B}C_{b}=$?
(b)$\bigcap\limits_{b\in B}C_{b}=$?
(b)$\bigcap\limits_{b\in B}C_{b}=\emptyset$
Proof: $\bigcap\limits_{b\in B}C_{b}\supset\emptyset$
True vacuously since assuming an element in $\emptyset$ is false.
$\bigcap\limits_{b\in B}C_{b}\subset\emptyset$
Assume $a\in \bigcap\limits_{b\in B}C_{b}$.
Then $a \in C_{b}$ for all $b \in B$
$a \in B-\{b\}$ forall $b \in B$
$\uparrow$ Having trouble deciding what to do here.
part(a)If $|B|=0$ or $|B|=1$
then $\bigcup\limits_{b\in B}C_{b}= \emptyset$
$\bigcup\limits_{b\in B}C_{b}\supset \emptyset$ is trivial
$\bigcup\limits_{b\in B}C_{b}\subset \emptyset$
Assume $x \in \bigcup\limits_{b\in B}C_{b}$
then $x \in C_{b}$ for some $b \in B$
$\uparrow$ Having trouble deciding what to do here.
If $|B|>1$
then $\bigcup\limits_{b\in B}C_{b}=B$
Now I do not know how to finish the last part of this proof because I do not know where to start. Can I get some pointers?
 A: (b)
If $a\in B\setminus \{b\}$, it follows that $a\in B$.
But $a\in B - \{b\}$ for all $b\in B$, thus if $a\in B$, then $a\in B - \{a\}$. A contradiction.
(a)
If $|B|=0$ then $B=\emptyset$, since that's the only set without elements. Since there are no elements, there are also no $C_b$, and thus the union of those sets is the empty union (the union over the elements of the empty set). But the empty union gives the empty set.
If $|B|=1$, then $B$ has exactly one element $b_1$, that is, $B=\{b_1\}$.
Then there exists exactly one $C_b$, which is $C_{b_1} = B-\{b_1\} = \{b_1\}-\{b_1\} = \emptyset$. The union of just one set is the set itself, which is $C_{b_1}=\emptyset.$
If $|B|>1$, there exist at least two elements $b_1\ne b_2\in B$. Then we have
$\bigcup_{b\in B} C_b \supset C_{b_1}\cup C_{b_2}$. Now prove that the union on the right hand side is $B$. Together with $\bigcup_{b\in B} C_b \subset B$, which is easy to prove, you get the desired relation.
A: A.  If x in $\bigcap\limits_{b\in B}C_{b}$,
for all b in B, x in C$_b$.
Thus x in C$_x$, a contradiction.
Consequently $\bigcap\limits_{b\in B}C_{b}$ is empty.
B. For multipoint B, clearly $\bigcup\limits_{b\in B}C_{b}$ subset B.
If x in B, exists b in B with x in C$_b$,
Thus x in $\bigcup\limits_{b\in B}C_{b}\ $.
Consequently B subset $\bigcup\limits_{b\in B}C_{b}$.
