Help with a proof please.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

Help with a proof please.

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

##### 1 Attachment

- anonymous

She lied, no solution in the book...grrrr

- perl

I think you can use the definition of general union of sets

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

okay, but not sure what to do with that. Tbh im not even sure exactly what the problem states

- anonymous

the union of a to some i is not a subset of the union of b to some i: not to sure what this means

- perl

Let's do a concrete example. Let \( I= \{1,2,3\} \) , that is the indexing set

- perl

Let \( I= \{1,2,3\} \)
hypothesis:
$$A_1 \cup A_2 \cup A_3 \nsubseteq B_1 \cup B_2 \cup B_3$$

- anonymous

okay

- perl

now let's use use the definition of \( \nsubseteq\)

- perl

\(\Large X \nsubseteq Y \iff \exists x \in X , x \cancel \in Y \)

- anonymous

okay

- anonymous

so how does that imply there exists some j in I

- perl

sorry I am experiencing lag

- anonymous

its okay

- perl

Let \( I= \{1,2,3\} \)
hypothesis:
\[\large{
A_1 \cup A_2 \cup A_3 \nsubseteq B_1 \cup B_2 \cup B_3
\\ \iff\\
\exists x \in A_1 \cup A_2 \cup A_3 ~,~ x \not\in B_1 \cup B_2 \cup B_3
\\ \iff\\
\exists x \in A_1 \mathrm{~ or~ } \exists x \in A_2 \mathrm{~or~}
\exists x \in A_3 \mathrm{~and~}
\\~ x \not\in B_1 \mathrm{~and~}
x \not\in B_2 \mathrm{~and~}
x \not\in B_3
}
\]

- anonymous

okay

- perl

but its probably easier to use the definition of general union

- anonymous

what ever works, but how does this imply the j in I or did I miss something

- anonymous

is it because x in all A is not in any B therefore any j in I will satisfy the conclusion

- anonymous

for any j, x will not be in B right? that's what you showed correct?

- anonymous

okay

- anonymous

okay

- perl

We can use the definition of union of indexed sets
\[
\displaystyle \bigcup_{i \mathop \in I} A_i
\nsubseteq \bigcup_{i \mathop \in I} B_i
\\ \iff \\
\exists x \in \bigcup_{i \mathop \in I} A_i ~, ~ x \not \in \bigcup_{i \mathop \in I} B_i
\\ \iff \\
\exists x \in \left\{{y: \exists i \in I: y \in A_i}\right\} ~, ~ x \not \in
\left\{{z: \exists i \in I: z \in B_i}\right\}
\]

- anonymous

hm that notation kinda confuses me. give me a sec while I try to think about it

- perl

yes its a bit odd. let me try to simplify it

- anonymous

okay

- perl

we can use the idea here http://en.wikipedia.org/wiki/Union_%28set_theory%29#Arbitrary_unions

- anonymous

the idea of the union or the notation?

- anonymous

oh that really helps

- perl

i used different variables since they are independent of each other

- perl

This is a simpler definition
\[
\displaystyle \bigcup_{i \mathop \in I} A_i=\left\{{\textrm{x in at least one } A_i}\right\}
\\~\\
\bigcup_{i \mathop \in I} B_i = \left\{{\textrm{x in at least one } B_i}\right\}
\]

- perl

\[
\displaystyle \bigcup_{i \mathop \in I} A_i
\nsubseteq \bigcup_{i \mathop \in I} B_i
\\ \iff \\
\exists x \in \bigcup_{i \mathop \in I} A_i ~, ~ x \not \in \bigcup_{i \mathop \in I} B_i
\\ \iff \\
\exists x \textrm{ such that} ~x \in \left\{{\textrm{y in at least one } A_i}\right\} ~, ~ x \not \in
\left\{ \textrm{z in at least one } B_i \right\}
\]

- anonymous

so this implies that since for all i we have this then there must be some j where the same thing is true rihgt?

- perl

right

- perl

if x is not in at least one Bi, then it is not in any Bi

- anonymous

okay sweet I get it now!!

- anonymous

I would give you more than one medal if I could lol
Thanks for your time perl

- perl

I think this is fine though
We know the definition of big union:
$$\displaystyle \bigcup_{i \mathop \in I} A_i := \left\{{x~ | ~ \exists i \in I: x \in A_i}\right\}$$Therefore it follows
\[
\displaystyle \bigcup_{i \mathop \in I} A_i
\nsubseteq \bigcup_{i \mathop \in I} B_i
\\ \iff \\
\exists x \in \bigcup_{i \mathop \in I} A_i ~, ~ x \not \in \bigcup_{i \mathop \in I} B_i
\\ \iff \\
\exists j \in I: x \in A_j ~, \neg ~\left( \exists i \in I: x \in B_i \right )
\\ \iff \\
\exists j \in I: x \in A_j ~, ~ \forall i \in I ~ x \not \in B_i
\]

- perl

note that `:` means `such that` here

- perl

let me make this even more clear

- anonymous

lol okay

- Loser66

@perl why can't we label them as \( \displaystyle \bigcup_{i \mathop \in I} A_i := A\)
and \(\displaystyle \bigcup_{i \mathop \in I} B_i := B\)

- Loser66

That simplifies the proof a lot.

- perl

\[\displaystyle \bigcup_{i \mathop \in I} A_i := \left\{{x~ | ~ \exists i \in I: x \in A_i}\right\}\]Therefore it follows
\[
\displaystyle \bigcup_{i \mathop \in I} A_i
\nsubseteq \bigcup_{i \mathop \in I} B_i
\\ \iff \\
\exists x \in \bigcup_{i \mathop \in I} A_i ~, ~ x \not \in \bigcup_{i \mathop \in I} B_i
\\ \iff \\
\exists j \in I: x \in A_j ~, \neg ~\left( \exists i \in I: x \in B_i \right )
\\ \iff \\
\exists j \in I: x \in A_j ~, ~ \forall i \in I ~ x \not \in B_i
\\ \iff \\
\forall i \in I ~ \left( ~\exists j \in I: x \in A_j ~ , x \not \in B_i ~\right)
\\ \iff \\
\forall i \in I ~ \left( A_j \nsubseteq B_i ~\right)
\]

- perl

@Loser66 you can try , see if you can complete a proof that way.

- perl

the only comment i might add is , I set i = j, since you are guaranteed the existence of an i

- anonymous

haha thanks a million for this.
okay so im going to keep this question open so that I can refer to it later when I try the problem again on my own

- perl

Someone might find a shorter proof , however I think this is a solid proof.

- anonymous

I was just worried about understanding it since my final is tomorrow lol

- Loser66

\(x\in A, x\notin B\)
\(x\in A\) that is \(x \in A_i\) for some \(i\in I\)
\(x\notin B\) that is \(x\notin B_i\) \(\forall i\in I\)

- Loser66

I don't know why but it's quite trivial to me.

Looking for something else?

Not the answer you are looking for? Search for more explanations.