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

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1. anonymous

2. anonymous

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

3. perl

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

4. anonymous

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

5. 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

6. perl

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

7. perl

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

8. anonymous

okay

9. perl

now let's use use the definition of $$\nsubseteq$$

10. perl

$$\Large X \nsubseteq Y \iff \exists x \in X , x \cancel \in Y$$

11. anonymous

okay

12. anonymous

so how does that imply there exists some j in I

13. perl

sorry I am experiencing lag

14. anonymous

its okay

15. perl

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

16. anonymous

okay

17. perl

but its probably easier to use the definition of general union

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

okay

22. anonymous

okay

23. perl

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

24. anonymous

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

25. perl

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

26. anonymous

okay

27. perl

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

28. anonymous

the idea of the union or the notation?

29. anonymous

oh that really helps

30. perl

i used different variables since they are independent of each other

31. perl

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

32. perl

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

33. 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?

34. perl

right

35. perl

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

36. anonymous

okay sweet I get it now!!

37. anonymous

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

38. 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 $\displaystyle \bigcup_{i \mathop \in I} A_i \nsubseteq \bigcup_{i \mathop \in I} B_i \\ \iff \\ \exists x \in \bigcup_{i \mathop \in I} A_i ~, ~ x \not \in \bigcup_{i \mathop \in I} B_i \\ \iff \\ \exists j \in I: x \in A_j ~, \neg ~\left( \exists i \in I: x \in B_i \right ) \\ \iff \\ \exists j \in I: x \in A_j ~, ~ \forall i \in I ~ x \not \in B_i$

39. perl

note that : means such that here

40. perl

let me make this even more clear

41. anonymous

lol okay

42. Loser66

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

43. Loser66

That simplifies the proof a lot.

44. perl

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

45. perl

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

46. perl

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

47. 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

48. perl

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

49. anonymous

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

50. 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 $$x\notin B_i$$ $$\forall i\in I$$

51. Loser66

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