Turn on thread page Beta
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by generalebriety)
    If A is a proper subset of B, then there is an element of B that is not in A, but there is not necessarily a unique element of B that is not in A. "Unique" means "one, and only one".
    Yeah, I know my mistake of using unique as it means only one. I guess I need to be careful with my language.

    Thanks for the contrapositon.

    Couldn't you write
    A^c \subseteq B^c \Rightarrow A \cap B \not= A
    =>
    A^c \subseteq B^c \Rightarrow (A \cap B \subset A) or (A \cap B \supset A)

    Is this correct? it seems correct.
    Offline

    12
    ReputationRep:
    (Original post by Simplicity)
    Yeah, I know my mistake of using unique as it means only one. I guess I need to be careful with my language.

    Thanks for the contrapositon.

    Couldn't you write
    A^c \subseteq B^c \Rightarrow A \cap B \not= A
    =>
    A^c \subseteq B^c \Rightarrow (A \cap B \subset A) or (A \cap B \supset A)

    Is this correct? it seems correct.
    [On the second statement] For all sets  A \subset A \cap B is not possible (as all x in  A \cap B are in A, by definition, I think), but it's true for all sets that  A \cap B \subseteq A .
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    (Original post by Simplicity)
    A^c \subseteq B^c \Rightarrow A \cap B \not= A
    No. What does A^c even mean here? And even if we're in a system where taking complements makes sense, the opposite of A \subseteq B is not A^c \subseteq B^c. For example, the sets A = {1, 2}, B = {2, 3}.

    (Original post by Simplicity)
    =>
    A^c \subseteq B^c \Rightarrow (A \cap B \subset A) or (A \cap B \supset A)
    No. Where on earth are you getting this from? It's just not right, and I can't explain why, because it seems to have come from nowhere.

    You seem to be having a very big problem with set inclusion. It's not a total order relation. Given any two real numbers x and y, you know that x < y, x = y or x > y; you don't have a similar property with sets. Given two non-empty sets A and B, you know that A = B, \, A\subset B,\, B\subset A, A and B are disjoint, or A and B have non-empty intersection but A \neq A\cap B \neq B. And it gets worse when you start playing about with the empty set, because you can then have several of those holding true at once.

    So, your implication that "A is not equal to B" is the same thing as "A\cap B strictly contains A or is strictly contained in A" is simply false. It will never strictly contain A (because all of the elements in it are also in A); it might be strictly contained in A, but it might also be A itself.

    Edit: typo.
    Offline

    0
    ReputationRep:
    I have a large amount of respect for your patience.
    Offline

    16
    ReputationRep:
    (Original post by generalebriety)
    No. What does A^c even mean here? And even if we're in a system where taking complements makes sense, the opposite of A \subseteq B is not A^c \subseteq B^c. For example, the sets A = {1, 2}, B = {2, 3}.


    No. Where on earth are you getting this from? It's just not right, and I can't explain why, because it seems to have come from nowhere.

    You seem to be having a very big problem with set inclusion. It's not a total order relation. Given any two real numbers x and y, you know that x < y, x = y or x > y; you don't have a similar property with sets. Given two non-empty sets A and B, you know that A = B, \, A\subset B,\, B\subset A, A and B are disjoint, or A and B have non-empty intersection but A \neq A\cap B \neq B. And it gets worse when you start playing about with the empty set, because you can then have several of those holding true at once.

    So, your implication that "A is not equal to B" is the same thing as "A\cap B strictly contains A or is strictly contained in A" is simply false. It will never strictly contain A (because all of the elements in it are also in A); it might be strictly contained in A, but it might also be A itself.

    Edit: typo.
    wtf your example?
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    (Original post by Totally Tom)
    wtf your example?
    There are situations where A \not \subseteq B and A^c \not \subseteq B^c.

    Not very well worded, I agree. But Simplicity attempted to negate the sentence A \subseteq B with A^c \subseteq B^c.
    • Thread Starter
    Offline

    15
    ReputationRep:
    Yeah, I see the error in thinking.

    I was trying to generalise this.
    A=B
    So to prove that generally you show
    A \subseteq B and A \supseteq B
    =>
     A=B

    However, yeah I see you can't generalise it to what I'm doing.

    (Original post by generalebriety)
    No. What does even mean here? And even if we're in a system where taking complements makes sense, the opposite of is not . For example, the sets A = {1, 2}, B = {2, 3}.
    I don't know what you mean. The book has x \not \in A \Rightarrow x \in A^c, yeah its the complement. So I don't see why this is wrong
     x \not \in A \Rightarrow x \not \in B
    =>
     A^c \subseteq B^c

    Surely, the book has
    A \subseteq B \Leftrightarrow B^c \subseteq A^c.

    So I don't see the problem with that. Although, I do see the problem of splitting not equal into two subsets as that doesn't make sense.
    • Wiki Support Team
    Offline

    14
    ReputationRep:
    Wiki Support Team
    (Original post by Simplicity)
    I don't know what you mean. The book has x \not \in A \Rightarrow x \in A^c, yeah its the complement.
    The complement doesn't always make sense, though - we have to be working in a given system for it to make sense. Like in a probabilistic situation, for instance. What's the complement of {12, pi, Q, monkey}? Or is this work on set theory purely limited to stuff we already know? Because if so, it's a bit pointless.

    (Original post by Simplicity)
     x \not \in A \Rightarrow x \not \in B
    This isn't true, though. You were trying to negate the sentence "A is contained in B", i.e. "for all x in A, x is in B". The negation is simply "A is not contained in B", i.e. "there is some x in A such that x is not in B".

    (Original post by Simplicity)
    Surely, the book has
    A \subseteq B \Leftrightarrow B^c \subseteq A^c.
    That's correct.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by generalebriety)
    The complement doesn't always make sense, though - we have to be working in a given system for it to make sense. Like in a probabilistic situation, for instance. What's the complement of {12, pi, Q, monkey}? Or is this work on set theory purely limited to stuff we already know? Because if so, it's a bit pointless.
    Yeah, I do get what your saying. Okay. I guess you would have to specify a set U.

    (Original post by generalebriety)
    This isn't true, though. You were trying to negate the sentence "A is contained in B", i.e. "for all x in A, x is in B". The negation is simply "A is not contained in B", i.e. "there is some x in A such that x is not in B".
    Yeah, I get what your saying. I should really put on a dunce hat for being so stupid.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: July 24, 2009

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.