Results 1 to 10 of 10

Math Help - A set in a set?

  1. #1
    Newbie
    Joined
    Jan 2006
    Posts
    6

    A set in a set?

    Hi!

    I've just taken on a course in discrete mathematics, i think im doing fairly well but there is one thing i really don't understand about sets and i thought maybe someone here could help me out with it.

    A= {a,b,c}
    B= {a,b,g}
    C= {c,d,e,f}
    G= {a,b,c,d,e,f,g}

    {a,b,{g}} = B true/false?

    What does this last line mean? (The left side of the =)

    I can't find it in my litterature either. :/

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    538
    Thanks
    11
    (Well, sets may contain other sets, that's not very upsetting - you 'll be used to it in no time)


    When writing {a,b,{g}}, you have formed a set with elements a,b and {g} - careful, not just g, but the one-element set {g} (called a "singleton set").

    So, writing "a belongs to {a,b,{g}}" is right, but "g belongs to {a,b,{g}}" is wrong; Actually {g} belongs to {a,b,{g}}.

    Try now to show that {a,b,{g}} is not equal to B, keeping the singleton in mind!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2006
    Posts
    6
    Uhhm, ok. Here goes nothing.

    Since {g} is not equal to g, {a,b,g} is not equal to {a,b,{g}} and therefore, {a,b,{g}} is not equal to B?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Equality of sets means they share the same elements,
    and as you said \{g\}=g is false. The sets are not equal.

    Another method to use is given two sets S,S', if you can show S\subseteq S' \mbox{ and }S'\subseteq S then S=S'.
    Thus, show that each set is a subset of each other. Which is not true here.
    Last edited by ThePerfectHacker; January 27th 2006 at 07:25 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2006
    Posts
    6
    B= {a,b,g}


    {{b}} belongs to B
    {{b}} is a subset of B
    {{b}} is a subset of the powerset of B

    I hope i understood everything correctly.
    According to "what i know" i would say that all of the above statements are false, is that correct?
    If it would be {b} instead of {{b}}, they would all be true?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    538
    Thanks
    11
    B= {a,b,g}


    {{b}} belongs to B
    {{b}} is a subset of B
    {{b}} is a subset of the powerset of B
    For this set B,

    - b belongs to B (drop curly brackets once to obtain elements)

    - {b} is a subset of B (enclosing elements in brackets, forms a subset)

    - {{b}} is a subset of the powerset
    P(B)={ {a},{b},{g},{a,b},{a,g}{b,g}{a,b,g} }

    and, what is the relation of {b} tp P(B)?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2006
    Posts
    6
    {b} is a proper subset of P(B) as far as i know.

    So {{b}} is a subset of P(B) because the element of {{b}}, {b} that is, is included in P(B) and therefore, {{b}} is a subset of P(B)?

    Just checking, P(B) stands for "the powerset of B" right?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2005
    From
    Wethersfield, CT
    Posts
    92
    Hi Rebesques:

    I suspect this was just an oversight (you obviously know your stuff) but,
    nP(S) = 2^(nP(S)) for any set, S, where n denotes "cardinality of". Thus
    nP(B) = 2^nB = 2^3 = 8. You omitted {} from the power set.

    Regards,

    Rich B.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    538
    Thanks
    11
    {b} is a proper subset of P(B) as far as i know.

    So {{b}} is a subset of P(B) because the element of {{b}}, {b} that is, is included in P(B) and therefore, {{b}} is a subset of P(B)?

    Just checking, P(B) stands for "the powerset of B" right?
    Everything is ok, except from the first sentence. it is {{b}} that is a proper (containing less!) subset of P(B).

    =========================================
    You omitted {} from the power set.
    Yeah you are right Rich. I forget stuff all the time...
    (better it be the empty set than my wallet )
    Last edited by Rebesques; January 27th 2006 at 05:41 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,966
    Thanks
    350
    Awards
    1
    In relation to all this, just WHY can we include {g} in a set along with the elements a and b? I realize we CAN do this, but doesn't it create some sort of logical equivalence of the "nature" (for lack of a better word) of the element a to the nature of the set {g}? To say an element and a set have some feature in common bothers me. It would seem to me that all the elements of a set would need to have the same "nature." (And that way we don't run into things like Russell's paradox...we couldn't form the set of all sets and call it a set...it would need to be called something else.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum