Proof verification of union/intersection of family of sets

by user707991   Last Updated October 17, 2019 18:20 PM

Can I please get a proof verification or a showing of how to prove these set equalities correctly?

Let $B$ be an arbitrary set.$C_{b}=B-\{b\}$


(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 $\forall B$, $\bigcap\limits_{b\in B}C_{b}=\emptyset$


If $B=\emptyset$

Then $\bigcup\limits_{b\in B}\emptyset= \emptyset$

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 $b\in B$.


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 $y\in \bigcup\limits_{b\in B}C_{b}$


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 $\bigcap\limits_{b\in B}C_{b}= \emptyset$

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