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


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


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 $b\in B$.


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


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.

  • $\begingroup$ Thank you so much for the response $\endgroup$ – user707991 Oct 18 '19 at 0:29

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