anonymous
  • anonymous
Help with a proof please.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
She lied, no solution in the book...grrrr
perl
  • perl
I think you can use the definition of general union of sets

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anonymous
  • anonymous
okay, but not sure what to do with that. Tbh im not even sure exactly what the problem states
anonymous
  • 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
  • perl
Let's do a concrete example. Let \( I= \{1,2,3\} \) , that is the indexing set
perl
  • 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
  • anonymous
okay
perl
  • perl
now let's use use the definition of \( \nsubseteq\)
perl
  • perl
\(\Large X \nsubseteq Y \iff \exists x \in X , x \cancel \in Y \)
anonymous
  • anonymous
okay
anonymous
  • anonymous
so how does that imply there exists some j in I
perl
  • perl
sorry I am experiencing lag
anonymous
  • anonymous
its okay
perl
  • 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
  • anonymous
okay
perl
  • perl
but its probably easier to use the definition of general union
anonymous
  • anonymous
what ever works, but how does this imply the j in I or did I miss something
anonymous
  • anonymous
is it because x in all A is not in any B therefore any j in I will satisfy the conclusion
anonymous
  • anonymous
for any j, x will not be in B right? that's what you showed correct?
anonymous
  • anonymous
okay
anonymous
  • anonymous
okay
perl
  • 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
  • anonymous
hm that notation kinda confuses me. give me a sec while I try to think about it
perl
  • perl
yes its a bit odd. let me try to simplify it
anonymous
  • anonymous
okay
perl
  • perl
we can use the idea here http://en.wikipedia.org/wiki/Union_%28set_theory%29#Arbitrary_unions
anonymous
  • anonymous
the idea of the union or the notation?
anonymous
  • anonymous
oh that really helps
perl
  • perl
i used different variables since they are independent of each other
perl
  • 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
  • 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
  • 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
  • perl
right
perl
  • perl
if x is not in at least one Bi, then it is not in any Bi
anonymous
  • anonymous
okay sweet I get it now!!
anonymous
  • anonymous
I would give you more than one medal if I could lol Thanks for your time perl
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
  • perl
note that `:` means `such that` here
perl
  • perl
let me make this even more clear
anonymous
  • anonymous
lol okay
Loser66
  • 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
  • Loser66
That simplifies the proof a lot.
perl
  • 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
  • perl
@Loser66 you can try , see if you can complete a proof that way.
perl
  • perl
the only comment i might add is , I set i = j, since you are guaranteed the existence of an i
anonymous
  • 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
  • perl
Someone might find a shorter proof , however I think this is a solid proof.
anonymous
  • anonymous
I was just worried about understanding it since my final is tomorrow lol
Loser66
  • 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
  • Loser66
I don't know why but it's quite trivial to me.

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