omarathon
Badges: 13
Rep:
?
#1
Report Thread starter 2 years ago
#1
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.

Name:  wat.jpg
Views: 42
Size:  87.8 KB

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
0
reply
univ4464
Badges: 10
Rep:
?
#2
Report 2 years ago
#2
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.
0
reply
omarathon
Badges: 13
Rep:
?
#3
Report Thread starter 2 years ago
#3
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?
0
reply
omarathon
Badges: 13
Rep:
?
#4
Report Thread starter 2 years ago
#4
I'm not sure about the other way round
0
reply
univ4464
Badges: 10
Rep:
?
#5
Report 2 years ago
#5
(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 = B or A = Cdoes not imply that A = B \cup C where A B and C are sets
0
reply
omarathon
Badges: 13
Rep:
?
#6
Report Thread starter 2 years ago
#6
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.
0
reply
omarathon
Badges: 13
Rep:
?
#7
Report Thread starter 2 years ago
#7
(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 = B or A = Cdoes not imply that A = 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
0
reply
univ4464
Badges: 10
Rep:
?
#8
Report 2 years ago
#8
(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 x \in \cup S \Rightarrow x \in A \in S_1 \text{ or } x \in B \in S_2 and then if it is in A then x \in \cup S_1 \Rightarrow x \in (\cup S_1) \cup (\cup S_2) and then the same thing for B
0
reply
omarathon
Badges: 13
Rep:
?
#9
Report Thread starter 2 years ago
#9
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?
0
reply
univ4464
Badges: 10
Rep:
?
#10
Report 2 years ago
#10
(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 x \in S_1 \Rightarrow x \in \cup S_1 Since \{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 x \in \cup S_1 then there exists an A \in S_1 such that x \in A. I think that is the trick you need to do to answer the question.

x \in A \in B does not imply that x \in B
2
reply
omarathon
Badges: 13
Rep:
?
#11
Report Thread starter 2 years ago
#11
(Original post by univ4464)
I don't think x \in S_1 \Rightarrow x \in \cup S_1 Since \{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 x \in \cup S_1 then there exists an A \in S_1 such that x \in A. I think that is the trick you need to do to answer the question.

x \in A \in B does not imply that x \in B
Thanks a lot, this makes perfect sense now!
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (644)
33.47%
Yes, I like the idea of applying to uni after I received my grades (PQA) (816)
42.41%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (378)
19.65%
I think there is a better option than the ones suggested (let us know in the thread!) (86)
4.47%

Watched Threads

View All