# 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.

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