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Homework Help: Probability Question

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose that for two events A and B. A [itex]\subseteq[/itex] B. Show that [itex]B^{c} \subseteq A^{c}[/itex].

    2. Relevant equations

    3. The attempt at a solution

    [itex]1- B \subseteq 1- A[/itex]

    I'm not sure where to go from here. Thanks for any help you can provide.
  2. jcsd
  3. Jan 29, 2013 #2


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    That's really a "set" problem rather than a probability problem. I presume your "1" represents the "universal set" rather than a number?

    The "standard" way to prove "[itex]X\subset Y[/itex]" is to start "let [itex]x\in X[/itex]" and, using the definitions and information you are given about X and Y, end with "therefore [itex]x\in Y[/itex]". Here, "X" is the set [itex]B^c[/itex], the complement of B, the set of all objects that are not in B.

    So, if [itex]x\in B^c[/itex], then x is not in B. Since A is a subset of B (every member of A is also in B), x is not in A and so [itex]x\in A^c[/itex].
  4. Jan 29, 2013 #3
    Hi HallsofIvy,

    Is universal set and null set the same?


  5. Jan 29, 2013 #4


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    No. Not the same.
  6. Jan 29, 2013 #5
    You mean by universal set Ω?

    Also how does showing that
    x[itex]\in B^{c}[/itex] and x[itex]\in A^{c}[/itex]

    show that

    [itex]B^{c} \subseteq A^{c}[/itex]?

    I'm not seeing this
  7. Jan 29, 2013 #6


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    Okay, it sounds like you are using [itex]\Omega[/itex] to mean the "universal set", the set of all things we are allowing in our sets.

    It is NOT true that "[itex]x\in B^c[/itex] and [itex]x\in A^c[/itex]" implies [itex]B^c\subseteq A^c[/itex]" and I did not claim it was. What I showed was that if[itex]x\in B^c[/itex] implies [itex]x\in A^c[/itex], then [itex]B^c\subseteq A^c[/itex].

    That follows directly from the definition of "[itex]\subset[/itex]". If you are doing problems like that you surely should know that definition.
  8. Jan 29, 2013 #7
    [itex]B^{c} \subseteq A^{c}[/itex]

    This means that B complement is a subset or equal to A complement. If x is an element of B complement and A complement than why must B complement be a subset or equal to A complement? How do I know that A complement is not a subset of B complement?

    Since x is an element of A complement and B complement then...

    B complement must be a subset or equal to A complement
    A complement must be a subset or equal to B complement

    how do I know which one is true?
  9. Jan 29, 2013 #8


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    Homework Helper
    Gold Member

    None of that makes sense. If some x is an element of B complement and A complement then all that proves is that their intersection is non-empty. You need to think in terms "if x is an element of ... then x is an element of ..." You're trying to prove that Bc ⊆ Ac. Can you fill in the "..." ?
  10. Jan 30, 2013 #9


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    I didn't say that! A single "x" tells us nothing. The point was that this is true for all members of B complement. That is the definition of "[itex]X\subseteq Y[/itex]": for all [itex]x\in X[/itex], [itex]x\in Y[/itex].

    By knowing what "subset" and "member" mean! Go back and look at my argument, showing that any element of B complement must be a member of A complement.

    And that is the definition of "B complement is a subset of A complement".

    If your are asking, as you appear to be, "How can I prove this without knowing any of the definitions", you can't!
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