A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Help with a proof please.

  • This Question is Open
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  7. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  9. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  10. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so how does that imply there exists some j in I

  13. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    sorry I am experiencing lag

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    its okay

  15. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  17. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    but its probably easier to use the definition of general union

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  23. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  25. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

  27. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the idea of the union or the notation?

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh that really helps

  30. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i used different variables since they are independent of each other

  31. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  33. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    right

  35. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay sweet I get it now!!

  37. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  38. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    note that `:` means `such that` here

  40. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    let me make this even more clear

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol okay

  42. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  43. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That simplifies the proof a lot.

  44. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  45. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  46. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  47. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  49. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  50. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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 \(x\notin B_i\) \(\forall i\in I\)

  51. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.