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anonymous
 one year ago
Help with a proof please.
anonymous
 one year ago
Help with a proof please.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0She lied, no solution in the book...grrrr

perl
 one year ago
Best ResponseYou've already chosen the best response.2I think you can use the definition of general union of sets

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, but not sure what to do with that. Tbh im not even sure exactly what the problem states

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.2Let's do a concrete example. Let \( I= \{1,2,3\} \) , that is the indexing set

perl
 one year ago
Best ResponseYou've already chosen the best response.2Let \( I= \{1,2,3\} \) hypothesis: $$A_1 \cup A_2 \cup A_3 \nsubseteq B_1 \cup B_2 \cup B_3$$

perl
 one year ago
Best ResponseYou've already chosen the best response.2now let's use use the definition of \( \nsubseteq\)

perl
 one year ago
Best ResponseYou've already chosen the best response.2\(\Large X \nsubseteq Y \iff \exists x \in X , x \cancel \in Y \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so how does that imply there exists some j in I

perl
 one year ago
Best ResponseYou've already chosen the best response.2sorry I am experiencing lag

perl
 one year ago
Best ResponseYou've already chosen the best response.2Let \( I= \{1,2,3\} \) hypothesis: \[\large{ A_1 \cup A_2 \cup A_3 \nsubseteq B_1 \cup B_2 \cup B_3 \\ \iff\\ \exists x \in A_1 \cup A_2 \cup A_3 ~,~ x \not\in B_1 \cup B_2 \cup B_3 \\ \iff\\ \exists x \in A_1 \mathrm{~ or~ } \exists x \in A_2 \mathrm{~or~} \exists x \in A_3 \mathrm{~and~} \\~ x \not\in B_1 \mathrm{~and~} x \not\in B_2 \mathrm{~and~} x \not\in B_3 } \]

perl
 one year ago
Best ResponseYou've already chosen the best response.2but its probably easier to use the definition of general union

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what ever works, but how does this imply the j in I or did I miss something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it because x in all A is not in any B therefore any j in I will satisfy the conclusion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for any j, x will not be in B right? that's what you showed correct?

perl
 one year ago
Best ResponseYou've already chosen the best response.2We 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
 one year ago
Best ResponseYou've already chosen the best response.0hm that notation kinda confuses me. give me a sec while I try to think about it

perl
 one year ago
Best ResponseYou've already chosen the best response.2yes its a bit odd. let me try to simplify it

perl
 one year ago
Best ResponseYou've already chosen the best response.2we can use the idea here http://en.wikipedia.org/wiki/Union_%28set_theory%29#Arbitrary_unions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the idea of the union or the notation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh that really helps

perl
 one year ago
Best ResponseYou've already chosen the best response.2i used different variables since they are independent of each other

perl
 one year ago
Best ResponseYou've already chosen the best response.2This 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
 one year ago
Best ResponseYou've already chosen the best response.2\[ \displaystyle \bigcup_{i \mathop \in I} A_i \nsubseteq \bigcup_{i \mathop \in I} B_i \\ \iff \\ \exists x \in \bigcup_{i \mathop \in I} A_i ~, ~ x \not \in \bigcup_{i \mathop \in I} B_i \\ \iff \\ \exists x \textrm{ such that} ~x \in \left\{{\textrm{y in at least one } A_i}\right\} ~, ~ x \not \in \left\{ \textrm{z in at least one } B_i \right\} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so this implies that since for all i we have this then there must be some j where the same thing is true rihgt?

perl
 one year ago
Best ResponseYou've already chosen the best response.2if x is not in at least one Bi, then it is not in any Bi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay sweet I get it now!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would give you more than one medal if I could lol Thanks for your time perl

perl
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2note that `:` means `such that` here

perl
 one year ago
Best ResponseYou've already chosen the best response.2let me make this even more clear

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0That simplifies the proof a lot.

perl
 one year ago
Best ResponseYou've already chosen the best response.2\[\displaystyle \bigcup_{i \mathop \in I} A_i := \left\{{x~  ~ \exists i \in I: x \in A_i}\right\}\]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
 one year ago
Best ResponseYou've already chosen the best response.2@Loser66 you can try , see if you can complete a proof that way.

perl
 one year ago
Best ResponseYou've already chosen the best response.2the only comment i might add is , I set i = j, since you are guaranteed the existence of an i

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha 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
 one year ago
Best ResponseYou've already chosen the best response.2Someone might find a shorter proof , however I think this is a solid proof.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was just worried about understanding it since my final is tomorrow lol

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(x\in A, x\notin B\) \(x\in A\) 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
 one year ago
Best ResponseYou've already chosen the best response.0I don't know why but it's quite trivial to me.
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