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Sets of sets question help

Hi,

I have been doing questions from this Foundations of Mathematics book and I have absolutely no idea how to do a lot of them. Anyhow question 6 is the one I want to know how to do as I find sets interesting.

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can anyone give me some hints in the right direction? Are s1 and s2 both sets of sets? It hasn't specified but is this to be assumed?

thanks
(edited 5 years ago)
Prove that if an element is in the left hand side then it is in the right hand side so, the left hand side is contained in the right hand side, and then do the same thing the other way around. These questions are usually just looking at the definitions.
Reply 2
Let x be an element of S

so x is an element of S1 OR x is an element of S2 because S = S1 u S2

So, U(S) = U(S1) OR U(S) = U(S2) therefore U(S) = U(S1) u U(S2)


Is this sufficient... I feel like this isn't right?
(edited 5 years ago)
Reply 3
I'm not sure about the other way round
Original post by omarathon
Let x be an element of S

so x is an element of S1 OR x is an element of S2 because S = S1 u S2

So, U(S) = U(S1) OR U(S) = U(S2) therefore U(S) = U(S1) u U(S2)


Is this sufficient... I feel like this isn't right?


I think the U(S) notation is specific to the book you're using, so I don't know what it means, what does it mean?

If you want to be super pedantic you have to make sure the sets are not empty, but if its quite easy to check the result is true if s1 and s2 are empty, since the empty set is unique.

I don't think your last line correct. You are trying to prove that U(S) = U(S1) U (US2) but you have stated something that is not true in general, A=BA = B or A=CA = Cdoes not imply that A=BCA = B \cup C where A B and C are sets
Reply 5
let x be element of U(S1) u U(S2)

so x is an element of U(S1) or U(S2)

how do I go from that if x is an element of U(S1), x is an element of S1? Because the elements of U(S1) differ from the elements of S1.
Reply 6
Original post by univ4464
I think the U(S) notation is specific to the book you're using, so I don't know what it means, what does it mean?

If you want to be super pedantic you have to make sure the sets are not empty, but if its quite easy to check the result is true if s1 and s2 are empty, since the empty set is unique.

I don't think your last line correct. You are trying to prove that U(S) = U(S1) U (US2) but you have stated something that is not true in general, A=BA = B or A=CA = Cdoes not imply that A=BCA = B \cup C where A B and C are sets


The U notation is the union of all sets within a set, so if S = {{1,2,3}, {3,4,5}, {5,6,7}} then U(S) = {1,2,3,4,5,6,7}.

ok so I am not sure what to do if the last line isn't true
Original post by omarathon
The U notation is the union of all sets within a set, so if S = {{1,2,3}, {3,4,5}, {5,6,7}} then U(S) = {1,2,3,4,5,6,7}.

ok so I am not sure what to do if the last line isn't true


I think you could start by doing something like this xSxAS1 or xBS2x \in \cup S \Rightarrow x \in A \in S_1 \text{ or } x \in B \in S_2 and then if it is in AA then xS1x(S1)(S2)x \in \cup S_1 \Rightarrow x \in (\cup S_1) \cup (\cup S_2) and then the same thing for BB
Reply 8
thanks but can you explain that if x belongs to S1 this means that x belongs to U(S1)? S1 is a set of sets and U(S1) is a set of elements which are in the sets embedded in S1, so S1 doesn't exist in U(S1) since U(S1) doesn't contain sets? I'm probably over-complicating it since it's obvious why if x belongs so a set Y then x also belongs to u(Y) but the capital U makes it strange. Also why do you introduce A and B, if they just translate to S1 and S2 respectively? Is this a convention?
(edited 5 years ago)
Original post by omarathon
thanks but can you explain that if x belongs to S1 this means that x belongs to U(S1)? S1 is a set of sets and U(S1) is a set of elements which are in the sets embedded in S1, so S1 doesn't exist in U(S1) since U(S1) doesn't contain sets? I'm probably over-complicating it since it's obvious why if x belongs so a set Y then x also belongs to u(Y) but the capital U makes it strange. Also why do you introduce A and B, if they just translate to S1 and S2 respectively? Is this a convention?


I don't think xS1xS1x \in S_1 \Rightarrow x \in \cup S_1 Since {1}{{1},{2,3}} but {1}{1,2,3}={{1},{2,3}}\{1\} \in \{\{1\}, \{2, 3\}\} \text{ but } \{1\} \notin \{1, 2, 3\} = \cup \{\{1\}, \{2, 3\}\}

I should have probably been more clear with the sets A and B. What I meant was that if xS1x \in \cup S_1 then there exists an AS1A \in S_1 such that xAx \in A. I think that is the trick you need to do to answer the question.

xABx \in A \in B does not imply that xBx \in B
(edited 5 years ago)
Reply 10
Original post by univ4464
I don't think xS1xS1x \in S_1 \Rightarrow x \in \cup S_1 Since {1}{{1},{2,3}} but {1}{1,2,3}={{1},{2,3}}\{1\} \in \{\{1\}, \{2, 3\}\} \text{ but } \{1\} \notin \{1, 2, 3\} = \cup \{\{1\}, \{2, 3\}\}

I should have probably been more clear with the sets A and B. What I meant was that if xS1x \in \cup S_1 then there exists an AS1A \in S_1 such that xAx \in A. I think that is the trick you need to do to answer the question.

xABx \in A \in B does not imply that xBx \in B


Thanks a lot, this makes perfect sense now!

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