Question

Help with a proof please.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

okay

Answer 8

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

Answer 9

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

Answer 10

okay

Answer 11

so how does that imply there exists some j in I

Answer 12

sorry I am experiencing lag

Answer 13

its okay

Answer 14

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


}

\]

Answer 15

okay

Answer 16

but its probably easier to use the definition of general union

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

okay

Answer 21

okay

Answer 22

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\}




\]

Answer 23

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

Answer 24

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

Answer 25

okay

Answer 26

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

Answer 27

the idea of the union or the notation?

Answer 28

oh that really helps

Answer 29

i used different variables since they are independent of each other

Answer 30

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\}





\]

Answer 31

\[
\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\}

\]

Answer 32

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

Answer 33

right

Answer 34

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

Answer 35

okay sweet I get it now!!

Answer 36

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

Answer 37

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


\]

Answer 38

note that `:` means `such that` here

Answer 39

let me make this even more clear

Answer 40

lol okay

Answer 41

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

Answer 42

That simplifies the proof a lot.

Answer 43

\[\displaystyle \bigcup_{i \mathop \in I} A_i := \left\{{x~ | ~ \exists i \in I: x \in A_i}\right\}\]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)


\]

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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